Ibstedt Computer

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Computer nalysis o Number SequencesHenry Ibstedt

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- - - - - - - - - - - - - - - - - - - ~

2 3 4 5 7 a

merican Research Pressup to US

998

9 1 11 12


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Computer nalysis o Number SequencesHenry Ibstedt

GlimllliDge 936280 60 robySwedea(May-October)

7 ne du Sergent lando92130 Jssy les Moulineaux

Fraau(November-April)

E-mail: [email protected]

Americaa Research PressLuptoa. US1998

C Hellry Ibstedt Americall Research Press


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The am i pidDle n Cers to the last article in this book. It illustrates bowrapidly the sums 1:. .. 1: ~ 0 2 1: and 1 :_ (1 where S n) is then n S(n} Sr nJSmarandacbe function and Sr n) is the sequences of square residues, growbeyond 1 2 12. The y-axis scale is the natural logarithm of he number ofterms required. A detailed discussion is given in chapter VI

Referents:Dr. R Muller Ir 1. Castillo American Research PressandProt: Mihaly Bcnczc. 6 Hannanului Street. 2212 Sacele 3. Jud Brasov.RomaniaPrinted in the United States ofAmericaby American ResearchPress140 Ir Wmdow Rock RoadLupton. Box 199AZ 86508. USAE-mail: [email protected]

ISBN I-879585-S9-6Standard AddressNumber 297-5092


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Preface

This is a book on empirical number theory concentrating on the analysis onumber sequences. Its focus is on a small part o a very large number ointeger sequences defined y Florentin Smarandache. The author has.however. when appropriate included some other o his research results whichorganically belongs to this area. The content is organized into chaptersaccording to the main considerations to make when programming theanalysis. They are not mutually exclusive.As in my previous book Surfing on the ocean o numbers - a fewSmarandache Notions and Similar Topics an attempt has been made topresent results so that they are easy to understand and not to burdensome toread. There are many tables. some o which may be used for reference. but ingeneral they are there to show the overall results obtained at a glance. n mostcases it is the way in which sequences behave. not the individual figures.which is o importance. Some graphs have been included to illustrateimportant findings.Some o the results in this book were presented by the author at the FirstInternational Conference on Smarandache Notions in Number Theory.August 21-23, 1997. Craiova. Romania.References have been given after some o the chapters. However, constant usehas been made o the following Smarandache source materials:Only Problems. Not Solutions , Florentin Smarandache.ome Notions and Questions in Number Theory, C Dumitrescu V Seleacu.

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mustratioos graphics layout and final editing up to camera ready form basbeen done by the author. Most tables have been created by direct tr.msfer fromcomputer files establisbed at the time ofcomputing to the mamJSCript so as toavoid typing errors. The results often involve very large numbers which aredifficult to 3CC \JDmodate n pre-designed formats. t bas therefore often been1'If'O 'SSUY to use severalliDes to represent a number n tables as well as text.All calulations have been canied out using UBASIC veT 8.87.I am grat.cfu1 to Dr. R Muller who bas given all possible help andencouragement during the work on this book. He and his colleagues at the

merican Research rus have at all times facilitated the work through rapide mail communications.Finally it is thanks to the patience and understanding ofmy wife Anne-Mariethat this book bas come about But the winter months have passed. the book isfinished and I am DO longer going to be lost among manuscript pagesSummer is around the comer and the Swedish nature is waiting for us

Paris, April 1998Henry Ibstedt

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ContentsPartition SequencesIntroduction 7

1 2 The Smarandache Prime-Partial Digital Sequence1 3 The Smarandache Square-Partial Digital Sequence 161 4 The Smarandache Cube-Partial Digital Sequence 181 5 Partition of {1,2, ... 2n} into Trigrades using Pythagorean Triples 201 6 Smarandache Non-Null Squares 261 7 Smarandache Non-Null Cubes 27

ecunive Integer Sequences11 1 The Non-Arithmetic Progression 29II.2 The Prime-Product Sequence 34II.3 The Square-Product Sequence 39II.4 The Smarandache Prime-Digital SulrSequence 42i l l Non Recunive SequencesIII 1 Smarandache Primitive Numbers 48III.2 The Smarandache Function S n) 53III 3 Smarandache m-Power Residues 58V Periodic Sequences

IV 1 Introduction 601V 2 The ~ i g i t Smarandache Periodic Sequence 62IV. 3 The Smarandache n-Digit Periodic Sequence 631V 4 The Smarandache Subtraction Periodic Sequence 66IV 5 The Smarandache Multiplication Periodic Sequence 691V 6 The Smarandache Mixed Composition Periodic Sequence 72V Concatenated SequencesV l Introduction 75V.2 The Smarandache Odd Sequence 75V 3 The Smarandache Even Sequence 77V.4 The Smarandache Prime Sequence 78V On the Harmonic SeriesVI 1 Comparison of a few sequences 80VI.2 Integers represented as sums of terms of the hannonic series 82VI.3 Partial sums of the hannonic series as rational numbers 85

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Chapter IPartition Sequences

l ntroductionWe will study the partition of the sequence of digitsof positive integers into two or more groups of digits whose correspondingintegers mayor may not possess a certain property.Example:

17 17 1101is a partition of the 6-digit integer 177101 into three groups of digits parts)each ofwhich represents a prime, whereas

117711011is a partition of the same integer into four parts none ofwhich is a prime.Obviously there are many ways in which and integer can be partitioned intogroups of digits. The kind ofpartition described above may also be referred toas a de-concatenation in order to distinguish it from the classical concept ofpartition of an integer as first developed by Euler and later illustrated byFerrer [1]. However, the kind of partition we are dealing with will be clearfrom the context. We will now use the classical partition of an integer as ameans to help us formulate a strategy for partitioning sequences of integers inthe sense defined above.The partitions of 4 are: 1+ 1+ 1+ 1=2 1+ 1=2+2=3+ 1=4 with the correspondingFerrer diagram:

D Do1 2 3 4 5iagram 1 The Ferrer diagram or 4

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The last partition corresponds to 4 itself which is of no interest to us.Furthermore, we will link the vertical representations to form the patternshown in diagram 2:DODD lOD l l 102 3

Diagram 2. The Ferrer diagram for 4 1'f a1Tfl1Iged and with partition IUI1M1 O 5achltkdn our applications we will also need to consider the order in which thepartition elements occur. Diagram 2 shows the seven di1Iercnt ordered

partitions possible for 4 and bow they form a partition pattern for thepartition of the -kligit integer 9164 into groups of ntegers

1 2 3 5 fiDiagram J. The partitionp ttem for a 4 digit integer.From this we see that the integer 9164 can be partitioned into perfect squaresin exactly two ways. 9, 16, 4 (partition numero 3) and 9 1 64 (partitionnumero 4). This is how the mapping of and integer onto a partition patternhelps us study the properties of each partition elementOnly partitions 2 and 4 in diagram 2 contain partition elements of unequalsizes. These give rise to an increase of the number partition patterns throughanangement of the partition elements. t is easily understood that in apartition of 8 we have 2 parts of the type J and 4 parts of the type 0then the number of arrangements will be ~ IS . The integer 8 has 21

2 ~ 4partitions of the type shown in diagram 2 giving rise to 127 orderedpartitions. This could be calculated by considering each arrangement as wedid above. However we will soon arrive at this in a different way.

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The above considerations will be useful in order to express the partitions ofintegers in explicit form. When we consider the partition of a very largenumber of integers to find out how many of them possess a certain propertywe will be able to use a different approach.Let's represent n n-digit integer t in the form t=al+az10+a3IOz+ .. , auIon-l.We introduce the following definitions and computer analysis relatedconcepts in UBASIC.

8(t)= I i possesses a certain property otherwise 8(t)=O.\ symbolizes integer division, example 23678\100=23res symbolizes the remainder of the last performed integerdivision, in the example above res=678.

We will now take a look t the question of leading zeros. Is 3 I07 a partitionof 307 into two primes? Is 8 I 027 a partition of 8027 into two cubes? Theauthor prefers to have a unique representation of an integer, 7 is a I-digitinteger written 7 not 07 and 33=27 is written 27 - not 027. The computer,however, interprets 07 as 7 and 027 as 27. To avoid integers with leadingzeros the function (j,r) is defined as follows.

(j,r)=O i r


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31figit integers: I ~ In the algorithm below use is made of the factthat we know P t) for t-1-1 resulting in thefollowing table ofpanition patterns.

Table 1. Pa1i1ion pattems N number of patterns.n-digit integer

N o2 3

35 6 7 815 31 63 127

The algorithm has been i.mpJemented in UBASIC on a Pentium 100 Mhzportable computer to examine various panition sequences.

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L2 The Smarandache Prime-Partial Digital SequenceDermitioa: The prime-partial-digital sequence is the sequence of primenumbers with the property that each member of the sequence can bepartitioned into groups of digits such that each group is a prime.Table 2 shows the first one hundred terms of this sequence.We will see in Chapter IT that the Smarandache Prime-Digital SubSequence is infinite. The Prime-Partial Digital Sequence is in fact a subset ofthe Prime-Partial Digital Sub-Sequence and is therefore a fortiori infinitewhich settles affirmatively a conjecture by Sylvester Smith. [2]. We will makeobservations on the number of ways in which prime integers can bepartitioned into primes. How close is the maximum number of ways this canbe done to the theoretical maximum given in table 1 and what do thesepartitions look like?

Table 2 The first one hundred members o the prime partiak:igit sequence.23 37 53 73 113 137 173 193 197 211

223 227 229 233 241 257 271 277 283 293311 313 317 331 337 347 353 359 367 373379 383 389 397 433 523 541 547 557 571577 593 613 617 673 677 719 727 733 743757 761 773 797 977 1013 1033 1093 1097 1117

1123 1129 1153 1171 1277 1319 1327 1361 1367 13731493 1637 1723 1733 1741 1747 1753 1759 1777 17831789 1913 1931 1933 1973 1979 1993 1997 2113 21312137 2179 2213 2237 2239 2243 2251 2267 2269 2273

The property that we will examine with the function 8(t) is whether t is aprime or not. In UBASIC this is done in the following program.8(t): 1 .D(t) .D(t) is equivalent to 8(t)

20 ift=O then z=O :goto 4030 ifnxtprm(t-l)=t then z=1 else z=O40 return(z) z is the value ofD(t)

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The function e j,r) takes the following form in UBASICe j,r):e j,r)

10.IqI.,r) LO '.,r) is equivalent to20 i


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Table 4. The number of n-digit primes that can be partitioned into primes in pdifferent ways. The tail for n=8 is interposed.2 n=2 n=3 n 4 n=5 n=6 n 7 n=80 17 92 678 5044 41795 351852 30567061 4 40 251 1986 14789 118379 972ffi52 10 83 769 6306 54102 4360093 1 40 2811 25892 2284284 5 135 1445 14205 1397555 4 67 758 7676 822656 33 424 5012 557937 15 250 2922 353268 3 131 1877 244019 1 78 1258 1713110 0 46 873 12424

0 22 550 872412 27 481 667113 E n=8 12 277 484114 38 6 6 212 357315 9 17 3 147 278116 4 8 0 115 205217 4 3 0 62 157418 42 5 3 59 124419 43 4 33 99220 44 8 34 8112 45 4 18 61722 Ie 3 12 50023 47 9 6 41724 48 6 10 31125 49 2 4 26426 5 2 4 21427 51 4 2 15428 52 I 3 15029 53 0 9430 54 4 1 833 55 2 8532 5 3 0 6533 57 0 5134 58 0 4235 59 0 536 6 I 0 3037 32

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Table 5 Continued42 2 37 3 373 743 2 3 73 373 744 2 373 373 745 2 37337 3 746 2 337 37 3747 23 733 73 748 23 733 7 3749 23 7 337 3750 2 37 337 3751 23 73 373 752 2 3733 73 753 2 3733 7 3754 23 7 3373 755 2 37 3373 756 2 37337 3757 23 73373 758 23 7 3373759 2 37 3373760 2373373 7

With a slight exception for ~ g i t integers the ratio between primes that canbe partitioned into primes and the number of primes is increasing andindicates strongly that the prime-partial digital sequences is infinite.o list the actual partitions would be impossible but for each n ~ g i t integersequence of primes there are a few primes that are record holders in thenumber of ways in which they can be partitioned into primes p ways). Hereis a list of those primes.

n=3 p=3 373n=4 p=5 3137 3373 3733 and 3797n=5 p=12 37337n=6 p=18 237373 537373 and 733373n=7 p=37 2373373n=8 p=60 23733737We see that of the prime digits 2,3,5 and 7 the primes 3 and 7 play adominant role in the composition of those primes that lend themselves to themaximum number of partitions into primes. All partitions can be displayed in

S


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explicit form by executing a UBASIC program based on the partition patternswhich were subject to a detailed discussion in the introduction. Table Sshows the partitions for 23733737.

