On a Ramanujan-Type Congruence for Partition Triples with 5-Cores

N. V. Majid and S. N. Fathima Ramanujan School of Mathematical Science Department of Mathematics Pondicherry University Puducherry India

                     nvmajidtgi@gmail.com
                    dr.fathima.sn@gmail.com

Abstract Let B_{5}^{(3)}(n) denote the number of partition triples with 5-cores. In this short note, we prove a Ramanujan-type congruence modulo 5^{α} (α ≥ 1) for B_{5}^{(3)}(n) by using recurrence relations for the coefficients of B_{5}^{(3)}(n).

1 Introduction

A partition of a positive integer n is a finite non-increasing sequence of positive integers whose sum equals n. Let p(n) be the number of partitions of n. The generating function for p(n) is given by

∑_{n=0}^{∞} p(n)q^{n} = \frac{1}{(q; q)_{∞}} ,

where (a; q)_{0} = 1, (a; q)_{n} = ∏_{k=0}^{n−1}(1 − aq^{k}), and (a; q)_{∞} = ∏_{k=0}^{∞}(1 − aq^{k}).

The most inspiring congruences of p(n) discovered by Ramanujan for n ≥ 0 are

p(5n + 4) ≡ 0 (mod 5),
p(7n + 5) ≡ 0 (mod 7),
p(11n + 6) ≡ 0 (mod 11).

Ramanujan’s discovery inspired researchers to study the arithmetic properties of the restricted partitions, such as the t-core partition. A Ferrers-Young diagram of a partition

n = ∑_{i=1}^{k} λ_{i}, where λ_{i} ≥ λ_{k}, i ≤ k, is a pattern of dots, with λ_{i} dots in the i^{th} row. The hook

number of a dot is defined by the number of dots directly below, together with the number of dots directly to the right, as well as the dot itself. For a positive integer t ≥ 2, a partition is said to be t-core, if it has no hook numbers that are multiples of t.

Example 1. The Ferrers-Young diagram of the partition λ = 4 + 3 + 2 of 9 with the corresponding hook number is as follows:

•^{6}  •^{5}  •^{3}  •^{1}
•^{4}  •^{3}  •^{1}
•^{2} •^{1}

Here λ is 7-core and λ is t-core for t ≥ 7.

We let B_{t}(n) denote the number of t-core partitions of n. The generating function for the B_{t}(n) is given by [7, Eq. (2.1)]

∑_{n=0}^{∞} B_{t}(n)q^{n} = \frac{f_{t}^{t}}{f_{1}} ,

where f_{t} = (q^{t}; q^{t})_{∞}, for any integer t ≥ 2. Arithmetic properties of t-core partitions have been studied by several mathematicians (see, for example [1, 3, 4, 5, 8, 9, 12, 13]).

A partition triple or a tripartition of n is a triple of partitions (λ_{1}, λ_{2}, λ_{3}) such that the sum of all the parts of λ_{1}, λ_{2}, and λ_{3} equals n. A tripartition with t-core of n is a partition triples (λ_{1}, λ_{2}, λ_{3}) of n such that λ_{1}, λ_{2}, and λ_{3} are t-cores. Let B_{t}^{(3)}(n) denote the number of partition triples with t-cores of n. The generating function for B_{t}^{(3)}(n) is given by

∑_{n=0}^{∞} B_{t}^{(3)}(n)q^{n} = \frac{f_{t}^{3t}}{f_{1}^{3}} .  (1)

Dasappa [6] proved a Ramanujan-type congruence modulo 5^{α} for bipartition with 5-cores, which motivated establishing a Ramanujan-type congruence for tripartitions with 5-core.

The ultimate aim of this note is to prove the following Ramanujan-type congruence for B_{5}^{(3)}(n):

B_{5}^{(3)}(5^{α}n + 5^{α} − 3) ≡ 0 (mod 5^{α}), α ≥ 1.

The following are crucial lemmas that help prove our main congruence for B_{5}^{(3)}(n): Lemma 2. Let ∑_{n=0}^{∞} P_{4}(n)q^{n} = qf_{5}^{9}f_{1}^{3}. Then

∑_{n=0}^{∞} P_{4}(5n + 4)q^{n} = 5f_{5}^{3}f_{1}^{9}.  (2)

Proof. The following 5-dissection formula was first stated by Ramanujan [10, p. 212] without any proof:

f_{1} = f_{25}(\frac{1}{R(q^{5})} − q − q^{2}R(q^{5})),  (3)

where R(q) = \frac{(q;q^{5})_{∞}(q^{4};q^{5})_{∞}}{(q^{2};q^{5})_{∞}(q^{3};q^{5})_{∞}}. Watson [11] provided a proof of (3).

Using (3), we obtain

∑_{n=0}^{∞} P_{4}(n)q^{n} = f_{5}^{9}f_{25}^{3}(\frac{q}{R(q^{5})^{3}} − \frac{3q^{2}}{R(q^{5})^{2}} + 5q^{4} − 3q^{6}R(q^{5})^{2} − q^{7}R(q^{5})^{3}).  (4)

Extracting the terms involving q^{5n+4} from (4), dividing by q^{4}, and replacing q^{5} by q, we obtain (2). Lemma 3. Let ∑_{n=0}^{∞} P_{5}(n)q^{n} = f_{5}^{3}f_{1}^{9}. Then

∑_{n=0}^{∞} P_{5}(5n + 4)q^{n} = −90f_{5}^{3}f_{1}^{9} − 625qf_{5}^{9}f_{1}^{3}.

