Families of Congruences of Fractional Partition Functions ...Families of Congruences of Fractional Partition Functions Modulo Powers of Primes Hirakjyoti Das A joint work with Prof.

Families of Congruences of Fractional Partition Functions Modulo

Powers of Primes

Hirakjyoti Das

A joint work with Prof. Nayandeep Deka Baruah

Tezpur University, Napaam 784028, Assam, India

July, 2020

Hirakjyoti Das July, 2020 1 / 26

Contents

1 Partitions and the generating functions of the number of partitions

2 Fractional partition functions

3 Fractional 2-color partition functions


Partitions

4 = 4, 4 = 3 + 1, 4 = 2 + 2, 4 = 2 + 1 + 1, 4 = 1 + 1 + 1 + 1.

For each of the sums,

(4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

These above sequences are called the partitions of 4 and the summands/terms are called theparts of the partitions of 4. In general,

A partition λ := (λ1, λ2, λ3, . . . , λk ) of a positive integer n, is a finite non-increasing sequenceof positive integers (the λi s) such that n = λ1 + λ2 + λ3 + · · ·+ λk .


Partitions

4 = 4, 4 = 3 + 1, 4 = 2 + 2, 4 = 2 + 1 + 1, 4 = 1 + 1 + 1 + 1.

For each of the sums,

(4), (3, 1), (2, 2), (2, 1, 1), (1, 1, 1, 1).

These above sequences are called the partitions of 4 and the summands/terms are called theparts of the partitions of 4. In general,

A partition λ := (λ1, λ2, λ3, . . . , λk ) of a positive integer n, is a finite non-increasing sequenceof positive integers (the λi s) such that n = λ1 + λ2 + λ3 + · · ·+ λk .


The partition function and the generating function

p(n):= Number of partitions of a positive integer n.

The generating function for the number of partitions of n,p(n) was given by L. Euler (1707–1783).

Look at the following binomial expansions,

1

1− q= 1 + q1 + q2 + q3 + · · · = 1 + q1 + q1+1 + q1+1+1 + · · ·

1

1− q2= 1 + q2 + q4 + q6 + · · · = 1 + q2 + q2+2 + q2+2+2 + · · ·

So that

∞∏j=1

1

1− qj= (1 + q1 + q1+1 + · · · ) · (1 + q2 + q2+2 + · · · ) · (1 + q3 + q3+3 + · · · ) · · ·


The partition function and the generating function

Now, if we want get the coefficient of q3 from the series expansion of

∞∏j=1

1

1− qj= (1 + q1 + q1+1 + · · · ) · (1 + q2 + q2+2 + · · · ) · (1 + q3 + q3+3 + · · · ) · · · ,

then the contributors are

q3, q2+1, and q1+1+1.

Therefore,Coefficient(q3) = p(3).

In general,Coefficient(qn) = p(n).

So,

∞∑n=0

p(n)qn =∞∏j=1

1

1− qj; p(0) = 1.


A very very brief review

S. Ramanujan (1887–1920) first found

p(5n + 4) ≡ 0 (mod 5),

p(7n + 5) ≡ 0 (mod 7),

p(11n + 6) ≡ 0 (mod 11).

N.B. Integer power of the generating function of p(n), viz.

(∏ 1

1− qj

)n

generates the

n-colored partitions.

Thought: What if we raise∏ 1

1− qjto a rational number t!

Question: Can we interpret the coefficients in the series expansion of

(∏ 1

1− qj

)t

combinatorially? Is it worth studying the coefficients?


A few notation

For complex numbers a, and q such that |q| < 1,

(a; q)∞ = limn→∞

(a; q)n := limn→∞

n∏j=0

(1− aqj ) =∞∏j=0

(1− aqj ),

and

En := (qn; qn)∞ =∞∏j=1

(1− qnj ).

For example,

E1 := (1− q) · (1− q2) · (1− q3) · · · .


Fractional partition functions: Chan and Wang’s work

For any non-zero rational number t, we define

∞∑n=0

pt(n)qn = E t1 .

E−1/61 = 1 +

1

2 · 3q +

19

23 · 32q2 +

343

24 · 34q3 +

11305

27 · 35q4 + · · ·

S. T. Ng [Undergraduate Thesis, Singapore, 2003] proved, for all n ≥ 0,

p−2/3(19n + 9) ≡ 0 (mod 19).

Theorem (Chan and Wang [Acta Arith., Vol. 187(1), 2019])

For any integer n and prime `, let ord`(n) denote the integer k such that `k | n and `k+1 - n.Let t = a/b, where a, b ∈ Z, b ≥ 1 and gcd(a, b) = 1. Then

denom(pt(n)) = bn∏`|b

òrd`(n!).