U Tbe Saaarudadle Square-Partial Digital SequeaceDef uaitioD: The square-partial digital sequence is the sequence of perfectsquares which can be partitioned into two or more groups ofdigits which arcalso perfect scpwesThe first one hUDdred terms of his sequence is shown in table 6.

Table 6. The first one hln:hd tetms f the squae pafiol 1 sequence49 100 144 169 361 Xl 441 900 1225 13691444 1600 1681 1936 2500 3249 3600 4225 4900 6lO8100 902S 9 10000 1 ).4 )4 11025 11449 11664 12100 1254414161 14400 14641 15625 16641 16900 19044 19600 22500 25600

28900 32Xl 36100 36481 cOCXXl 40401 41616 42025 43681 441009 4S4OO 497 B 52900 57600 62500 6lO9 67600 72900 7a.coo81225 84100 9OCOO 93636 96100 9922S 102Xl 105625 108900115600 116964 117649 119025 121104 122500 1'B6OO 136161 136900 140625~ 144400 152100 1576 9 160000 1616 4 164025 166464 168100 170569176400 184900 193600 194481 202500 211600 220900 225625 230400 237169

This sequence is infinite. f all the infinitely many squares of the form 51021YI and write 1.1) as:

Substituting this and simplifying 2.1) and 3.1) results in:2d Y2 - YI -d) = n2/2YI +Y2= 2n + 1

Assumej E {I 2, ... nl2} and write 3.2) asY = n 2j; YI = n 1 - 2j

Insert these expressions in 2.2)n2 = 4d 4j - d -1)

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1.1)2.1)3.1)

1.2)

2.2)3.2)

3.3)

2.3)


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Integer solutions for d require the discriminant of this equation to be a squarer.(4)

Let a b aDd c be a PytbagoJean triple with 2 b and let k be an odd integer,(4) is then satisfied by 4j-l=kc, n=kb and z=ka It remains to evaluate XI, Xl.YI aDd z in terms of a b, c aDd It. To begin with we have j= kc+1 14. Since jz.. and c=1 mod 4) we can write k=4e-l where e z This gives d= 4el) c a) and the final solution:

Xl,2 = 4e - 1)(2b a) + 1 12YI,2 = 4e - 1)(2b c) + 1 12

(S)(6)

The conditions 2b>a aDd 2b>c restrict the number of Pythagorean tripleswhich can be used to generate trigrade partitions of the first N positiveintegers where N is given by:

N = 2b 4e-l), e z.. 7)A further condition is that XI and X2 gn.-en by (5) must belong to Ao i.c. x= )mod 2) ifO


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Table 9 Pythagorean triples nd their corresponding smaUest Na b c N a b c N a b c N3 4 5 24 55 48 73 288 17 144 1455 12 13 77 36 85 51 140 149 84015 8 17 13 84 85 85 132 1577 24 25 144 39 80 89 480 119 120 169 72021 20 29 65 72 97 165 52 17335 12 37 99 20 101 19 180 181 10809 40 41 91 60 109 360 57 176 18545 28 53 15 112 113 672 153 104 18511 60 61 360 117 44 125 95 168 193 100833 56 65 105 88 137 195 28 19763 16 65 143 24 145 133 156 205

Since the sets o and So are defined from the outset the trigrade partitions canbe unambigously described in terms of these sets and the exchange terms asshown in table 10. From this table it is seen that the trigrades correspondingto the last two Pythagorean triples have the same number of terms. Thismeans that we can obtain trigrades with only eight terms by subtraction. Forexample we see from table 10 that

3197,164,344,17 =317,44,272,89 (8)A k-grade aI,a2, .. .a b I ~ ...b,. is invariant under translation. To prove thisconsider the sets A={x+aI, x+a2, ... X+a,,} and B={x+b), x ~ .. . ,x+b,.}.Expand each term to the power m where ~

where the assumed k-grade property has been used in exchanging a; and bi We can consequently reduce all terms in the above trigrade by 17 if wereduced by 16 instead the eight terms would be positive and realively primebut would prevent further reductions) to obtain

30,147,180,327 =27,72,255,300

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8.1)


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Tbe k-grade property is evidently invariant under division by a commonfactor. Hence

30 49 60 109 = 9 24 85 100 8.2)Tbe steps shown above carried out for Pythagorean triples c


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11 A={l, 3,_., 419)v{ 422, 424, 840)v{382, 4591-{ 207,634 I840 B= {2, 4, ,420 )v{ 421 ,423 , ._,839 Iv{ 207,634 I - 382, 459 I12 91,60,109 A={ 1, 3,._, 179)v{ 182, 184 360)v{ 44,317) -{17 , 344 I360 B={2, 4 ,_ , 180)v{ 181, 183, 359v{l7 , 344 I -{ 44,317 I12 A={ 1,3, ._ ,419 )v{ 422 ,424 , ,840 v{ l02 , 739 I - 39, 802)

840 B={ 2 ,4 , ,420 Iv{ 421 ,423 , 8391v{ 39 ,802) - {102, 739 I

Table 11 Trigrades

l a2 a3 a4 b l b 2 b 3 b 40 49 60 109 9 24 85 1000 67 102 169 22 25 144 1470 85 136 221 25 36 185 1960 103 122 225 22 45 180 2030 107 242 349 25 62 287 3240 125 638 763 50 63 700 7130 129 244 373 48 49 324 3250 151 382 533 49 74 459 4840 155 230 385 34 77 308 3510 165 184 349 25 84 265 3240 181 384 565 60 81 484 5050 185 262 447 10 147 300 4370 203 554 757 81 86 671 6760 215 370 585 46 117 468 5390 229 304 533 49 108 425 4840 233 278 511 63 86 425 4480 241 556 797 72 121 676 7250 271 678 949 102 117 832 8470 283 542 825 58 165 660 7670 293 370 663 75 118 545 5880 301 456 757 81 132 625 6760 305 458 763 63 158 605 7000 331 994 1325 110 169 1156 12150 343 1074 1417 117 174 1243 1300

Table 11 Trigrades continued

a l a2 a3 a4 b l b 2 b 3 b 40 349 376 725 49 180 545 6760 359 506 865 81 170 695 7840 365 400 765 76 153 612 6890 379 522 901 54 225 676 8470 381 640 1021 121 156 865 900

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With these precisions we can write the algorithm for calculation of T\(n) inthe following way:

This algorithm was implemented for ns;IOO. s is shown in table 12 thissequence, after a slow start, grows very rapidly.Table 12. The Non Nul Square Representations.

1 1 1 2 2 2 2 3 4 44 5 7 7 7 9 12 13 13 162 23 23 27 35 41 42 47 61 71

75 82 1 4 124 134 146 178 217 237 2583 7 377 419 456 535 651 739 8 4 933 1126

l3 1422 1629 1955 2275 2513 2846 3397 3972 4435499 59 4 6933 78 7 8766 1 268 12 97 13718 154 9 17895

21 87 24 76 27 76 31248 36736 42214 47568 54636 64 17 7392483554 95596 111637 1293 6 146714 167379 1948 7 226 21 257447 293255

34 1 6 394953 4514 8 514 25 5941 3 69 35 79 967 9 1118 1 3845 12 5451

L7 Smarandache Non Null Cubes.In how many ways t(n) can n be written as a sum of non-null cubes? In thiscase there are much fewer representations. For non-null squares we have1 \(9)=4 whereas for non-null cubes we have t(9)=2. Calculation of the first100 terms follows the same method as for the non-null squares withcorresponding changes in our algorithms.

0(n) = {I i nis a. cubeoOtheIWlse

Table 13 The Non NuU Cube Representations.

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1 1 1 1 1 1 1 2 22 2 2 2 2 3 3 3 33 3 3 4 4 4 5 5 55 6 6 6 8 8 8 8 89 9 12 12 12 12 12 13 13

17 17 17 18 18 19 19 19 2424 27 27 29 29 29 35 35 3541 44 44 44 51 51 51 61 6166 66 74 74 74 89 89 94 981 9 1 9 1 9 13 13 136 146 146 162

eferences

[ ] Hardy and Wright n Introduction to the Theory ofNumbers OxfordUnivenilyPren

23591324416598

162

[ ] Sylvester Smith A Set of Conjectures on Srnarandache Sequences BulletinofPure andApplied Sciences Vol 15 1996.

[3] Charles Ashbacher Collection ofProblems on Srnarandache NotionsEmus Univenily Press Vail 1996.[4] Albert H. Beiler Recreations in the Theory ofNumbers DoverPublications Inc. New York.

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ChapterllRecunive Integer Sequences

II 1 The Non-Arithmetic ProgressionWe consider an ascending sequence of positive integers ai, a2 ... lin such thateach element is as small as possible and no t-term arithmetic progression is inthe sequence. In order to attack th problem of building such sequences weneed a more operational definition.Definition: The t-term non-arithmetic progression is defined as the set:{a; :a; is the smallest integer such that a;>


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~ = ac + d. d=1, 2, 3, .. ,Our solution is the smallest d for which none of the sets

{Ilt, a2 . ac a.:+d. a.:+ck, a.:+d-2e, ... ac+d- t-l)-e :C2d}contains a t-term arithmetic progression.We are certain that 3k l exists because in the worst case we may have tocoDtinue constructing sets until the tenD a.c+d- t-l)e is less than 1 in whichcase all possibilities have been tried with no t terms in arithmetic progression.The method is illustrated with an example in diagram 1.In the computer application of the above method the known terms of a no ttenD arithmetic progression were stored in an anay. The trial terms were ineach case added to this anay. In the example we have for d=1, e=1 the anay:1,2,3,5,6,8,9,10,11,10,9,8. Tbe terms are arranged in asa:nding order:1,2,3,5,6,8,8,9,9,10,10,11. Three terms 8,9 and 10 are duplicated and 11therefore bas to be rejected. For d=3, e=3 we have 1,2,3,5,6,8,9,10,13,10,7,4or in urencting order: 1,2,3,4,5,6,7,8,9,10,10,13 this is acceptable but wehave to cbeck for all values of e that produce terms which may fann a 4-termarithmetic progression and as we can see from diagram 1 this happens ford=3, e=4, so 13 bas to be rejected However, for d=5, e=5 no 4-termarithmetic progression is formed and e=6 does DOt produce terms that need tobe cbecked. hence 39 = IS.

