Let $t\in\Bbb{N}$ and consider the sequences $p\_t(n)$ defined by
$$\sum\_{n\geq0}p\_t(n)x^n=\prod\_{i\geq1}\frac1{(1-x^i)^t}=(x;x)\_{\infty}^{-t}.$$
The numbers $p\_t(n)$ can be regarded as enumerating *partitions of $n$ into parts of $t$ colors*.
Furthermore, $p\_t(n)=\sum\_{\lambda\vdash n}\prod\_{j\geq1}\binom{k\_j+t-1}{t-1}$ where
$\lambda=1^{k\_1}2^{k\_2}\cdots$ and each $k\_j\geq0$.


Note also that $p\_1(n)=p(n)$ is the
usual unrestricted partition of $n$. Ramanujan's famous congruences state
$$\begin{cases}
 p(5n+4)\equiv0\mod 5, \\ p(7n+5)\equiv0\mod 7, \\ p(11n+6)\equiv0\mod 11. 
\end{cases}$$
Following this tradition, I ask:



> 
> **QUESTION.** Do these congruences hold true?
> $$\begin{cases}
>  p\_4(5n+4)\equiv0\mod 5, \\ p\_4(7n+5)\equiv0\mod 7, \\ p\_3(11n+7)\equiv0\mod 11. 
> \end{cases}$$
> 
> 
> 


**POSTSCRIPT.** I thank Gjergji Zaimi for his response shown below. I now exhibit all the congruences that I have found. Let's see how many are already covered in the literature.
$$\begin{cases} p\_t(5n+1)\equiv0\mod 5 \qquad {} \text{if $t\equiv0\mod5$}, \\
p\_t(5n+2)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,2\mod5$}, \\
p\_t(5n+3)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,2,4\mod5$}, \\
p\_t(5n+4)\equiv0\mod 5 \qquad {} \text{if $t\equiv0,1,2,4\mod5$}. \end{cases}$$


$$\begin{cases}
p\_t(7n+1)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,\mod7$}, \\
p\_t(7n+2)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4\mod7$}, \\
p\_t(7n+3)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,6\mod7$}, \\
p\_t(7n+4)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4,6\mod7$}, \\
p\_t(7n+5)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,1,4\mod7$}, \\
p\_t(7n+6)\equiv0\mod 7 \qquad {} \text{if $t\equiv0,4,6\mod7$}. \end{cases}$$


$$\begin{cases}
p\_t(11n+1)\equiv0\mod 11 \qquad\text{if $t\equiv0\mod11$}, \\
p\_t(11n+2)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\
p\_t(11n+3)\equiv0\mod 11 \qquad\text{if $t\equiv0,10\mod11$}, \\
p\_t(11n+4)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\
p\_t(11n+5)\equiv0\mod 11 \qquad\text{if $t\equiv0,8\mod11$}, \\
p\_t(11n+6)\equiv0\mod 11 \qquad\text{if $t\equiv0,1,10\mod11$}, \\
p\_t(11n+7)\equiv0\mod 11 \qquad\text{if $t\equiv0,3,8\mod11$}, \\
p\_t(11n+8)\equiv0\mod 11 \qquad\text{if $t\equiv0,5,8,10\mod11$}, \\
p\_t(11n+9)\equiv0\mod 11 \qquad\text{if $t\equiv0,7,8,10\mod11$}, \\
p\_t(11n+10)\equiv0\mod 11 \qquad\text{if $t\equiv0,10\mod11$}. \end{cases}$$



Congruences for "colored partitions" a la Ramanujan
Yes, these are all true and they are all in the literature. The first two congruences are part of infinite families of the form
$$\begin{cases}
 p\_{\ell-1}(\ell n+a)\equiv 0\mod \ell, \\ p\_{\ell-3}(\ell n+b)\equiv0\mod \ell,
\end{cases}$$
where $\ell \geq 5$ is a prime and $a,b$ are such that: $24a+1$ is a quadratic nonresidue, $8b+1$ is either a quadratic nonresidue or divisible by $\ell$. Both congruences are immediate consequences of the Euler pentagonal theorem and Jacobi triple product respectively. This is explained in section 2 of the paper: I. Kiming and J. Olsson, *Congruences like Ramanujan's for powers of the partition function* (Arch. Math., 59(4):348-360, 1992) [doi: 10.1007/BF01197051](https://doi.org/10.1007/BF01197051), where they are called the *non-exceptional* case.


The third identity is a bit more recent. It was proven in Lin, B.L.S. *Ramanujan-style proof of $p\_{−3}(11n+7)≡0 \pmod{11}$* (Ramanujan J 42, 223–231 (2017)) [doi:10.1007/s11139-015-9733-5](https://doi.org/10.1007/s11139-015-9733-5). It was generalized to the following infinite family
$$p\_{\ell-8}\left(\ell n+\frac{\ell^2-1}{3}-\ell \left\lfloor\frac{\ell^2-1}{3\ell}\right\rfloor\right)\equiv 0 \mod \ell.$$
For this last congruence and many other *exceptional* infinite families see the paper Locus, M., Wagner, I. *Congruences for Powers of the Partition Function* (Ann. Comb. 21, 83–93 (2017)) [doi:10.1007/s00026-017-0342-4](https://doi.org/10.1007/s00026-017-0342-4) arXiv:[1604.07495](https://arxiv.org/abs/1604.07495).


