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Let $a_c(n)$ be the number of ways to partition a positive integer $n$ where each even part comes in $c$ colors. Then, we can supply the generating function $$\sum_{n\geq0}a_c(n)q^n=\prod_{k\geq1}\frac1{(1-q^k)(1-q^{2k})^{c-1}}.$$ In particular, $a_1(n)=p(n)$ is the usual number of (unrestricted) partitions of $n$; $a_2(n)$ is the so-called cubic partitions of $n$.

Example. $a_2(2)=\#\{{\color{red}2},2,11\}=3, a_2(4)=\#\{{\color{red}4},4,31,{\color{red}{22}},{\color{red}2}2,22,{\color{red}2}11,211,1111\}=9$.

Experiments suggest the below congruences.

QUESTION. Are these statements true?

$\bullet$ $a_2(3n+2)\equiv0\pmod 3$

$\bullet$ $a_4(5n+4)\equiv0\pmod 5$

$\bullet$ $a_3(7n+4)\equiv0\pmod 7$

$\bullet$ $a_5(11n+10)\equiv0\pmod{11}$.

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    $\begingroup$ This is very similar to Ramanujan's congruences. Have you tried to adapt a proof of those? $\endgroup$ Commented Jan 8, 2024 at 4:11
  • $\begingroup$ You are right, it seems the natural course but it is easier said than done. $\endgroup$ Commented Jan 8, 2024 at 13:30
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    $\begingroup$ The modulo 3 result is proven in H.-C. Chan, Ramanujan’s cubic continued fraction and ananalog of his “most beautiful identity”, Int. J. Number Theory 6 no.3 (2010), 673–680. $\endgroup$ Commented Jan 17, 2024 at 3:00
  • $\begingroup$ @BrianHopkins: thank you. $\endgroup$ Commented Jan 27, 2024 at 15:57

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Daniel Weber's comment provides a nice place to look for an answer to your question. I will provide an alternate approach.

One way to view congruences is through the lens of cranks. Essentially, for $p(n)$, the standard partition function, the crank splits the partitions of $5n+4$ into $5$ equally sized sets (also for the other two Ramanujan congruences.)

A pre-print from Rolan, Tripp, and Wagner, provides a starting point for a crank for the partitions you are describing. As Brian Hopkins pointed out, the mod $3$ result has been proven. Looking at Tables 2 and 3 of the pre-print, we see the mod $5$ and mod $7$ results.

For the mod $11$ result, one may consider the following Jacobi Form $$ \prod_{n=1}^\infty\frac{(1-q^n)}{(1-\zeta q^n)(1-\zeta^{-1} q^n)(1-\zeta^5 q^{2n})(1-\zeta^{-5} q^{2n})(1-\zeta^7 q^{2n})(1-\zeta^{-7} q^{2n})} $$ Where $q = e^{2 \pi i \tau}$ and $\zeta = e^{2 \pi i z}$ for $\tau,z \in \mathbb{H}$, the upper half complex plane. Can you (using some computer assistance) see that $$ \Phi_{11}(\zeta) = 1+\zeta+\zeta^2+\ldots+\zeta^{10} $$ divides the coefficient of $q^{11n+10}$?

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