Proof. Using (3), we obtain

∑_{n=0}^{∞} P_{5}(n)q^{n} = f_{5}^{3}f_{25}^{9}(\frac{1}{R(q^{5})^{9}} − \frac{9q}{R(q^{5})^{8}} + \frac{27q^{2}}{R(q^{5})^{7}} − \frac{12q^{3}}{R(q^{5})^{6}} − \frac{90q^{4}}{R(q^{5})^{5}} + \frac{126q^{5}}{R(q^{5})^{4}} + \frac{126q^{6}}{R(q^{5})^{3}}

− \frac{288q^{7}}{R(q^{5})^{2}} − \frac{117q^{8}}{R(q^{5})} + 365q^{6} + 117q^{10}R(q^{5}) − 288q^{11}R(q^{5})^{2} − 126q^{12}R(q^{5})^{3}

+ 126q^{13}R(q^{5})^{4} + 90q^{14}R(q^{5})^{5} − 12q^{15}R(q^{5})^{6} − 27q^{16}R(q^{5})^{7} − 9q^{17}R(q^{5})^{8}

− q^{18}R(q^{5})^{9}).  (5)

Extracting the terms involving q^{5n+4} from (5), dividing by q^{4}, and replacing q^{5} by q, we obtain

∑_{n=0}^{∞} P_{5}(5n + 4)q^{n} = f_{5}^{9}f_{1}^{3}(90q^{2}R(q)^{5} + 365q − \frac{90}{R(q)^{5}} ).  (6)

Berndt [2, Thm. 7.4.4] gave the following identity:

\frac{1}{R(q)^{5}} − 11q − q^{2}R(q)^{5} = \frac{(q; q)_{∞}^{6}}{(q^{5}; q^{5})_{∞}^{6}} .  (7)

Employing (7) in (6), we obtain

∑_{n=0}^{∞} P_{5}(5n + 4)q^{n} = f_{5}^{9}f_{1}^{3}(−90\frac{f_{1}^{6}}{f_{5}^{6}} − 625q) = −90f_{5}^{3}f_{1}^{9} − 625qf_{5}^{9}f_{1}^{3}.

Lemma 4. Let ∑_{n=0}^{∞} P_{3}(n)q^{n} = \frac{1}{f_{1}^{3}} . Then

∑_{n=0}^{∞} P_{3}(5n + 2)q^{n} = 9\frac{f_{5}^{3}}{f_{1}^{6}} + 375q \frac{f_{5}^{9}}{f_{1}^{12}} + 3125q^{2} \frac{f_{5}^{15}}{f_{1}^{18}} .

Proof. [2, Eq. (7.4.14), p. 165] We have

\frac{1}{f_{1}} =\frac{f_{25}^{5}}{f_{5}^{6}} (\frac{1}{R(q^{5})^{4}} + \frac{q}{R(q^{5})^{3}} + \frac{2q^{2}}{R(q^{5})^{2}} + \frac{3q^{3}}{R(q^{5})} + 5q^{4} − 3q^{5}R(q^{5})
+ 2q^{6}R(q^{5})^{2} − q^{7}R(q^{5})^{3} + q^{8}R(q^{5})^{4}).  (8)

Using (8), we obtain

∑_{n=0}^{∞} P_{3}(n)q^{n} = \frac{f_{25}^{15}}{f_{5}^{18}} (\frac{1}{R(q^{5})^{12}} + \frac{3q}{R(q^{5})^{11}} + \frac{9q^{2}}{R(q^{5})^{10}} + \frac{22q^{3}}{R(q^{5})^{9}} + \frac{51q^{4}}{R(q^{5})^{8}} + \frac{78q^{5}}{R(q^{5})^{7}} + \frac{134q^{6}}{R(q^{5})^{6}}

+ \frac{177q^{7}}{R(q^{5})^{5}} + \frac{216q^{8}}{R(q^{5})^{4}} + \frac{153q^{9}}{R(q^{5})^{3}} + \frac{219q^{10}}{R(q^{5})^{2}} + \frac{57q^{11}}{R(q^{5})} + 71q^{12} − 57q^{13}R(q^{5})

+ 219q^{14}R(q^{5})^{2} − 153q^{15}R(q^{5})^{3} + 216q^{16}R(q^{5})^{4} − 177q^{17}R(q^{5})^{5} + 134q^{18}R(q^{5})^{6} − 78q^{19}R(q^{5})^{7} + 51q^{20}R(q^{5})^{8} − 22q^{21}R(q^{5})^{9} + 9q^{22}R(q^{5})^{10} − 3q^{23}R(q^{5})^{11}

+ q^{24}R(q^{5})^{12}).  (9)

Extracting the terms involving q^{5n+2} from (9), dividing by q^{2}, and replacing q^{5} by q, we obtain