1 2 3 4 5 7 8 9 10 11 12 13 14 15Knownt8lmS 1 2 3 5 6 8 9 10rials

d=1 e=1 8 9 10 11 reject 11d=2 e=2 6 8 10 12 reject 12d=3 e=3 7 10 13 by nexte

e=4 1 5 9 13 reject 13d=4 e=3 2 6 10 14 reject 14d=5 s 5 10 15 accept 15

Diagram 1. To lind he 9 erm of he 4-term non-aif tvnetic progression.

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In table 1 there is an interesting leap for t=3 between the 64 and the 65terms in that i64 =365 and =730.lMking a little closer at such leaps wefiDdtbat:Table 1. The 65 first terms of the norHlrittYnefic progessions for t=3 to 15., t-3 t=5 F t l t=8 t=9 t =IO t-=IJ t=12 1=13 t - I fal5

1 1 I I I I 1 1 1 1 1 I 1 12 2 2 2 2 2 2 2 2 2 2 2 2 23 3 3 3 3 3 3 3 3 3 3 3 34 5 5 4 4 4 4 4 4 4 45 10 6 6 5 5 5 5 5 5 5 5 5 56 11 8 7 7 6 6 6 6 6 6 6 6 67 13 9 8 8 8 7 7 7 7 7 7 7 78 14 10 9 9 9 9 8 8 8 8 8 8 89 28 15 11 10 10 10 10 9 9 9 9 9 9

10 29 16 12 12 11 11 10 10 10 10 1011 31 17 13 13 12 12 12 12 12 11 I I 11 1112 32 19 '4 '4 13 13 13 13 13 13 12 12 1213 37 26 '6 15 15 14 14 14 14 14 14 13 1314 38 27 17 17 16 16 15 15 15 15 15 15 1415 40 29 18 18 17 17 16 16 16 16 16 16 1616 41 30 19 19 18 18 17 17 17 17 17 17 1717 82 31 26 20 19 19 19 18 18 18 18 18 1818 83 34 27 22 20 20 20 20 19 19 19 19 1919 as 37 28 23 22 2 21 21 20 20 20 20 2020 86 49 29 24 23 23 22 22 21 2 21 21 2121 91 50 31 2S 24 24 23 23 23 22 22 22 2222 92 51 32 26 2S 25 24 24 24 24 23 23 2323 94 53 33 33 26 26 27 2S 2S 2S 24 24 2424 95 54 34 34 27 27 28 26 26 26 2S 2S 2S2S 109 56 36 3S 29 28 29 27 27 27 27 26 2626 110 57 37 36 30 30 30 28 28 28 28 28 2727 112 58 38 37 31 31 31 31 29 29 29 29 2828 113 63 39 39 32 32 32 32 30 30 30 30 2929 118 6S 41 43 33 33 33 33 31 31 31 31 3130 119 66 42 44 34 34 34 34 32 32 32 32 3231 121 67 43 45 36 3S 37 3S 34 33 33 3332 122 80 44 46 37 37 38 36 3S 3S 34 34 3433 244 87 51 47 38 38 39 37 36 36 3S 3S 3S34 245 88 52 49 39 39 40 38 37 37 36 36 363S 2 7 89 53 50 4 4 4' 39 38 38 37 37 3736 248 91 54 51 41 41 43 41 39 39 38 38 3837 2S3 94 56 52 50 42 44 42 40 4 40 39 39

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Table lcontinued t 3 t=4 t=S t=6 t=7 t=8 t=9 t=10 t=11 t=12 t=13 t=14 t=IS38 254 99 57 59 51 45 4 41 41 41 439 256 1 2 58 6 52 45 46 42 42 42 42 4140 257 lOS 59 62 53 46 47 45 424 271 1 6 61 63 54 47 48 49 45 4542 272 1 9 62 64 55 48 49 50 46 46 45 45 46274 11 63 65 57 49 50 51 47 47 46 46 47 275 III 64 66 58 50 53 52 48 48 47 47 4845 280 122 66 68 59 59 55 53 49 49 48 48 4946 281 126 67 69 6 6 56 54 50 50 49 49 5047 283 136 68 71 61 61 57 55 51 51 50 50 5148 284 145 69 73 62 62 58 58 52 52 51 51 5249 325 149 76 77 64 63 59 59 53 53 53 52 5350 326 151 77 85 65 64 6 6 54 54 54 54 545 328 152 78 87 66 65 64 61 56 55 55 55 5552 329 16 79 88 67 67 65 62 57 57 56 56 5653 334 163 81 89 68 69 66 63 58 58 57 57 5854 335 167 82 9 69 7 67 64 59 59 58 58 5955 337 169 83 91 7 71 68 65 6 6 59 59 656 338 17 84 93 72 72 69 66 61 61 6 6 6157 352 171 86 96 73 74 7 68 62 62 61 61 6258 353 174 87 97 74 75 7 69 63 63 62 62 6359 355 176 88 98 75 76 78 7 64 64 63 63 646 356 177 89 99 76 77 79 7 65 65 64 64 656 361 183 91 1 78 78 8 72 67 66 66 65 6662 362 187 92 1 3 79 79 81 73 68 68 67 67 6763 364 188 93 1 4 8 81 82 74 69 69 68 68 6864 365 194 94 1 7 81 84 83 75 7 7 69 69 6965 73 196 126 III 82 85 84 77 7 7 7 7 7'1 977 36 179 183 13 139 138 126 1 9 1 9 1 8 1 8 113

Leap starts at Leap finishes at5 114 =35 1 28 =21441 =314 1 82 =241122 =341 1 244 =2122365 =3122 1 730 =2365

Does this chain of regularity continue indefinitely?

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Sometimes it is easier to look at what is missing than to look at what we have.Here are some observations on the only excluded integers when forming the:first 100 terms for t= 11, 12 ,Band 14.Fort=l1: 11,22,33,44,55,66, n 88, 99is l1nFort=12: 12,23,34,45,56,67,78,89,100is l1n+lFort=13: 13,26,39,52,65,78,91,104is 13nFort=14: 14,27,40,53,66,79,92,105is 13n+l

The nt missing int.cgeI

The nt missing integer



Do these regularities of mi t9ng integers continue indefinitely? What aboutsimilar observations for other values of t?

ll.2 The Prime-Product SequeaceDefinitioa: The terms of the prime product sequence are defined through {fa :fa = p.. + 1, p.. is the nth prime number}, where p.. denotes the product of allprime nnmbers which are less than or equal to p...The sequence begins {3, 7, 31, 211, 2311, 30031, ... }. n the initialdefinition of this sequence t\ was defined to be equal to 2. However, thereseems to be no reason fOT this exception.Questioe: How many members of this sequence are prime numbers?The question is in the same category as questions like How many prime twinsare there?, How many armichael numbers are there?, etc. So we may haveto contend ourselves by finding how frequently we find prime numbers whenexamjning a fairly large number of terms of this sequence.

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From the definition it is clear that the smallest prime number which divides 1is larger than p . The terms of this sequence grow rapidly. The prime numberfunctions prmdiv n) and nxtprm n) built into the Ubasic programminglanguage were used to construct a prime factorization program for n


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~ Prime foctorizafion of prime product tem lS P l N=p. l and its factors1 2 1 3 Prime number2 3 1 7 Prime number3 5 2 31 Prime number 7 3 211 Prime number5 11 2311 Prime number6 13 5 30031 = fRo fq7 17 6 510511-19 972778 19 7 9699691 0:347279539 23 9 223092871 -317 70376310 29 10 6469693231 =331 571 3423111 31 12 2005600490131 Prime number12 37 13 7420738134811 -181 . 60611 . 67642113 41 15 3G4250263527211 =61 450451 . 11072701I 43 17 13082761331670031= I 677S339888213fR315 7 18 614889782.588491.11 = 953 46727 1380818118116 53 20 32589158477190044731 = 73 139 . 1731856476186030117 fR 22 1922760350154212639071 =2773467105229 1902637726118 61 2 1172883813fR406970983271 =223 52S956867082S42.70m19 67 2S 7 8 5 8 3 2 1 5 5 1 ~ 6 7 0 5 5 8 7 9 0 9 1 =54731 72929714358152452960320 71 27 557940830126698960967.15391 = 1063 3J)3 ).49 598841289221448967321 73 29 4072968059924902.1S062132,3471 = 2521 . 161561604915704181.780695122 79 31 13217644767340672907899084554131 = 22093 1503181961 . 96888-41.2027982.723 83 33 26706451568927585135562..017992791 = 265739 l00498803S9648973291674312692. 89 35 237687.18963C5550770650537601358311 = 131 1039 2719642258918842943733718061.15 97 37 230556796394551842.7531021.7331756071 = 2336993 lJ8.488037123743602..091007.73549

26 101 39 2328623643584973609000633168805073631)71 = 9607032.23871t64553038099079fR.I2731 105727 103 .1 23984823528925228172706521638692258396211 = 2979700398839179365737007600131544333453128 107 43 2S66376117fR4999.14479fR78153.cOO71648394471 = -49 13203797

3 5 1 2 6 4 4 9 1 6 3 1 3 7 ~ 2 7 6 7 8 4 3 6 5 1 8 3 7 1 129 109 45 27973-4996817854936178276161872067809674997231 = 33450712904336480464 4 423429971462317755..03-470131 113 7 3161 OQ5.46.4O.417607788145206291543662.9327-4686991 = 5122427 .202S4367860073J)46707fR50695..024715705581931 127 9 40144769393331)3618909 11990260451366458852.7731 =

154349999552001 57900988201093 1 62801 O52999907396723132 131 51 5258964790S2627740771371797072.11912900610967452631 =

1951 2 2 9 9 3 1 1 7 2 3 2 3 1 8 f R - 4 7 3 1 1 . 1 4 4 ~ 1 5 1 4 3 7 2 8 2 6 6 6 1 7

36


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Table 2 Prime factorization of o d u c t terms. continued.P L ~ 1 and its facton

33 137 53 72047817630210000485677936198920432067383702541010311 =881 . 1657 . 32633677 . 160823938621 . 5330099340103 176429175930323334 139 56 I O O I ~ I 9 0 0 6 7 S 0 9 2 3 3 1 3 1 6 0 4 9 9 ~ ~ I678279959005528882498681 W . 147647686145oU245139224580493

35 149 sa 1492182350939279320058875736615841068S475838633268645JC)411 ' '8754952439965018161573521013453262140076844134219184937113

36 151 60 2 2 5 3 1 9 5 3 4 9 9 1 8 3 1 1 7 l 3 2 8 8 9 0 2 3 6 2 2 8 9 1 3 5 0 6 8 5 1 6 ~ 1 9 1 1 =23269086799UIOIW7 . 9683213481319911991636641541802024271084713

37 157 62 35375166993717494840635767087951 74421205757060478899n4224298711381 . 18678311930927 38893867968570583 2 4 4 2 1 8 7 ~ 1 1 3 3 4 7 5 3

38 16:3 64 57661522199759516590236300353361 J430656538.401560606631985606881113612141147272105608293226701926741)221051761322886563OSU231S38210739 161 66 96294742073S98392705694621590113442919641913060621307541S963491271205590139 53252429ln 70604576339566763

w 173 69 1665a99037873252193808516953508962S62S098050959487-4862046961683989711 =62614127 . 26605801 S60936115e035233319392756615852S098772260689062181793