N.B. The denominators of both pt(n) and t have the same prime divisors.




∞∑n=0

pt(n)qn = E t1 .

E−1/61 = 1 +

1

2 · 3q +

19

23 · 32q2 +

343

24 · 34q3 +

11305

27 · 35q4 + · · ·


p−2/3(19n + 9) ≡ 0 (mod 19).




òrd`(n!).





∞∑n=0

pt(n)qn = E t1 .

E−1/61 = 1 +

1

2 · 3q +

19

23 · 32q2 +

343

24 · 34q3 +

11305

27 · 35q4 + · · ·


p−2/3(19n + 9) ≡ 0 (mod 19).




òrd`(n!).





∞∑n=0

pt(n)qn = E t1 .

E−1/61 = 1 +

1

2 · 3q +

19

23 · 32q2 +

343

24 · 34q3 +

11305

27 · 35q4 + · · ·


p−2/3(19n + 9) ≡ 0 (mod 19).




òrd`(n!).





Suppose a, b, d ∈ Z, b ≥ 1 and gcd(a, b) = 1. Let ` be a prime divisor of a + db and 0 ≤ r < `.Suppose d , ` and r satisfy any of the following conditions:

1. d = 1 and 24r + 1 is a quadratic non-residue modulo `;

2. d = 3 and 8r + 1 is a quadratic non-residue modulo ` or 8r + 1 ≡ 0 (mod `);

3. d ∈ {4, 8, 14}, ` ≡ 5 (mod 6) and 24r + d ≡ 0 (mod `);

4. d ∈ {6, 10}, ` ≥ 5 and ` ≡ 3 (mod 4) and 24r + d ≡ 0 (mod `);

5. d = 26, ` ≡ 11 (mod 12) and 24r + d ≡ 0 (mod `).

Then, for n ≥ 0,

p−a/b(`n + r) ≡ 0 (mod `).

For example,

p−1/3(5n + r) ≡ 0 (mod 5), r ∈ {2, 3, 4},

p−2/3(5n + r) ≡ 0 (mod 5), r ∈ {3, 4},

p−1/2(7n + r) ≡ 0 (mod 7), r ∈ {2, 4, 5, 6}.



Chan and Wang conjectured 17 congruences for elementary proofs. Some of them are in thefollowing theorem.


For n ≥ 0, we have

p1/2(125n + r) ≡ 0 (mod 25), r ∈ {38, 63, 88, 113},

p2/3(25n + r) ≡ 0 (mod 25), r ∈ {19, 24},

p1/4(25n + r) ≡ 0 (mod 25), r ∈ {14, 24},

p1/4(25n + 19) ≡ 0 (mod 125),

p−1/3(25n + r) ≡ 0 (mod 125), r ∈ {18, 23},

p−3/4(25n + r) ≡ 0 (mod 25), r ∈ {13, 23},

p−3/4(25n + 18) ≡ 0 (mod 125),

and

p−3/4(125n + r) ≡ 0 (mod 3125), r ∈ {93, 118}.


The Rogers-Ramanujan continued fraction

The Rogers-Ramanujan continued fraction is defined as

R(q) :=q1/5

1 +

q

1 +

q2

1 +

q3

1 + · · · , |q| < 1,

which has the following well-known q-product representation

R(q) = q1/5 (q; q5)∞(q4; q5)∞

(q2; q5)∞(q3; q5)∞.

In some of the next slides

R(q) :=q1/5

R(q).


A very important dissection

n-dissection of E1 (Berndt, [Ramanujan’s notebook part III, Springer, 1991])

For integer n ≥ 1 with n ≡ ±1 (mod 6), if n = 6g + 1, where g ≥ 1, then

E1 = En2

((−1)gq(n2−1)/24 +

(n−1)/2∑j=1

(−1)j+gq(j−g)(3j−3g−1)/2 (q2jn; qn2)∞(qn

2−2jn; qn2)∞

(qjn; qn2 )∞(qn2−jn; qn2 )∞

),

while if n = 6g − 1, where g ≥ 1, then

E1 = En2

((−1)gq(n2−1)/24 +

(n−1)/2∑j=1

(−1)j+gq(j−g)(3j−3g+1)/2 (q2jn; qn2)∞(qn

2−2jn; qn2)∞


).

For example, when n = 5

E1 = E25

(R(q5)− q −

q2

R(q5)

).


