Higher-power congruence conjecture for partition coefficients

Both families of Conjecture 6.6 are false, including the first family at its smallest positive admissible color.

Write

$$F_s(q)=\prod_{j\ge1}(1-q^j)^{-s}=\sum_{N\ge0}p_s(N)q^N.$$

Its logarithmic derivative gives the exact integer recurrence

$$p_s(0)=1,\qquad N p_s(N)=s\sum_{j=1}^{N}\sigma_1(j)p_s(N-j).$$

For the first claimed family, take $s=2$, $\lambda=3$, $n=0$, and $\mu_3=73$. Indeed $2\equiv2\pmod5$ and $12\cdot73=876\equiv1\pmod{125}$, but the recurrence gives

$$p_2(73)=13\,733\,486\,100\equiv100\not\equiv0\pmod{125}.$$

For the second claimed family, take $s=6$, $\lambda=3$, $n=0$, and $\delta_3=99$. Here $6\equiv1\pmod5$ and $24\cdot99=2376\equiv1\pmod{125}$, whereas

$$p_6(99)=2\,341\,063\,750\,232\,023\,186\,800 \equiv50\not\equiv0\pmod{125}.$$

Moreover, the first family has infinitely many counterexamples already modulo $25$: for every integer $t\ge0$,

$$p_{12+25t}(23)\equiv20\pmod{25}.$$

To prove this, the recurrence yields

$$\begin{array}{c|rrrrr} N&3&8&13&18&23\\ \hline p_{12}(N)\bmod25&20&10&5&15&20. \end{array}$$

The lifted freshman's dream gives

$$(1-x)^{25}\equiv(1-x^5)^5\pmod{25}, \qquad F_{25t}(q)\equiv F_{5t}(q^5)\pmod{25}.$$

Thus

$$p_{12+25t}(23) \equiv\sum_{j=0}^{4}p_{12}(23-5j)p_{5t}(j)\pmod{25}.$$

For $1\le j\le4$, the congruence $F_{5t}(q)\equiv F_t(q^5)\pmod5$ implies $5\mid p_{5t}(j)$; the displayed coefficient table also gives $5\mid p_{12}(23-5j)$. Hence every term with $j\ge1$ vanishes modulo $25$, leaving

$$p_{12+25t}(23)\equiv p_{12}(23) =711\,028\,604\,520\equiv20\pmod{25}.$$

All these colors satisfy $12+25t\equiv2\pmod5$, and $12\cdot23\equiv1\pmod{25}$.

Source: T. Huber, “A theory of theta functions to the quintic base,” Journal of Number Theory 134 (2014), 49–92, equation (1.25) and Conjecture 6.6, doi:10.1016/j.jnt.2013.06.004.