41 179 71 2 9 8 1 9 ~ 1 2 1 4 2 6 9 1 7 2 4 5 3 4 6 7 8 1 4 ~ 1 1 2 1 7 4 8 2 ~ 1 4 1 4 3 4 1 S 8 91 =601 . 1651781 . 856041n 3589959471525310189119 ~ 6 6 4 ) 5 1 6 6 1 8 9 5 4 8 0 9029861252251

42 181 73 539734629280554978272021 wn673687806275517530J6 4 350655459511599582614291 =107453 5634838141 -89141572809641011233448913965712571636329746284031704028667

-43 191 76 1 0 3 0 8 9 3 1 4 1 ~ 5 6 0 8 8 8 8 3 5 6 7 - 4 3 7 0 9 9 8 6 2 3 8 4 8 2 9 9 5 9 ) 5 1 7 5 1 9 2 7 6 6 7 1 5 5 2 0 2 7 9329391 =32999 . 17560J.474759 . n l 4854151 3247 2305961466-437323959598530415862423316227152033

44 19 , 78 198962376391 690981 6040415251 5452851536027J.4402721821058212203976095413910572271 =21639-4960447 . 7979125905967339495018 Sn115230lnl6279758 )4.40201621 1ln4536159()51

5 197 80 3 9 1 9 5 5 8 8 1 4 9 1 6 3 1 2 3 3 8 3 1 6 1 8 0 4 5 5 4 4 2 1 1 7 5 2 5 9 7 3 8 6 n 3 3 6 1 9 8 7 ~ 7 8 0 4 1 8 3 2 9 0 7 9 6 5 4~ 3 7 1 9 1 = 5 2 1 8 3 1 5 0 2 S 7 7 2 3 1601684368321 -39081170243262541027 238759139583699nl58572653160969521

< 6 199 82 779992204168346155324919910632981 387668799678990355094505'303247.c8685 1 1536164700811 =.w 10723 57622771 5876645549 9458145520867-48632S95443062609609719222040521494786S503lW7

47 211 85 1645783550795210387735581011435S90727981167322669649249414629852197255934130751870911 =1051 2179103334328369975311908487 292812710684839-46096596672/366.4692934303J40L4872907J8.C889

48 223 87 367009731827331916046503456S5501 3673233980031 2955331 782619462457039988073311157667212931=13867889468159 264647142357166086767915984928967035604888100036053342930619468037572880509

49 227 89 83311209124804345037562 W637988103824113


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~ . Is ~ ~ ~t E E321o

Term F IIAli:>er

Diogan 3. The ruri:>er of prime fodors in the first 49 terms of the prime-productsequence.This product expression for a -l is used in the following basic program tocany out the reduction of at - l modu1us t,.. Terms for which S k)=O areignored in the expaDSion were the exponents k are contained in the armyEO/.o. The residue modulus t is stored in F. In the program. below thereduction is done to base A=7.

1 cim 1 C X X 11 M=N-l:1 J.=O120 T = l ~1 wtlIe (M-T}>=OUO i n c ~ T = ~ISO wend16 dec : M = M - l \ 2 : i n c ' : ~ r x . ) = ~170 if M>O then goto 12018 F=I190 for to rx2 A=7210 f o r l ( ~ l t o ~ ~2 040 A=(A 2)@N250 next26 F=PA:F=F@N270 next

38


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This program revealed that there are t most three terms t,. of the sequence inthe interval 5OSnS200 which could be prime numbers These are:Term 75. N=379 +]. N is a 154 c:I2465986881376139M 1383761530961031-408346655636-467 .a1602797552123175013568630Q3638612390661668406235422311783742390510526587257026501:XJ)2696834793248526734305B016341659418Z02506367176Z0123329a06.46166635537169754290487515755971504173810639342556891244860294929a 966644747931Term 172 N=I021 +1. N s a 428 q a t rvnber.N=20832SSU4186971 2 6 2 Z 8 5 5 9 2 0 4 0 2 8 7 "-457268652856889007473M900Z84018145718728623019158728631608857214186313893i'93Q928.47430169:l88598087188708302659775388131 m2605885038331625282052311121 m 1 9 J 5 4 0 4 8 3 3 2 1 7 Q 3 6 4 5 6 3 0 0 7177616888535712671502325086556344276636618033120098(7112476455894240568)9053468323906745795726223468483433625259anl8741195919732397361348834503191 7 7 5 3 5 8 6 8 4 6 9 0 S 7 6 4 :AJ66276875058596100236112260054944287636531The last two primes or pseudo primes are remaIkable in that they aregenerated by the prime twins 1019 and 1021.Sumaaary of results: The number of primes q among the first 200 terms ofthe prime-prodnct sequence is given by 6Sq$9. The six confirmed primes areterms numero 1.2.3.4.5 and 11. The three terms which are either primes orpseudo primes are terms numero 15, 111 and 112. The latter two are theterms 101911 1 and 1021#+1.

1L3 The Square Produd eqaeacc

Definition: The terms of the square-product sequence are defined through {t,.: t = n )2+I}39


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Table 3 Prime fodorizafion of e - f a c t o r i a l termsn l f N= n1)4\ and its foeton\ 1 1 22 1 1 53 2 1 374 3 1 Sl75 5 1 144016 6 2 518401=13-398777 8 2 2S401601a 101 1515 18 10 2 162S702)1= 17 9S6295539 12 1 13168189440110 14 1 131 681894XlO111 16 1 15933S09222lOO11218 2 229-442Sl2802S6OOO1,. I 01 22717082


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Tobie 3 ccnlrluedn L f N=lnlJ2+1 and i i i facton34 n 4 871 62-4056287592837104338371913676165092719:117fH61 :11359414779904000000000

01=193- 3217 866100731693-:114521434-43145645231476890464)9690129119741062428681657619735 81 3 1OOn:1l4689523012254S2814S05594253302238S812620552707152831 053S21OOOOOOOOO0 0 0 0 0 1 = 3 1 7 . J 7 3 - 9 0 3 0 1 9 ~ 7 8 2 8 5 4 5 5 6 3 2 3 S S 3 6 0 8 6 3 4 7 4 8 2 0 1 1 7 6 1 6 3 2 1 9 t 8 0 m16189815715361

36 84 3 13837903517621823881868475992S0152279701201315623630847006904S7SS90400000000000001-73- 57986941373 -3269017431698277804505207628624951781788248063136614675)116377341498693JI 87 3 1894l899156242768942n943633734584709109446010887S06295524S23639832S760000000000000001-127


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The first terms of his sequence are {2, 3, 5, 7, 23, 37, 53, 73, ... }. SylvesterSmith [1] conjectured that this sequence is jnfinjte In this paper we willprove that this sequence is in fact infinite. Let s first calculate some moreterms of the sequence and at the same time find how many terms theIe is inthe sequence in a given intemIl, say between let and 1 ~ 1 .The programbelow is writtm in Ubasic One version of the program bas been used toproduce table 4 showing the first 100 terms of the sequence. The output of theactual vemon bas been used to produce the calculated part of table 5 whichwe are going to compare with the theoretically esimatcd part in the sametable.

10 point 220 cim ~ 6 ) . B ~ }3) for l to 6:read ( ~ } : n e x t-iO data 1.4.6.8.9.050 for 1 to :read ( ~ ) : n e x t6 ) data 2.3.5.770 for K ~ to 7eo M ~ . N = O

'Digits not aIowed stored in )'Digits aIowed stored in R )'Calc. for 7 separate intervals

90 for 1 to 4 'Only 2.3.5.7 aIowed as first dgit100 P = B ~ ~ ) l O A J ( ~ : P O = P : S = ( B ~ ~ } + 1 1 OAK :gosub 150110 next120 print ~ . N . M I N13) nextl-iO end150 while P


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Table 4. The first 100 terms in he prirne ciOtai sub seq.JeI lCe.2 3 5 7 23 37 53 73 223 227233 257 277 337 353 373 523 557 577 727

733 757 773 2237 2273 2333 2357 2377 2557 27532777 3253 3257 3323 3373 3527 3533 3557 3727 37335227 5233 5237 5273 5323 5333 5527 5557 5573 57377237 7253 7333 7523 7537 7573 7577 7723 7727 77537757 22273 22277 22573 22727 22777 23227 23327 23333 2335723537 23557 23753 23773 25237 25253 25357 25373 25523 2553725577 25733 27253 27277 27337 27527 27733 27737 27773 3223332237 32257 32323 32327 32353 32377 32533 32537 32573 33223Table S. Corrpaison of results.

Ie 1 2 3 5 6 7omputer cOUlt

m 4 15 38 128 389 1325 4b43Iog m) 0.6021 1.1761 1.5798 21072 2.5899 3.1222 3.6668

n 13 64 472 3771 30848 261682 2275350mIn 0:XJ769 0.23438 0.oaos1 0.03394 0.01261 0.00506 0.00204

Theoretical es1imates:m 4 11 34 109 364 253 4395

Iog m) 0.5922 1.0430 1.5278 2.0365 2.5615 3.09S0 3.6430n 7 55 421 3399 28464 244745 2146190

min 0.50000 O.2XXX) O oeooo 0.03200 0.012BO 0.00512 0.00205

Tbeorem:The Smarandacbe prime-digital sub sequence is infinite.Proof:We recall the prime counting function 1t(x}. The number of primes pg isdenoted 1t(x}. For sufficiently large values of x the order of magnitude of

44


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x x) is given by 1f x :;:s I . Let a and b be digits such that a>b; () andogxn a,b.k) the approximate number of primes in the interval boHf a-I


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values shows a VeJY close fit as can be seen from diagram where loglo m isplotted against k

The prime-digital sub sequence4.00003.50003.00002.50002.00001.50001.00000.50000.0000

2 3 5 6 7e

Diog-ern 5. agIO m as a fundion of Ie lhe upper urve COlTesponds to fhecomputer count.toga 10gbor large values of k we can ignore the terms and k in

comparison with log 10 in (1). For large k we therefore havea-b lOkn a,b,k) kloglO

and (2) becomes

46

(1 )


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4lOkn k) llt klO8lOCombining this with 3) we g

522km k) llt klog10

(2 )

4)

From which we see apply for in.ctana: l Hospital s rule) that m k)-+co ask-+a:). A foniori the prime-digital su sequen e is infinite.

47


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ChaptermNon-Recursive Sequences

Dd 1Ilitioa: For a given prime p and positive integer n the Smarandachcprimitive number Sp(n) is the smallest positive integer such that Sp(n) isdivisible by pDThis sequence is important for the caJqdation of the Smarandache functionS(n) which is defined as the smallest integer such that S(n) is divisible y n.We note th t Sp n)=S pj.It follows jmmectiate1y from the definition that the sequence Sp(n) for a givenprime P consists ofmultiples ofp. Furthermore the definition implies that thesequence is DOJHbcending, i.e. S p n ~ Sp(n+I).It is evident from the definition th t

Sp(n)=np for n p (1)Upper and lower bounds for Sp(n) have been established by PM Gmnh [I):

p - l ) n + l ~ S p n ~ n (2)Let's asssume Sp(n-l)


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Algorithms for the calculation of Spen are important as a stepping stone forthe calculation of the Smanmdache function. The lack of a closed formulamakes this a inteIesting topic. Three methods will be described,implemeDU:d aud compared.MdJaodLLet s deDote Spen =m. We want to calculate the smallest value of m for which123 (mod If). This is carried out in the following Ubasic program:

10 input p.n-;P.N20 if NO50 incM60 Y=Y M@Z70 wend80 print P.N.M90 end

'The factorial is reduced mod z each timethe loop is executed.