A very important dissection

n-dissection of E1 (Berndt, [Ramanujan’s notebook part III, Springer, 1991])

For integer n ≥ 1 with n ≡ ±1 (mod 6), if n = 6g + 1, where g ≥ 1, then

E1 = En2

((−1)gq(n2−1)/24 +

(n−1)/2∑j=1

(−1)j+gq(j−g)(3j−3g−1)/2 (q2jn; qn2)∞(qn

2−2jn; qn2)∞


),

while if n = 6g − 1, where g ≥ 1, then

E1 = En2

((−1)gq(n2−1)/24 +

(n−1)/2∑j=1

(−1)j+gq(j−g)(3j−3g+1)/2 (q2jn; qn2)∞(qn

2−2jn; qn2)∞


).

For example, when n = 5

E1 = E25

(R(q5)− q −

q2

R(q5)

).


Fractional partition functions: Some congruences modulo powers of 5

Theorem (Baruah and Das)

For all n ≥ 0, we have

p−1/6(25n + r) ≡ 0 (mod 25), r ∈ {9, 14, 19, 24},

p1/6(125n + r) ≡ 0 (mod 25), r ∈ {96, 121},

p−5/6(125n + r) ≡ 0 (mod 25), r ∈ {95, 120},

p5/6(25n + r) ≡ 0 (mod 125), r ∈ {15, 20},

and

p5/6(125n + r) ≡ 0 (mod 625), r ∈ {65, 70}.

N.B. The method used to find the above theorem lets us prove all the conjectural congruences

modulo powers of 5 by Chan and Wang.


Fractional partition functions: A general family of congruences


Let ` ≥ 5 be a prime and k > 1 and s be positive integers such that s ≤ bk/2c. Then, for alln ≥ 0, we have

p−(`k−b)/b

(`2s · n + `2s−1 · r +

(`− 24b`/24c)`2s−1 − 1

24

)≡ 0 (mod `k−2s+1),

where 0 ≤ r < `, r 6= b`/24c, and (`, b) = 1.

Some cases, when ` = 5, b = 1567, and k = 5

p−1558/1567(52n + 5r + 1) ≡ 0 (mod 54), r ∈ {1, 2, 3, 4},

p−1558/1567(54n + 53r + 26) ≡ 0 (mod 52), r ∈ {1, 2, 3, 4}.

N.B. The sequences (52n + 5r + 1) and (54n + 53r + 26) do not have common terms.


Fractional partition functions: A general family of congruences


Let ` ≥ 5 be a prime and k > 1 and s be positive integers such that s ≤ bk/2c. Then, for alln ≥ 0, we have

p−(`k−b)/b

(`2s · n + `2s−1 · r +

(`− 24b`/24c)`2s−1 − 1

24

)≡ 0 (mod `k−2s+1),

where 0 ≤ r < `, r 6= b`/24c, and (`, b) = 1.

Some cases, when ` = 5, b = 1567, and k = 5

p−1558/1567(52n + 5r + 1) ≡ 0 (mod 54), r ∈ {1, 2, 3, 4},

p−1558/1567(54n + 53r + 26) ≡ 0 (mod 52), r ∈ {1, 2, 3, 4}.

N.B. The sequences (52n + 5r + 1) and (54n + 53r + 26) do not have common terms.


Fractional partition functions: Congruences modulo powers of 5, 7, and 11

Using dissections of E21 , we find


Let k > 1 and s be positive integers such that s ≤ bk/2c. Then, for all n ≥ 0, we have

p−(5k−2b)/b

(52s · n + 52s−1 · r +

52s − 1

12

)≡ 0 (mod 5k−2s+1), r ∈ {1, 2, 3, 4},

p−(7k−2b)/b

(72s · n + 72s−1 · r +

72s − 1

12

)≡ 0 (mod 7k−2s+1), r ∈ {1, 2, . . . , 6},

and

p−(11k−2b)/b

(112s · n + 112s−1 · r +

112s − 1

12

)≡ 0 (mod 11k−2s+1), r ∈ {1, 2, . . . , 11},

where b’s in the above congruences are co-prime to the moduli.

N.B. Similar congruences do not hold true for prime 13.


