Higher-power congruence conjecture for partition coefficients

At least 12 years old · documented by 2013

Let $p_r(n)$ denote the coefficients in the partition-generating series used in the paper. For a positive integer $\lambda$ , let $\mu_{\lambda}$ and $\delta_{\lambda}$ satisfy

12\mu_{\lambda}\equiv1\pmod{5^{\lambda}},\qquad24\delta_{\lambda}\equiv1\pmod{5^{\lambda}}.

Higher-power congruence conjecture. For integers $r$ and $k$ with $r\equiv2\pmod5$ and $k\equiv1\pmod5$ , respectively,

p_r(5^{\lambda}n+\mu_{\lambda})\equiv0\pmod{5^{\lambda}},\qquad p_k(5^{\lambda}n+\delta_{\lambda})\equiv0\pmod{5^{\lambda}}.

This is presented as a generalization of Ramanujan's conjectures using the techniques developed in the paper. The supplied text gives no proof or resolution.

References

Primary source

Tim Huber, “A theory of theta functions to the quintic base”, arXiv:1304.0684 (2013).

modular-forms partition-congruences

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader claims explicit calculations disprove both congruence families, but no independent source has checked the alleged counterexamples.

Huber’s 2013 paper formulates the higher-power congruence conjecture for partition coefficients, extending Ramanujan-type divisibility to every positive power of $5$ . The retrieved primary source gives no proof or resolution.

Posted attempt

A reader claims the first family fails at $r=2$ , $\lambda=3$ , $n=0$ , with $p_2(73)\equiv100\pmod{125}$ , and the second fails at $r=6$ , $\lambda=3$ , $n=0$ , with $p_6(99)\equiv50\pmod{125}$ . The same calculation claims infinitely many first-family failures modulo $25$ . This is an alleged complete refutation, but it has not been independently verified.

Current status (as of August 2026): The conjecture is not verified; a reader-written calculation claims both families are false, while no independently checked proof or counterexample is recorded.

Sources

arxiv.org

CounterexampleThis solution needs a summarySee full solutionHide full solution

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.