This program is effective for small values of n. As we move on to largeexponents the increasingly large modulus slows down the program. A study ofthe compexity of this short aud elegant program, which can equally well beused to calculate S(n), has been carried out by S.and T. Tabirca [3].

ethod 1When we have Spen =np. For n>p our potential solutions are multiples ofp. The first multiple we need to examine is p2. For a given multiple m of p wedetermine first the largest power of P which is less than m i.e. we determinek so that p k ~ l We then continue by counting the number of factors pinm . This is given by S = [mIp]+ [mlp2) + [mlp1 + ... + [m/pk). IfS


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10 input p.n-;P.N20 if N


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110 C(I)=N\B(I)120 N=res130 M=M+C(I)PAi140 next150 print M160 end

ReaIts.

Calculate the coefficientsCI, C:z, 'Express $p(n) using theseI coefficients

Which of these three programs is the most effective' Since Sp(n)=np whenDSp all three programs will be equally efficient for DSp. The programs weretherefore tested by caJadating Sp(n) for n=p+1, p+2, ... p SO forp=2 3 S 7 ... 229 (the first SO primes).The largest number manipulated in tileprograms is therefore 2 2 ~ which is a S S ~ g i t number. The programswere fitted as routines in the same in- and output program.. Short integerswere used whenever possible. To caJotlate these 2500 values of Sp(n) methodI took 31 m 26 S, method n 22 s and method l l 11 5 on a Pentium 100 Mhzlaptop. The time t was measured in millisecnods. In figure 1 In t is plottedagainst the number of primes for which the programs have been executed Itis seen that method l l is the most effective. It runs about twice as fast asmethod n Method I is effectM: for small primes but slows down considerablywhen the modulus p increases.

nt15

5I I I I I I I ,I iI I I II i

r . , I I IV I I I i

1

oQ ~

Prime number runero

Figure 1. COIIIJKl ison ofuecJllion times or methods I to Ill.

Sl

r=Jg


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n p,234567891011121314151617181920212223242526V282930313233343536373839~4243...4546.7

~

8101212141616

20222424

26282830323232323234363638~

~~42..4648484848SO52

18182124V

510152025253035

~45SOSO55

Table 1. n to n=1. 2. ... 48 a1d 77

212835

~

112233...5566n8899

13263952657891

~

17 19 23 2934 38 46 5851 57 69 8768 76 92 11685 95 115 ~

102 114 138 17.119 133 161 203136 152 1 C 232153 171 207 261170 190 230 290187 209 2S3 319~ Z2B 276 304822 24 299 377

31 :P 43 4762 82 86 9493 111 123 129 14112. 148 ~ 172 188

155 185 205 215 235186 222 246 258 282217 259 287 301 329248 296 328 ~ :P279 333 369 387310 370 .10 430341 ~ 7 451 Q3372 ..... 492 516~ 481 533 559

60 238 266 322 406 ~ 518 574 60265 98 154 255 285 345 435 465 555 6 15 ~70 98 165 ~ m ~ ~ ~ ~ ~ ~ ~

39 75 112 187 221 289 ~ .14 522 558 666 738 n4' ' I ; 3231' ' l l ' . 3>~ eo 119 198 234 361 4:P 551 589 703 779 81745 85 126 209 24 361 ~ 580 620 7 ~ 820 860~ 90 133 220 260 ~ 380 483 609 651 n 861 90348 95 I ~ 231 273 357 399 506 638 682 814 902 94651rJo01 147 242 286 :P4 41af5291 667 713 851 943 98954l.l2 1 147 242 299 391 4 3 7 ~ 6 7 888 984 103254 105 154 2S3 312 l8 456 552 725 n 925 1025 107554 110 161 2M 325 .25 .75 575 754 806 962 1066 111857 115 168 275 338 2 494 598 783 8:P 999 1107 116160 120 175 286 338 459 513 621 812 868 1036 1148 ~63 125 182 297 351 476 532 1073 1189 124763 125 189 308 ~ 493 551 667 1110 1230 129066 ~ 319 377 510 570 690 961 1147 1271 133369 ~ 330 390 527 589 713 961 11 C 1312 1:P672 135 203 341 l s... 6011 736 92tI 992 1221 1353 1.1972 ~ 210 352 416 561 627 759 957 1023 1258 1394 146275 145 217 363 429 578 ... 782 986 1()5.4 1295 1435 150578 ISO 2 2 ~ 363 442 578 665 805 1015 1085 1332 1476 154881 ISO 231 :P4 455 595 6 C 828 la 1 1 1 ~ 136911517 159181 155 238 385 468 612 703 851 1073 11.7 1369 1558 163481 160 245 396 481 629 722 87. 1102 1178 1406 159981 165 245 ~ 7 494 ... 722 897 1131 120'1 I ~ I ~C 170 252 .18 S07 663 741 920 1160 2 ~ 1480

87 175 259 m S07 680 760 943 1189 1271 1 5 1 7 ~ ~ - . - ' : : : : : ,90 175 266 ~ 520 697 779 966 1218 1302 155490 leo 273 451 533 714 798 989 12.7 1333 159193 185 280 462 s..s 731 817 1012 IV6 ~ 1628 1804

190 287 Q3 559 ~ ~ ~ ~ ~ 6 6 5 ~99 195 29 572 765 855 1058 1334 1426 1702 188699 200 29 585 782 87. 1058 1363 I ~ 1739 1927

52


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In table 1 c onsecutive squares have been emphasized by frames. The series inlight frames follow the rule s, p)= Sp(p+ l)=pz which is obvious from thedefinition The series in bold frames is less obvious and this study will beCODCluded by proving the following:

Sp(4p+4)= Sp(4p+3)=4pz or S p ~ } ~ S p 4 p + 3 ~ 2 p ) zProof: Applying the theorem ofDumitrescu and Seleacu we have

4p+4=4-(1+p), ie. c1=O and ez= from which

and4p+3=p-l+3 l+p), i.e. c}=p and c r from which

111.1 The Smarandache Function S(n)Ddinitioa: S(n) is the smallest integer such that S(n) is divisible by nThe properties of this function have been subject to detailed studies in manypapers. Some of these properties are listed here. When needed n isrepresented in the form n = tt .p ;: .. tcCltL S(n)Sn2. P is prime i S(p)=p3. f n ~ , then S(mn} ;S(n)-S(m)

S3


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4. f m,n)=l. then S(mn)=max{S(m).S(n)}S S(n)=max{ SPt ( ~ l ) SP2 ( ~ 2 ) ,SIIk ( ~ k } } where Sp(n) are

Smarandache primitive numbers as defined in section 11 16. Ifn is square free i.e. ~ l = a . r =at.. then S(n)=max


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130 for J =l to ~140 if ( J ~ S th n = F ( J ~ )150 nextThe largest value ofF{} s the value o SIn}160 return170 'Factorisation of N1. output i ) A e ~ i }

simple routine to factorize l180 K =O:for 1 =1 to 1 0 : F I ~ J = O : n e x t190 for 1 =1 to O : E ( I ~ ) = O : n e x t200 while N1>1210 P=prmdiv Nll220 if = F ( K ~ ) then inc ~ ( K ~ ) else inc ~ F ( K ~ ) = P : i n c E ( K ~ )230 N1=N1\P240 wend250 return260 'Calculation of Sp(n). In: p2.n2%. Out m. This routine s

do umented in section II270 1 2 - - O : S ~ o = O280 while S2


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Table 2 The Smarondoche functionl 0 1 2 3 4 a . 2 3 4 5 3 7 4 6

1 5 4 13 7 5 6 17 6 192 - 5 7 23 4 10 13 9 7 293 - 5 31 8 11 17 7 6 37 19 134 - 5 41 7 43 11 6 23 47 6 45- 10 17 13 53 9 11 7 19 29 59- 5 61 31 7 8 13 11 67 17 23- 7 71 6 73 37 10 19 13 79.- 6 9 41 83 7 17 43 29 11 89,- 6 13 23 31 47 19 8 97 4 1110 10 101 17 103 13 7 53 107 9 10911- 11 37 7 113 19 23 29 13 59 1712- 5 22 61 41 31 15 7 127 8 4313- 13 131 19 67 9 17 137 23 13914- 7 47 71 13 6 29 73 14 37 14915- 10 151 19 17 11 31 13 157 79 538 23 9 163 41 83 167 7 2617- 17 19 43 173 29 10 59 89 1791 . - 6 181 13 61 23 37 31 17 47 9

19 191 8 193 97 13 14 197 19920- 10 67 101 29 17 41 103 23 13 1921- 7 211 53 71 107 43 9 31 109 7322- 17 37 223 8 10 113 227 19 22923= 23 11 29 233 13 47 59 79 17 23924 6 241 22 12 61 4 41 19 31 8325- 15 251 7 23 127 17 10 257 43 372 = 13 29 131 263 53 19 89 67 26927 9 271 17 13 137 11 23 277 139 3128- 7 281 47 283 71 19 13 4 8 34

29 97 73 293 4 59 37 149 2330- 10 43 151 101 19 61 17 307 11 1033C 31 311 13 313 157 7 79 317 53 2932= 8 107 23 19 9 13 163 109 41 4733 331 83 37 167 67 7 337 26 11334- 17 31 19 21 43 23 173 347 29 34935- 10 13 11 353 59 71 89 17 179 3593 - 6 38 181 22 13 73 61 367 23 413 - 37 53 31 373 :7 15 47 29 9 37938- 19 127 191 383 8 193 43 97 3893 13 23 4 131 197 79 397 199 1940- 10 401 67 31 101 9 29 37 17 40941- 41 137 103 59 23 83 13 139 19 41942- 7 421 211 47 53 17 71 61 107 1343- 43 431 9 433 31 29 109 23 73 43944- 11 14 17 443 37 89 223 149 8 449u- 10 4 113 151 227 13 19 457 229 17. .- 23 461 11 463 29 31 233 467 13 6747- 47 157 59 43 79 19 17 53 239 4794 .- 8 37 241 23 22 97 12 487 61 163. , = 14 491 4 29 19 11 31 71 83 499

56


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50 15 167 251 503 7 101 23 26 127 509

Tabfe 2 cfd The Smorandache functionIi 0 1 2 3 , 5 7 8 ,51 17 73 12 19 257 103 43 47 37 173

52 - 13 521 29 523 131 10 263 31 11 4653 - 53 59 19 41 89 107 67 179 269 145 9 541 271 181 17 109 13 547 137 6155 - 11 29 23 79 277 37 139 557 31 435 . - 7 17 281 563 47 113 283 9 71 56957 - 19 571 13 191 41 23 8 577 34 19358 - 29 83 97 53 73 13 293 587 14 315 59 197 37 593 11 17 149 199 23 599ac- 10 601 43 67 151 22 101 607 19 2911- 61 47 17 613 307 41 11 617 103 619.2 - 31 23 311 89 13 20 313 19 157 37. 3 - 7 631 79 211 317 127 53 14 29 716 - 8 641 107 643 23 43 19 647 9 59e 13 31 163 653 109 131 41 73 47 659&a- II 661 331 17 83 19 37 29 167 223n 67 61 8 673 337 10 26 677 113 97A - 17 227 31 683 19 137 21 229 43 53.,- 23 691 173 11 347 139 29 41 349 23370- 10 701 13 37 11 47 353 101 59 70971- 71 79 89 31 17 13 179 239 359 71972- 6 103 38 241 181 29 22 727 13 1573- 73 43 61 733 367 14 23 67 41 7397 37 19 53 743 31 149 373 83 17 10775- 15 751 47 251 29 151 9 757 379 237 . - 19 761 127 109 191 17 383 59 10 76977- 11 257 193 773 43 31 97 37 389 417 . - 13 71 23 29 14 157 131 787 197 26375 : 79 113 11 61 397 53 199 797 19 47.0 10 89 401 73 67 23 31 269 1 1 80981 - 9 811 29 271 37 163 17 43 409 1382- 41 821 137 823 103 11 59 827 23 82983- 83 277 13 17 139 167 19 31 419 839u: 7 58 421 281 211 26 47 22 53 28385 17 37 71 853 61 19 1 7 857 13 859