∑_{n=0}^{∞} P_{3}(5n + 2)q^{n} = \frac{f_{5}^{15}}{f_{1}^{18}} (9q^{4}R(q)^{10} − 177q^{3}R(q)^{5} + 71q^{2} + 177\frac{q}{R(q)^{5}} + \frac{9}{R(q)^{10}} ). (10)

Employing (7) in (10), we deduce that

∑_{n=0}^{∞} P_{3}(5n + 2)q^{n} = \frac{f_{5}^{15}}{f_{1}^{18}} (9\frac{f_{1}^{12}}{f_{5}^{12}} + 375q \frac{f_{1}^{6}}{f_{5}^{6}} + 3125q^{2})

= 9\frac{f_{5}^{3}}{f_{1}^{6}} + 375q \frac{f_{5}^{9}}{f_{1}^{12}} + 3125q^{2} \frac{f_{5}^{15}}{f_{1}^{18}} .

Theorem 5. For all integers α ≥ 0, we have

∑_{n=0}^{∞} B_{5}^{(3)}(5^{α+1}n + 5^{α+1} − 3)q^{n} = A_{α}f_{5}^{3}f_{1}^{9} + B_{α}qf_{5}^{9}f_{1}^{3} + C_{α}q^{2} ∑_{n=0}^{∞} B_{5}^{(3)}(n)q^{n},  (11)

where A_{0} = 9, B_{0} = 375, C_{0} = 3125, and for any integer n ≥ 1

A_{n} = −90A_{n−1} + 5B_{n−1} + 9C_{n−1},  (12)
B_{n} = −625A_{n−1} + 5B_{n−1} + 375C_{n−1},  (13)
C_{n} = C_{0}^{n+1}.  (14)

Proof. From (1), we have

∑_{n=0}^{∞} B_{5}^{(3)}(n) = \frac{f_{5}^{15}}{f_{1}^{3}} .

Employing Lemma 4, we obtain

∑_{n=0}^{∞} B_{5}^{(3)}(5n + 2)q^{n} = 9f_{5}^{3}f_{1}^{9} + 375qf_{5}^{9}f_{1}^{3} + 3125q^{2} ∑_{n=0}^{∞} B_{5}^{(3)}(n)q^{n}.  (15)

Eq. (15) is the α = 0 case of Eq. (11). Now assume for α ≥ 0. Replacing n by 5n + 4 in (11), using Lemmas (2) and (3), and (15), we obtain

∑_{n=0}^{∞} B_{5}^{(3)}(5^{α+2}n + 5^{α+2} − 3)q^{n}

= A_{α}(−90f_{5}^{3}f_{1}^{9} − 625qf_{5}^{9}f_{1}^{3}) + B_{α}5f_{5}^{3}f_{1}^{9} + C_{α}(9f_{5}^{3}f_{1}^{9} + 375qf_{5}^{9}f_{1}^{3} + 3125q^{2} ∑_{n=0}^{∞} B_{5}^{(3)}(n)q^{n})

= (−90A_{α} + 5B_{α} + 9C_{α})f_{5}^{3}f_{1}^{9} + (−625A_{α} + 375C_{α})f_{5}^{9}f_{1}^{3} + 3125C_{α}q^{2} ∑_{n=0}^{∞} B_{5}^{(3)}(n)q^{n}

= A_{α+1}f_{5}^{3}f_{1}^{9} + B_{α+1}qf_{5}^{9}f_{1}^{3} + C_{α+1}q^{2} ∑_{n=0}^{∞} B_{5}^{(3)}(n)q^{n}.

That is, (11) holds for α + 1. This completes the proof by induction of (11).

Theorem 6. For all integers α ≥ 1 and n ≥ 0, we have

B_{5}^{(3)}(5^{α}n + 5^{α} − 3) ≡ 0 (mod 5^{α}).  (16)

Proof. From (12), (13), and (14), we see that

A_{1} ≡ 0 (mod 5), B_{1} ≡ 0 (mod 5^{3}), C_{1} ≡ 0 (mod 5^{5}),
A_{2} ≡ 0 (mod 5^{2}), B_{2} ≡ 0 (mod 5^{4}), C_{2} ≡ 0 (mod 5^{10}),
⋮  ⋮  ⋮
A_{α} ≡ 0 (mod 5^{α}), B_{α} ≡ 0 (mod 5^{α+2}), C_{α} ≡ 0 (mod 5^{5α}).

That is, A_{α}, B_{α}, and C_{α} are multiples of 5^{α}. Then (11) implies (16).

2 Acknowledgments

The first author is thankful to CSIR-UGC Junior Research Fellowship, Government of India, Council of Scientific & Industrial Research, CSIR Complex, Library Avenue, Pusa, New Delhi-110012. We would also like to extend our gratitude to the referee from the Journal of Integer Sequences for careful reading and comments on the manuscript.

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2020 Mathematics Subject Classification: Primary 05A17; Secondary 11P83. Keywords: congruence, partition, core partition.

Received December 29 2021; revised versions received December 30 2021; June 21 2022; June 23 2022. Published in Journal of Integer Sequences, June 23 2022.

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