Fractional partition functions: Congruences modulo powers of 3, 5 and 7

Using dissections of E31 and E4

1 , we find


Let k, m and s be positive integers such that s ≤ m + 1. Then, for all n ≥ 0, we have

p−(3k+m−3b)/b

(32s · n + 32s−1 · r +

32s − 1

8

)≡ 0 (mod 3k+m−s+1), r ∈ {1, 2},

p−(5k+m−3b)/b

(52s · n + 52s−1 · r +

52s − 1

8

)≡ 0 (mod 5k+m−s+1), r ∈ {1, 2, 3, 4},

p−(5k+m−4b)/b

(52s · n + 52s−1 · r +

52s − 1

6

)≡ 0 (mod 5k+m−s+1), r ∈ {1, 2, 3, 4},

and

p−(7k+m−3b)/b

(72s · n + 72s−1 · r +

72s − 1

8

)≡ 0 (mod 7k+m−s+1), r ∈ {1, 2, . . . , 6},



Fractional partition functions: Congruences modulo powers of 3, 5, and 7

Using dissections of E61 , E8

1 , and E141 , we find



p−(3k−6b)/b

(32s · n + 32s−1 · r +

32s − 1

4

)≡ 0 (mod 3k ), r ∈ {1, 2},

p−(5k−8b)/b

(52s · n + 52s−1 · r +

2 · 52s−1 − 1

3

)≡ 0 (mod 5k ), r ∈ {0, 2, 3, 4},

p−(5k−14b)/b

(52s · n + 52s−1 · r +

11 · 52s−1 − 7

12

)≡ 0 (mod 5k ), r ∈ {0, 1, 3, 4},

and

p−(7k−6b)/b

(72s · n + 72s−1 · r +

3 · 72s−1 − 1

4

)≡ 0 (mod 7k ), r ∈ {0, 2, 3, 4, 5, 6},



Fractional partition functions: Balanced congruences


Let k > 1 be an odd integer. Then, for all n ≥ 0, we have

p−(3k−6b)/b

(3k · n +

3k+1 − 1

4

)≡ 0 (mod 3k ),

p−(5k−8b)/b

(5k · n +

2 · 5k − 1

3

)≡ 0 (mod 5k ),

p−(5k−14b)/b

(5k · n +

11 · 5k − 7

12

)≡ 0 (mod 5k ),

and

p−(7k−6b)/b

(7k · n +

3 · 7k − 1

4

)≡ 0 (mod 7k ),



Fractional 2-color partition functions

For any non-zero rational number t and integer r > 1, we define

∞∑n=0

p[1,r ;t](n)qn = (E1Er )t .

For instance,

(E1E3)1/6 = 1−1

2 · 3q−

17

23 · 32q2 −

451

24 · 34q3 −

6191

27 · 35q4 −

12053

28 · 36q5 −

2845933

210 · 38q6 + O

(q7).

Therefore, it is also meaningful to explore congruences for p[1,r ;t](n) modulo powers of prime `

such that ` - denominator of t.


Fractional 2-color partition functions: Congruences modulo primes


Suppose a, b, d ∈ Z, b ≥ 1 and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , ` and r satisfy any of the following two conditions:

1. d = 2, ` ≡ 3 (mod 4), and 4r + 1 ≡ 0 (mod `),

2. d = 3, ` ≡ 5 or 7 (mod 8), and 8r + 3 ≡ 0 (mod `).

Then, for all n ≥ 0,

p[1,2;−a/b](`n + r) ≡ 0 (mod `).


Suppose a, b, d ∈ Z, b ≥ 1, and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , `, and r satisfy the following condition:

d = 3, ` ≡ 5 or 11 (mod 12) and 2r + 1 ≡ 0 (mod `).


p[1,3;−a/b](`n + r) ≡ 0 (mod `).




Suppose a, b, d ∈ Z, b ≥ 1 and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , ` and r satisfy any of the following two conditions:

1. d = 2, ` ≡ 3 (mod 4), and 4r + 1 ≡ 0 (mod `),

2. d = 3, ` ≡ 5 or 7 (mod 8), and 8r + 3 ≡ 0 (mod `).


p[1,2;−a/b](`n + r) ≡ 0 (mod `).


Suppose a, b, d ∈ Z, b ≥ 1, and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , `, and r satisfy the following condition:

d = 3, ` ≡ 5 or 11 (mod 12) and 2r + 1 ≡ 0 (mod `).


p[1,3;−a/b](`n + r) ≡ 0 (mod `).




Suppose a, b, d ∈ Z, b ≥ 1, and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , `, and r satisfy any of the following two conditions:

1. d = 2, ` ≡ 3 (mod 4), and 12r + 5 ≡ 0 (mod `),

2. d = 3, ` ≡ 3 (mod 4), and 8r + 5 ≡ 0 (mod `).


p[1,4;−a/b](`n + r) ≡ 0 (mod `).


For integer k ≥ 1 and all n ≥ 0, we have

p[1,4;−(5k−3b)/b](5n + r) ≡ 0 (mod 5), where (5, b) = 1 and r ∈ {2, 3}.