43 41 431 863 9 173 433 34 31 7987- 29 67 109 97 23 15 73 877 439 29388- 11 881 14 883 17 59 443 887 37 1278 89 11 223 47 149 179 8 23 449 31. 0 - 10 53 41 43 113 181 1' 9 7 227 1 1u 13 911 19 83 457 61 229 131 17 9192 - 23 307 461 71 11 37 463 103 29 9293 - 31 19 233 311 467 17 13 937 67 313

47 941 157 41 59 9 43 947 79 73- 19 317 17 953 53 191 239 29 479 1376- 8 62 37 107 241 193 23 967 22 197 - 97 971 12 139 487 13 61 977 163 8914 109 491 983 41 197 29 47 19 43

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11 991 31 331 71 199 83 997 499 37m J Smarudadle m-Power ResiduaDdiaitioa: The m-power R:Sidue of n is the largest m-power b e numberwhich divides n. Nocation M,(n).Let, . .1. - Cl1 Cl2 CIt then M )5 express n m u....lorm n - PI Pi ....Pk r n == PI P2 ....Pitwhere


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m Jb Square esiduesFor n in the form n = p ~ l _ ~ _ t the square residue Sr(n) is given bySr(nFPt-P2- -- otApart from the multiplidive property we add two JIlOIe properties of thissequence

f a >O and P any prime number then S r p ~ ) = p _Sr(nm)=Sr(n)-Sr(m) 1(Sr(n). Sr(m.

The first 100 terms of his sequence are given in table _Table . The first 100 terms of the squae residues sequence

11 0 1 2 3 5 6 7 8 91 2 3 2 5 6 7 2 31 10 6 13 14 15 2 17 6 192 10 21 22 23 6 5 26 3 14 293 30 31 2 33 34 35 6 37 38 3910 41 42 43 22 15 46 47 6 710 51 26 53 6 55 14 57 58 59

6 30 61 62 21 2 65 66 67 34 697 70 71 6 73 74 15 38 77 78 798 10 3 82 83 42 85 86 87 22 899 30 91 46 93 94 95 6 97 14 33In chapter VI we will return to the Smarandache function and the squareresidues.RefereBca:[1J PilGrmis: A Date I lII S(p ). SntorandocJw FiutctiOft JOfJlmoJ Vol 23. o 1 (1993)[2] c DumiaraI:u v Sdaaa, SomeNdiom aDd QueItiom in NUIlIbc:F Tbcuy, Erlrfu

Ufliwnityh n 1994)[3} Sabia.ad ~ TabiR:a, CmlpurafjgnaJ AIpcds ~ 1 Fuoction,

~ N C I l i O I u J O f l T 1 I l J l Vol 8 No. 123, Fall 1997[4] Hc:ary n.tcdI, The F10reaIin 8marmdadJc FIIMion S(n). Sntorandot;he FJOtctionJOflT1IlJl Vol 23, o 1 1993)

[5} c DumiIracu, v Sdaaa, The 8marmdadJc Fundioo, Erlrfu Ufliwnity rus 1996)59


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Chapter IVPeriodic Sequences

IV.llatroc1actioaIn Mathematical Spectntm, vol 29 o 1 [11 is an article on Smarandacbe'speriodic sequences which terminates with the statement:

There will always be a periodic setplena whenever we have aftmction fS-JS when S is a finite set of positive integers nd werepeat the ftmctionf

However consider the following trivial function f x 0 : S ~ S , where S is anascending set of integers {ah


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With these two remarks in mind let s look at these sequences. There are in allfour different ones reported in the above mentioned article in MathematicalSpectrum. The study of the first one will be carried out in much detail in viewof the above remarks.

IV.2 The Two-Digit Smarandacbe Periodic Sequencet has been assumed that the definition given below leads to a repetition

according to Dirichlet s box principle (or the statement made above .However, as we will see, this definition leads to a collapse of the sequence.Preliminary dermition. Let Nit be an integer of at most two digits and let Nk be its digital reverse. We define the element Nk+1 of the sequence through

where the sequence is initiated by an arbitrary two digit integer N1Let s write NI in the form N1=10a+b where a and b are digits. We then have

N2= 110a+b-l0b-a 1=91 a-b 1The 1a-b 1can only assume 10 different values 0,1,2, ... ,9. This means thatN3 is generated from only 10 different values of N2. Let s first find out whichtwo digit integers result in 1a-b 1= 0,1,2, and 9 respectively.Ia bl Corresponding two digit integers0 22 33 55 60 77 88 991 10 12 21 23 32 34 43


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It is DOW easy to follow the iteration of he sequence which invariablytenninatc:s in 0, tabJc 1.Table I Iteration of sequence occorc:ing to he prelminary defin1ionfo bl N2 N3 N. Ns Ns N,0 01 9 02 18 63 27 45 9 03 27 45 9 04 36 27 45 9 05 45 9 06 54 9 07 63 27 45 9 0

8 72 45 9 09 81 63 27 45 9 0The termiDation of the sequence is preceded by the one digit dement 9 whosen:vcrsc is 9. The foUowiDg dc::finition is therefore proposedDd 1Ditioa of SlDU'UdadIe'. two-digit perioclic sequence Let be aninteger ofat most two digits Nk is defined through

rhe reverse o f ~ ifNk is a two digit integer~ = i

l ~ l i Nk is a one digit integerWe define the element Nk -l of he sequence through

where the sequence is initiated by an arbitrary two digit integer Nl withunequal digitsModifying table 1 according to the above definition results in table 2.

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Let s consider the cases n=3. n=4. n=5 and n=6.F l

Domain 100SNl s999. . The iteration will lead to an invariant or a loop(periodic sequence)l. There are 9 symmetric integers in the domain, 101.HI. 121 202. 212. ... for which NrO (invariant). All other initialintegers itefate into various enuy points of the same periodic sequence. Thenumber of numbers in the domain resulting in each enuy of the loop isdenoted s in table 3.

sloop

Table 3 Smaandache 3 dgit periodc seqJenCe23999 1189 200693 240297 120495

It is easy to explain the relation between this loop and the loop found for n=2.Consider N = 3 0 I O a I I ~ . From this we have IN-N' =99IaTao1=H9IaraoIwhich is II times the corresponding expression for n=2 and as we can sec thisproduces a 9 as middle (or first) digit in the sequence for n=3.8=4.Domain l000SN1s9999. The largest number of iterations carried out in orderto reach the first member of he loop is 18 and it happened for N1=1019. Theiteration process ended up in the invariant 0 for 182 values ofN1 90 of theseare simply the symmetric integers in the domain like N1=4334. 1881. 7mete.. the other 92 are due to symmetric integers obtained after a couple ofiterations. Iterations of the other 8818 integers in the domain result in one ofthe following 4 loops or a cyclic permutation of one of these. The number ofnumbers in the domain resulting in each entIy of the loops is denoted s intable 4.

1 1lIiI s dIIbon ed ill ddail iIlSlUfing m dw cean o[Nvmben by the author. Vail Univ. Pn:s1997.

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Table 6. Smaraldoche - d ~ periodc sequencess 13667 13667l 1 0s 13542 12551 26093l2 13586 65340s 12685 12685 26271l3 219978 659934s 19107 2711 7127 123320 12446 164711l 900 8100 iOOO 2700 .c;oos 2S057 18 12259 20993 u49 62776l5 9090 81810 636X) 27270 -i504SOs 7931 1 799 2603 2229 1 29995 35B269l6 9990 89910 69930 29970 49950s 25375 11 12375 21266 4409 63436l7 90009 810081 630063 270027 450045s U88 2 1005 1033 237 3765L8 90905' 818181 636363 272727s 1809 11 1350 1570 510 52S(lL9 99099 891891 693693 ~ 7 8 7 95495s 19139 2648 7 82 123673 12 72 16522110 99999 899991 699993 299997 99995s 152 1254 972 92 111 826 485 29 725l l1 10989 97B021 8571 2 615384 131868 736263 373626 2527 7 4 14505s 623 64 156 796 377 36 525 140 19L12 43659 912681 726462 461835 76329 847341 703593 308286 374517ctd 596 117 156 793 327 65 530 139 179 5813ctd 340956 318087 462726 164538 670923 341847 4 0 6 ~ 6 286:DJ 517374

Table 7. The subfrodion periodc sequence. 1 , f f lt 12 20 1 9 89 97 78 86 67 75 56 64 45 53 34 42 23 3112 13 30 2 19 90 8 79 96 68 85 57 74 46 63 35 52 24 4113 14 40 3 29 91 18 80 7 69 95 58 84 47 73 35 62 2S 514 15 50 4 39 92 28 81 17 70 6 S9 94 48 83 37 72 26 6115 16 60 5 49 93 38 82 27 71

IV.4 TIle SlUI Udadae Sabtradioa Periodic SequenceDef uaitioa: Let Nk be a positive integer ofat most n digits and let c be itsdigital reverse. N., is defined through

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Wedefinetheelement ~ ofhesequencethrough~ l = I N c ~ 1

wherecisapositivein teger.Thesequence isin i tia tedbyanub i tr .uypositiven-digit in teger N l Itisobvious fromthe definitionthatONk


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1SCS9 0=2 ~ S 9 9 9A computer analysis revealed a number of interesting facts amc:eming theapplication of the iter.dive function.There are no periodic sequeua:s for c=l c=2 and c=S All iterations result inthe invariant 0 after, SOJDdimes a large number of iterations

Table 8 loop statistics. L=IengIh of loop f=first term of loopc t l- 11 22 33 50 1 167 189 2001 Nl 92 N. 93 NI 241 59 151 2 a2 214 Nl 9 421 3645 NI 96 N D 59 841 288

2 1697 Nl 1 9 535

1 1012 1013 144 145 136 13

8 N. 203 43 851 2522 3053 129 Nl 21 79 237 174 205 16 1617 121

8 81


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orthe other values of c there are always some values of NJ which do notproduce periodic sequences but terminate on 0 instead. Those values of NJwhich produce periodic sequences will either have NJ as the first term of thesequence or one of the values f determined by 1 ~ 1 as first term. There areonly eight different possible value for the length of the loops namely 11 2233 50 100 167 189 200. Table 8 shows how many of the 900 initiatingintegers in the interval l00sNJS999 result in each type of loop or invariant 0for each value of c.A few examples:

or c=2 and NJ=202 the sequence ends in the invariant 0 after only 2iterations:202200 0or c=9 and NJ =208 a loop is closed after only 11 iterations:208 793 388 874 469 955 550 46 631 127 712 208or c=7 and NJ=109 we have an example of the longest loop obtained. It has200 elements and the loop is closed after 286 iterations:

1 9 894 491 18777447 6775335 46 633 329 916 6122 9 895 591 188 874 47116775445 4773333 266133 989669118997447226775555 488333311266144 9 897791 19 84 473 367 756 65 49933 332 2266155 9 898 891 191184 474 467 757 75 5 43 333 326 616 6 9 899 991 192 284 475 567 758 85 51143 334 426 617 7 9 9 2 193384 476 667 759 959 52243 335 526 618 8 9 9 11 2194484 477 767 76 6 53343336 626 619 9Q9 9 2 2 2195584 478 86776116 54 443 337 726 62 19 9 3 3 2 196 684 479 967 762 26 55 543 338 826 6211199 4 4 2197784 48 77 76336 56643 339 926 622 219 9 5 5 2198884 481177764469 57743 34 36 623 319 9 6 692 199 984 482 277 765 56 58 843 341136 624 419 9 7 7 2 200 549338777667 69953352246635 529 918 812 2111 5 494487777 77 7 63353346 636 629 919 912 212 2 5 495 58777887 71163 354 446 637 729 92 22 213 3 5 496 687 779 97 72 263 355 546 638 829 921122 214 4 5 497 787 78 8 73363356 646 639 929 922 222 215 5 5 498887 78118 74463357746 64 39 923 322 216 6 5 499 987 78228 75563358846 641139 924 422 217 7 5 5 2

IV. 5 The Smarandacbe Multiplication Periodic SequenceDefinition: Let c>1 be a fixed integer and No and arbitrary positive integer.NIo -J is derived from Nil: by multiplying each digit x of Nil: by c retaining onlythe last digit of the product cx to become the corresponding digit of NIo -J.

