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We consider a class of generating functions analogous to the generating function of the partition function and establish a bound on the primes $\ell$ for which their coefficients $c(n)$ obey congruences of the form $c(\ell n + a) \equiv 0…

Number Theory · Mathematics 2009-04-24 Jonah Sinick

For the partition function $p(n)$, Ramanujan proved the striking identities $$ P_5(q):=\sum_{n\geq 0} p(5n+4)q^n =5\prod_{n\geq 1} \frac{\left(q^5;q^5\right)_{\infty}^5}{(q;q)_{\infty}^6}, $$ $$ P_7(q):=\sum_{n\geq 0} p(7n+5)q^n…

Number Theory · Mathematics 2025-10-08 Kathrin Bringmann , William Craig , Ken Ono

Let \pi be a partition. In [2] we defined BG-rank(\pi) as an alternating sum of parities of parts. This statistic was employed to generalize and refine the famous Ramanujan modulo 5 partition congruence. Let p_j(n)(a_{t,j}(n)) denote a…

Combinatorics · Mathematics 2007-05-23 Alexander Berkovich , Frank G. Garvan

It is well known that Ramanujan conjectured congruences modulo powers of 5, 7 and and 11 for the partition function. These were subsequently proved by Watson (1938) and Atkin (1967). In 2009 Choi, Kang, and Lovejoy proved congruences modulo…

Number Theory · Mathematics 2021-01-05 Dandan Chen , Rong Chen , Frank Garvan

We investigate Ramanujan congruences for the function which counts the overpartitions of n with restricted odd differences. In particular, we show that only one such congruence exists. Our method involves using the theory of modular forms…

Number Theory · Mathematics 2022-04-07 Michael Hanson , Jeremiah Smith

Let $f(q)$ denote Ramanujan's mock theta function \[f(q) = \sum_{n=0}^{\infty} a(n) q^{n} := 1+\sum_{n=1}^{\infty} \frac{q^{n^{2}}}{(1+q)^{2}(1+q^{2})^{2}\cdots(1+q^{n})^{2}}.\] It is known that there are many linear congruences for the…

Number Theory · Mathematics 2015-04-15 Nickolas Andersen

We prove that, if $m,n\geqslant 1$ and $a_1,\ldots,a_m$ are nonnegative integers, then \begin{align*} \frac{[a_1+\cdots+a_m+1]!}{[a_1]!\ldots[a_m]!}\sum^{n-1}_{h=0}q^h\prod_{i=1}^m{h\brack a_i} \equiv 0\pmod{[n]}, \end{align*} where…

Number Theory · Mathematics 2015-04-22 Victor J. W. Guo , Ji-Cai Liu

For rational $\alpha$, the fractional partition functions $p_\alpha(n)$ are given by the coefficients of the generating function $(q;q)^\alpha_\infty$. When $\alpha=-1$, one obtains the usual partition function. Congruences of the form…

Number Theory · Mathematics 2019-07-17 Erin Bevilacqua , Kapil Chandran , Yunseo Choi

We show that the series expansions of certain $q$-products have \textit{matching coefficients} with their reciprocals. Several of the results are associated to Ramanujan's continued fractions. For example, let $R(q)$ denote the…

Number Theory · Mathematics 2023-01-26 Nayandeep Deka Baruah , Hirakjyoti Das

It is well known that Ramanujan conjectured congruences modulo powers of $5$, $7$ and and $11$ for the partition function. These were subsequently proved by Watson (1938) and Atkin (1967). In 2009 Choi, Kang, and Lovejoy proved congruences…

Number Theory · Mathematics 2024-07-11 Dandan Chen , Rong Chen , Frank Garvan

Let $p_{-k}(n)$ enumerate the number of $k$-colored partitions of $n$. In this paper, we establish some infinite families of congruences modulo 25 for $k$-colored partitions. Furthermore, we prove some infinite families of Ramanujan-type…

Combinatorics · Mathematics 2017-11-08 Dazhao Tang

The partition function $ p_{[1^c\ell^d]}(n)$ can be defined using the generating function, \[\sum_{n=0}^{\infty}p_{[1^c{\ell}^d]}(n)q^n=\prod_{n=1}^{\infty}\dfrac{1}{(1-q^n)^c(1-q^{\ell n})^d}.\] In \cite{P}, we proved infinite family of…

Number Theory · Mathematics 2020-11-17 Shashika Petta Mestrige

Let $\lambda =\left( \lambda_{1},\lambda_{2},...,\lambda_{r}\right) $ be an integer partition, and $\left[p_{\lambda }\right] $ the $q$-analog of the symmetric power function $%p_{\lambda }$. This $q$-analogue has been defined as a special…

Combinatorics · Mathematics 2024-09-16 Vincent Brugidou

Denote by p_k the k-th power sum symmetric polynomial n variables. The interpretation of the q-analogue of the binomial coefficient as Hilbert function leads us to discover that n consecutive power sums in n variables form a regular…

Commutative Algebra · Mathematics 2018-08-30 Aldo Conca , Christian Krattenthaler , Junzo Watanabe

We obtain asymptotic formulas for sums over arithmetic progressions of coefficients of polynomials of the form $$\prod_{j=1}^n\prod_{k=1}^{p-1}(1-q^{pj-k})^s,$$ where $p$ is an odd prime and $n, s$ are positive integers. Let us denote by…

Number Theory · Mathematics 2021-04-08 Jiyou Li , Xiang Yu

The coefficients of the generating function $(q;q)^\alpha_\infty$ produce $p_\alpha(n)$ for $\alpha \in \mathbb{Q}$. In particular, when $\alpha = -1$, the partition function is obtained. Recently, Chan and Wang identified and proved…

Number Theory · Mathematics 2021-03-16 Yunseo Choi

Recently, several mathematicians have investigated various partition functions with the goal of discovering Ramanujan-type congruences. One such function is $\overline{B}_{2^\alpha}(n)$, which represents the number of $2^\alpha-$regular…

Number Theory · Mathematics 2025-02-25 Hemanthkumar B. , Sumanth Bharadwaj H. S

Let $A(q)=\sum_{n=0}^{\infty}a_n q^n$ and $B(q)=\sum_{n=0}^{\infty}b_n q^n$ be two eta quotients. Previously, we considered the problem of when \[ a_n=0 <=> b_n=0. \] Here we consider the ``mod $m$'' version of this problem, i.e. eta…

Number Theory · Mathematics 2025-07-29 Tim Huber , James McLaughlin , Dongxi Ye

Let $p_k(n)$ be given by the $k$-th power of the Euler Product $\prod _{n=1}^{\infty}(1-q^n)^k=\sum_{n=0}^{\infty}p_k(n)q^{n}$. By investigating the properties of the modular equations of the second and the third order under the Atkin…

Combinatorics · Mathematics 2018-03-14 Julia Q. D. Du , Edward Y. S. Liu , Jack C. D. Zhao

We give elementary proofs for the Apagodu-Zeilberger-Stanton-Amdeberhan-Tauraso congruences $$\sum\limits_{n=0}^{p-1}\dbinom{2n}{n} \equiv\eta_{p}\mod p^{2},$$ $$\sum\limits_{n=0}^{rp-1}\dbinom{2n}{n}…

Combinatorics · Mathematics 2019-01-31 Darij Grinberg
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