Suppose a, b, d ∈ Z, b ≥ 1, and (a, b) = 1. Let ` be an odd prime divisor of a + db and0 ≤ r < `. Suppose d , `, and r satisfy any of the following two conditions:

1. d = 2, ` ≡ 3 (mod 4), and 12r + 5 ≡ 0 (mod `),

2. d = 3, ` ≡ 3 (mod 4), and 8r + 5 ≡ 0 (mod `).


p[1,4;−a/b](`n + r) ≡ 0 (mod `).


For integer k ≥ 1 and all n ≥ 0, we have

p[1,4;−(5k−3b)/b](5n + r) ≡ 0 (mod 5), where (5, b) = 1 and r ∈ {2, 3}.


Fractional 2-color partition functions: Congruences modulo powers of 3, 5, 7, and 11



p[1,2;−(3k−b)/b]

(32s · n + 32s−1 · r +

32s − 1

8

)≡ 0 (mod 3k−2s+1), r ∈ {1, 2},

p[1,2;−(5k−b)/b]

(52s · n + 52s−1 · r +

52s − 1

8

)≡ 0 (mod 5k−2s+1), r ∈ {1, 2, 3, 4},

p[1,2;−(7k−b)/b]

(72s · n + 72s−1 · r +

72s − 1

8

)≡ 0 (mod 7k−2s+1), r ∈ {1, 2, . . . , 6},

p[1,3;−(5k−b)/b]

(52s · n + 52s−1 · r +

52s − 1

6

)≡ 0 (mod 5k−2s+1), r ∈ {1, 2, 3, 4},

p[1,4;−(7k−b)/b]

(72s · n + 72s−1 · r +

11 · 72s−1 − 5

24

)≡ 0 (mod 7k−2s+1), r ∈ {0, 2, 3, . . . , 6},

p[1,3;−(11k−b)/b]

(112s · n + 112s−1 · r +

5 · 112s−1 − 1

6

)≡ 0 (mod 11k−2s+1), r ∈ {0, 2, 3, . . . , 10},

p[1,4;−(11k−b)/b]

(112s · n + 112s−1 · r +

7 · 112s−1 − 5

24

)≡ 0 (mod 11k−2s+1), r ∈ {0, 1, 3, 4, . . . , 10},



Fractional 2-color partition functions: Congruences modulo powers of 3 and 5

Using dissections of (E1E2)2 and (E1E3)2, we find


Let k, m, and s be positive integers such that s ≤ m + 1. Then, for all n ≥ 0, we have

p[1,2;−(3k+m−2b)/b]

(32s · n + 32s−1 · r +

32s − 1

4

)≡ 0 (mod 3k+m−s+1), r ∈ {1, 2}

and

p[1,3;−(5k+m−2b)/b]

(52s · n + 52s−1 · r +

2 · 52s−1 − 1

3

)≡ 0 (mod 5k+m−s+1), r ∈ {0, 2, 3, 4},



Fractional 2-color partition functions: Congruences modulo powers of 3 and 5

Using dissections of (E1E2)2, (E1E2)5, and (E1E3)3, we find



p[1,2;−(3k−5b)/b]

(32s · n + 32s−1 · r +

7 · 32s−1 − 5

8

)≡ 0 (mod 3k ), r ∈ {0, 2},

p[1,2;−(5k−3b)/b]

(52s · n + 52s−1 · r +

7 · 52s−1 − 3

8

)≡ 0 (mod 5k ), r ∈ {0, 2, 3, 4},

and

p[1,3;−(5k−3b)/b]

(52s · n + 52s−1 · r +

52s−1 − 1

2

)≡ 0 (mod 5k ), r ∈ {0, 1, 3, 4},



Fractional 2-color partition functions: Balanced Congruences


Let k > 1 be an odd integer. Then, for all n ≥ 0, we have

p[1,2;−(3k−5b)/b]

(3k · n +

7 · 3k − 5

8

)≡ 0 (mod 3k ),

p[1,2;−(5k−3b)/b]

(5k · n +

7 · 5k − 3

8

)≡ 0 (mod 5k ),

and

p[1,3;−(5k−3b)/b]

(5k · n +

5k − 1

2

)≡ 0 (mod 5k ),



References

[1] Berndt, B. C. Ramanujan’s Notebooks Part III, Springer-Verlag, New York, 1991.

[2] Chan, H. H. and Wang, L. Fractional powers of the generating function for the partitionfunction. Acta Arithmatica, 187(1):59–80, 2019.

[3] Ng, S. T. The Ramanujan’s partition congruences. Undergraduate Thesis, NationalUniversity of Singapore, 2003.

Thanks


Families of Congruences of Fractional Partition Functions ...Families of Congruences of Fractional Partition Functions Modulo Powers of Primes Hirakjyoti Das A joint work with Prof.

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