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In this case each digit position goes through a separate development withoutinterference with the SUIl OUDding digits. Let s as an example consider thethird digit of a 6-digit integer for c=3. The iteration of the third digit followsthescbema

xx7yyy - the third digit has been aIbitrarily chosen to be 7.xxlyyyruyyyxx< yyyxx7yyy - which closes the loop for the third digit

Let s now consider all the digits ofa six-digit integer 237456:237456691258873654419852237456 - which closes the loop.The digits 5 and 0 are invariant under this iteration. All other digits have aperiod of 4 for c=3.

Cooclusioe: Integers whose digits are all equal to 5 are invariant under thegiven operation. All other integers iterate into a loop of length 4.

We have seen that the iteration process for each digit for a given value of ccompletely determines the iteration process for any n-digit integer. t istherefore of nterest to see these single digit iteration sequences:With the help of table 9 it is now easy to characterize the iteration process foreach value of c.Integers composed of he digit 5 result in an invariant after one iteration.Apart form this we have for:c=lo Four term loops starting on the first or second term.c=3. Four term loops starting with the first termc=4 Two term loops starting on the first or second term (could be called aswitch or pendulum).

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Table 9 One digit multiplication sequencesc=2 c=3 C=4

1 2 4 8 6 2 1 3 9 7 1 1 4 6 42 4 8 6 2 2 6 8 4 2 2 8 23 6 2 4 8 6 3 9 7 1 3 3 2 8 24 8 6 2 4 4 2 6 8 4 4 6 45 5 5 56 2 4 8 6 6 8 4 2 6 6 4 67 4 8 6 2 4 7 1 3 9 7 7 8 2 88 6 2 4 8 8 4 2 6 8 8 2 89 8 6 2 4 8 9 7 1 3 9 9 6 4 6c=5 c=6 c=7

1 5 5 1 6 6 1 7 9 3 12 2 2 2 4 8 6 23 5 5 3 8 8 3 1 7 9 34 4 4 4 8 6 2 45 5 5 5 56 6 6 6 2 4 8 67 5 5 7 2 2 7 9 3 1 78 8 8 8 6 2 4 89 5 5 9 4 4 9 3 1 7 9c=8 c=9

1 8 4 2 6 8 1 9 12 6 8 4 2 2 8 23 4 2 6 8 4 3 7 34 2 6 8 4 4 6 45 0 0 5 56 8 4 2 6 6 4 67 6 8 4 2 6 7 3 78 4 2 6 8 8 2 89 2 6 8 4 2 9 1 9

7


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25 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 7326 84 34 7 86 52 73 14 53 82 16 75 32 5 64 12 31 42 62 8427 95 54 91 18 97 72 9528 16 75 32 51 64 1231 42 62 84 34 7 86 52 73 14 53 82 1629 27 95 54 91 18 97 72 953 33 6 66 331 42 62 84 34 7 86 52 73 14 53 82 16 75 32 51 64 12 3132 51 64 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 3233 6 66 3 3334 7 86 52 73 453 82 16 75 325 64 12 31 42 62 84 3435 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 73 14 53 8236 93 3637 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 73 1438 25 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 7339 36 93364 44 8 88 7 77 5 55 1 2 22 44 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 73 14 5342 62 84 34 7 86 52 73 14 53 82 16 75 32 5 64 12 31 4243 7 86 52 73 14 53 82 16 75 32 5 64 12 31 42 62 84 34 744 8 88 7 77 5 55 1 2 22 4 4445 91 18 97 72 95 54 9146 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 51 64 1247 23 51 64 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 5148 34 7 86 52 73 14 53 82 16 75 32 5 64 12 31 42 62 84 3449 45 91 18 97 72 95 54 915 55 1 2 22 4 44 8 88 7 77 551 64 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 552 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 5253 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 73 14 5354 9 18 97 72 95 5455 1 2 22 4 44 8 88 7 77 5 5556 2 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 5 64 12 3157 32 51 64 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 3258 43 7 86 52 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 759 54 91 18 97 72 95 546 66 3 33 66 75 32 51 64 12 31 42 62 84 34 7 86 52 73 14 53 82 16 7562 84 34 7 86 52 73 14 53 82 16 75 32 5 64 12 31 42 6263 93 36 9364 12 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 5 6465 2 31 42 62 84 34 7 86 52 73 14 53 82 16 75 32 51 64 12 3166 3 33 6 6667 4 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 52 73 14 5368 52 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7 86 5269 63 93 36 93

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7 77 5 55 1 11 2 22 41 44 SO 88 771 86 52 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7172 95 54 91 18 97 7273 14 53 82 16 75 32 51 64 12 31 42 62 84 34 71 86 52 7374 23 51 64 12 31 42 62 84 34 71 86 52 73 14 53 82 16 75 32 5175 32 51 64 12 31 42 62 84 34 71 86 52 73 14 53 82 16 7576 41 53 82 16 75 32 51 64 12 31 42 628434 71 86 52 73 14 5377 5 55 1 11 2 22 0 44 SO 88 7 7778 61 75 32 51 64 12 31 42 62 84 34 71 86 52 73 14 53 82 16 7579 72 95 54 91 18 97 72SO 88 7 77 5 55 1 11 2 22 41 44 SO81 97 72 95 54 91 18 9782 16 75 32 51 64 12 31 42 62 84 34 71 86 52 73 14 53 8283 25 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 71 86 52 7384 34 71 86 52 73 14 53 82 16 75 32 51 64 12 31 42 62 8485 43 71 86 52 73 14 53 82 16 75 32 51 64 12 31 42 62 84 34 7186 52 73 4 53 82 16 75 32 51 64 12 31 42 62 84 34 71 8687 61 75 32 51 64 12 31 42 62 84 34 71 86 52 73 14 53 82 16 7588 7 77 5 55 1 11 2 22 0 44 SO 8889 81 97 72 95 54 91 18 979 99991 18 97 72 95 54 9192 27 95 54 91 18 97 72 9593369394 45 91 18 97 72 95 54 9195 54 91 18 97 72 9596 63 93 36 9397 72 95 54 91 18 9798 81 97 72 95 54 91 18 97999 99

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Chapter VSmarandaehe Concatenated Sequences

V.I IntroductionSmarandache formulated a series of very artificially conceived sequencesthrough concatenation. The sequences studied below are special cases of theSmarandache Concatenated S-sequence.Definition: Let G={gJ, ~ .... gc } be an ordered set of positive integerswith a given property G. The corresponding concatenated S.G sequence isdefined through

In table 1 the first 20 tenos are listed for three cases, which we will deal within some detail below.

V.2 The Smarandaehe Odd SequenceThe Smarandache Odd Sequence is generated by choosingG={1,3,5,7,9,U, ..... } Smarandache asks how many tenos in this sequenceare primes and as is often the case we have no answer. But for this and theother concatenated sequences we can take a look at a fairly large number oftenos and see how frequently we find primes or potential primes. As in thecase of prime-product sequence we will resort to Fermat s little theorem tofind all primes/pseudo-primes among the first 200 tenos. f they are not toobig we can then proceed to test i f they are primes. For the Smarandache OddSequence there are only five cases which all were confirmed to be primesusing the elliptic curve prime factorization program. In table 2 is the term

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number L is the num er of digits of N and N is a prime number member ofthe SmarandacbeOdd Sequence:

Table 1c The first 2 terms of the Smarcn::Jcche dd Sequence

l2 21 1516 2734 6349 93

113135

135713579

1357911135791113

135791113151357911131517

13579111315171913579111315171921

1357911131517192123135791113151719212325

135791113151719212325271357911131517192123252729

13579111315171921232527293113579111315171921232527293133

1357911131517192123252729313335135791113151719212325272931333537

135791113151719212325272931333537391357911131517192123252729313335373941

Table 2 Prime numbers in the Smcrcndcche dd SeQuence

N13135791113151719135791113151719212325272931135791113151719212325272931333537394143454749515355575961636567135791113151719212325272931333537394143454749515355575961636567697173757779818385878991939597

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Term #201 is a S48 digjt number.V.3 1 1Ie SIUrUIdadIe va SequaaceThe SIUJ 1IIIdache va Sequaace is generated by choosing G={2.4.6.8.10... ... }. The question here is : How many terms a le nth powers of a positiveinteger?A term which is a nth power must be of the form 1 'a where a is an odd nthpower. The first step is tben:fore to find the highest power of 2 which dividesa given member of the sequence, ie. to determine and at the same time wewill find a We then have to test i a is a nth power. The basic programbelow bas been implemented for the first 200 terms o the sequeDCC. No nthpowers were found.

Table 1b. The first 20 tefmS o the Smaa ldache Even2242462468

246810246810122468101214246810121416246810121416182468101214161820246810121416182022246810121416182022242468101214161820222426246810121416182022242628246810121416182022242628302468101214161820222426283032246810121416182022242628303234246810121416182022242628303234362468101214161820222426283032343638


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24681012141618202224262830323436384024681012141618202224262830323436384042Ubuic PI Op'Ul: (only the essential p t of the progan is isted)

60 N=270 fO ~ to we step 280 ~ t l o g ~ } J 1 o g 1 0 ) } + 190 N = N * I ~ + ~100 A : : N ~110 repeat120 Al =A;A=A\2:inc ~130 until resO132 dec = A lUO 8=:r0U'ld(AA(16JJ150 if A ~ A then print . N160 next170 end

'Detem1ine length ofA d d o n ~

'Detem1ine (::n)'Determine A (=o)Check if c is c nthpowec

So there is not even a perfect square among the first 200 terms of theSmarandacbe Even Sequence. Are there terms in this sequence which are 2-pwhere p is a prime (or pseudo prime). With a small change in the programused for the Smarandacbe Odd Sequence we can easily find out. Strangelyenough not a single termwas found to be of he form 2-p.V.4 The Smarudacbe Prime SequeDceThe Smarudadle PriJDe Seqraeace is generated by {2 3 S 7 11, ... } Againwe ask: - How many are primes' - and again we apply the method of findingthe number ofprimesIpseudo primes among the first 200 terms.There are only 4 cases to consider: Terms #2 and 4 are primes namely 23and 2357. The other two cases are: term #128 which is a 355 digit numberand term #174 which is a 499 digit number.'1282357111317192329313741434753596167717379838997101103107109113127131137139

1 4 9 1 5 1 1 5 7 1 6 3 1 6 7 1 7 3 1 7 9 1 8 1 1 9 1 1 9 3 1 9 7 1 9 9 2 1 1 ~ 4 1 2 5 1 2 5 7 2 6 3 2 6 9 2 7 1 27728128329330731131331733133734734935335936737337938338939740140941942143


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14334394434494574614634674794874914995 35 95215235415475575635695715775875935996 16 76136176196316416436476536596616736776836917 17 9719

Table 1c The first 2 terms of the marandoche Prime Sequence223

2352357

23571123571113

2357111317235711131719

235711131719232357111317192329

23571113171923293123571113171923293137

2357111317192329313741235711131719232931374143

235711131719232931374143472357111317192329313741434753

23571113171923293137414347535923571113171923293137414347535961

2357111317192329313741434753596167235711131719232931374143475359616771

23571113171923293137414347535961677173

17423571113171923293137414347535961677173798389971 11 31 71 91131271311371391491511571631671731791811911931971992112232272292332392412512572632692712772812832933 73113133173313373473493533593673733793833893974 14 9419421431433439443449457461463467479487491499S 3S 95215235415475575635695715775875935996 16 76136176196316416436476536596616736776836917 17 97197277337397437517577617697737877978 98118218238278298398538578598638778818838879 79119199299379419479539679719779839919971 91 131 191 211 311 33

Are these two numbers prime numbers?

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Chapter VIOn the Harmoaic Series

VLl CoaapariJoa 01 a few lequeacaThe harmonic 1 1 1 nd seIysenes 1+2 + 3 ... k+ ... IS so 1mportaDl so IDtmstudied that nothing JDOIe can be said about i t or can there? We may alwayspose a few questionsWe know that the series is divergent. ~ the series _1_ isk= lkClloonvecgent for a.>1 and divergent for ruol. Since it is the borderline it maybe interesting to examine the following questions:Questioa 1 What is the smallest number of terms m we need in order tomake the sum greater than a given positive integer w1 In other words the

m 1smallest m for which r - w.k=\k

Questioa 1. What is the smallest m for which f \1 2 w?k= \k

Using the notatiOllS for the Smarandache function and the square residuesintroduced in chapter II.2.3 we pose the correspouding questions:Questioa 3. What is the smallest m for which : ZI ) w ?

k= 8 kQuestioa 4. What is the smallest m for which : +( w?

k= 8r kTable 1 shows the values of m required in each case for w=1. 2. 3 12. Indiagram 1 In(m) is plotted against w. A simplified version of this diagram is

GO 1shown on the cover We see that the series a is very sensitive tok=1 k

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changes in the value of cx + s on the convergent side of the graph fork=1 Sr k11k and + ) n the divergent side.

k IS kTable 1 Comparison of sequences

w123456789

101112

6

Ser ie s 1:l / kValues of m

24113183

227616

16744550

123673361791380

Ser ies 2: Ser ie s 3:1/S2 k) l / S / k)Values of m Values of ID24 2

144 9462 54

1045 2431944 7293200 20484862 65616912 163849477 37746

12480 9331216008 20995220020 497664

5 - - - - - - - - - - - - - - - - - - - - - - - - - ~ ~ ~ ~

Ser ies 4:1/kl . 02Values of m24

1235

102311966

30821011534167

118971427698

3 - - - - - - - - : ~ = . . l . ~ - - _ _ : : l - - - - - - - - - - - SX-Series2

-+-Series3-o-Ser ies42 - ~ - - ~ ~ ~ ~ ~ - - - - - - - - - - - - - - - - -

0 - - - - - - - - - - - - - - - - - - - - - - - -12 3 4 5 6 7 8 9 10 11 12

iagram I. he behaviourofsome sequences: In m) plotted against w

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Given a positive integer w. Can we lepieseot w as a sum of a finite IlUDlber ofdifferent terms chosen from the harmonic series? For w=1 we have the simple1 1 1answer 1 = butwbataboutw=2 w=3etc.2 3 6 ,

MetIIod 1: k-perfed DllJDben. By definition a k-petfcct IlUDlber is a numbern which satisfies the equation (J(n)=kn, where o(n) is the sum of divisors of(1 and n included). k =a(n) therefore gives the desired Iq)IeSeIltation of theninteger k. The result of a computer search for k-perfect numbers is given intable 2. A column is included for the number of divisors m since this is equalto the number of terms required for the representatioa The numbers are givenin factor form as this is more interesting than the dcrimal lepiesentation. A 2perfect number is in the literature simply ldelred to a pe rfea number. Theyhave attracted a lot of interest T te problem o perfect monbers a favoritewit ancient Greeks owes its origin to the mrmber mysticism o thePythagoreans (quote from Elemenwy Number TbeoIy, Uspensky Heaslet).

Table 2. ~ e c t I 1lJITt)ers n. m = the rv.Jr'I ber of civisors.t n2 4 23=62 6 22.7=282 10 2 31=4962 14 2 6.127=81282 26 2 .8191=335503362 34 2 1 ~ 1 3 1 7 1 = 8 5 8 9 8 6 9 5 62 38 2 1 ~ 5 2 4 2 8 7 = 1 3 7 4 3 8 6 9 1 3 2 82 62 230 .2147483647=23058430081399521283 16 2 3.3.5=1203 24 25.3.7=6723 288 2 8.5.7.19.37.73=4598182403 80 2 9.3.11.31=5237763 224 213 .3.11.43.127=14763048963 480 2 H S71931151=SI001180160


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Table 2. continued.I t 114 216 22.3 2.5.72.13.19=21785404 96 23.3 2.5.7.13=327604 96 2533.5.7=302404 384 2 7.33.5 2.17.31=455328004 320 2 9.33.5.11.31=235699204 480 2 9.3 2.7.11.13.31=1429908484 1056 21.3 3.5 2.23.31.89=438614784004 9984 2 25 .3 3.5 2.19.31.683.2731.8191S_- 1 92027:j 5:7:i1:2: 17:19:i'4'182439-0-4-0---'5 2304 2 7.3 5.5.72.13.17.19=319983955205 5280 21.3 .5.7.112.19.23.895 6336 2 1.3 5.5.72.13.19.23.895 3456 2 11.3 3.5 2.7 213.19.315 9216 211 35.53.73 133 175 43008 22.33.53.73.133.17.127.3375 709632 2 21.3 6.5 2.7.19.23 2.31.79.89.137.547.683.10935 94208 2 22 .3 7.5.7.11.19.41.47.151.197.178481

i T 6 S e f i f s Z : j : S 3 : 7 3 : i i i : i j i : i 9 3 : j i : j 7 : 6 i ~ 7 j : 1 8 1 - - - ---For w=3 we have a representation by 16 terms:3=1 2 3 4 5 6 8 10 12 15 20 24 30 40 60 120The smallest number of terms required for w=5 is 1920 and for 6 a staggering1,658,880 terms. For w=7 we have no representation at all. We would firsthave to find 7-perfect number.No great results. It s like having used a sledge hammer to kill a mosquito andmissed the target. But k-perfect number are interesting in themselves.Method 2: Trivial Expansions. A term +n the harmonic series can alwaysbe replaced by two other different terms from the harmonic series:1 k+1 1 1 G . f .-=---= -- --- . lVen an expansIOn 0 an IOteger w IOto termsk k k + 1) k + 1 k k + 1

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of the harmonic series it is possible to construct infinitely many such trivialexpaDSioos ofw. Ifbowever, in a given expanqon ofw, each term is replacedby two other terms as above and furtllamore the replacemcot process iscarried out until all terms in the new rcprcseotation arc diffc:rcut from onea Otber and also from the tenns in the original repn:sentation then a newrepresentation bas been obtained Obviously such represr:naation for WI aadW2 with DO common terms can be added together to form a R:pJ CSCntation ofW=Wl+w2

Example:1 1 11; : 36

Expand k -+ k+l2 -+ 3 63 -+ 4 126 -+ 7 42

k k+l)Not to be used (repeated)

3 and 6 arc repeated and since their cxpaDSion gives 4, 12 aad 7,42 we haveto continue the expansion:

4 -+ 5 2012 13 1567 - 8 5642 - 43 1806From this we can write the following representation of 2:

2;: 1 5 7 8 12 13 20 42 43 56 156 1806Adding our original representation of 1we get the following rcprcseotation 3:3 = 12 3 4 5 6 7 8 12 13 20 42 43 56 156 1806s in the case of 3-perfect numbers we have a representation of 3 with 16

terms. The minimum number of tenns for which a representation is possibleis 11 as can be seen from table 1 Let s try to get closer to this.MetJaod 3. Maximum density. For a given value of w table 1 gives theminimum number of terms from the harmonic series required to form arepresentation. For a given value of lF3 this DUJDber is In. We can thereforetry to choose the first m-l terms as close toget1a as possible, i.e. we form

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m- 1 1s=w- k=l k

where 5 is a small fraction which we have to exp nd in harmonic terms. Asbefore we apply the idea to w=3. We get

3=1 2 3 4 5 6 7 8 9 10 15 230 57960Bravol Only 13 terms. We had 16 with the other two methods. But nopleasure l sts forever. The nasty little fraction s will give no end to problemwhen we try large values for w. Although small s will have large numeratorsand denominators as we shall soon see. Maybe better get the sledge hammerout again and go in search for the 7-perfect number.

VLJ Partial sums of the harmonic series u rational Dumbersn 1We consider consecutive partial sums Sn = L - expressed as rational

k=l knumbers reduced to lowest terms.Question 1. Can there be a repetition of numerators?

a aAssume Sn-l and sn where a,b)=l and a,b1)=1. We then haveb b1~ = . . + . . which is equivalent to an(1H>I), but a,bb1)=1 and the latterb1 b nequation is therefore impossible and the assumption is wrong. We conclude:Two consecutive sums 5.. 1 and s" cannot have equal numerators whenreduced to lowest terms.A necess ry condition for fidig two sums Sn =. B.. and Sm = am , m>n+1 withbn bmequal numerators a"=a,,. when in lowest terms is therefore that nl exists so thatn


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agg=360968703235711654233892612988250163157207alOO=14466636279520351160221518043104131447711etcNow, let P be a prime then 3p>ap.l hcause s is a moootonously increasingfunction ofn and

.p-l 1 p-lP+bp-lS p = - - - + - = ~ ~ ~ -bp_1 P bp-IPcannot be reduced to lower terms..Coajecture: Let p; and J j+l denote amsecutive primes. The Il11IDel3tOl S in theinterval l after reducing to lowest terms, are an greater than j -1 ,i.c. when a term of the type . . is added the numerator takes a leap to a higherpvalue from which, although it may decrease, i t will DC\ er drop to a valuebelow the value before the leap.Questioa 2. What is the most frequeDtIy occurring denominator in the

1 1 1sequence So = 1 - - '" - when reduced to lowest terms?2 3 nLet l __ b Then 1 an b ~ : a_ theSa_l =t; w u ~ (a, )=1. Sa =t;+;; Sa = ~ .greatest common divisor ofb and n, cF(b.n). Then it is seen jllJJlJf diately thatthe fraction can be reduced by d. Put d and n=n1d and consequeudy

51} ht~ = htn}dI f a further reduction is possible then it must be by a factor in d or by d itselfbecause nI,bt)=l and


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where i and Pi l are consecutive primes. In table 3 the number of equaldenominators is denoted q and the corresponding number of digits of thedenominator L

Table 3 Equal denominators for n


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Ibstedt Computer

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