Rademacher's infinite partial fraction conjecture for partition generating functions

Let pN(n)p_N(n) denote the number of partitions of nn into at most NN parts, and write

n0pN(n)xn=j=1N11xj\sum_{n\geq 0}p_N(n)x^n=\prod_{j=1}^N\frac{1}{1-x^j}

with partial-fraction coefficients Ch,k,l(N)C_{h,k,l}(N) defined by

j=1N11xj=k=1N0h<kgcd(h,k)=1l=1N/kCh,k,l(N)(xe2πih/k)l.\prod_{j=1}^N\frac{1}{1-x^j}=\sum_{k=1}^N\sum_{\substack{0\leq h<k\gcd(h,k)=1}}\sum_{l=1}^{\lfloor N/k\rfloor}\frac{C_{h,k,l}(N)}{(x-e^{2\pi i h/k})^l}.

Here s(h,k)s(h,k) is the Dedekind sum, Δα\Delta_\alpha is the forward difference operator, and L3/2L_{3/2} is given by L3/2(y2)=12πy2(2cos(2y)sin(2y)y)L_{3/2}(-y^2)=-\frac{1}{2\sqrt{\pi}y^2}\left(2\cos(2y)-\frac{\sin(2y)}{y}\right). Rademacher's conjecture. For all integers h,k,lh,k,l such that 0h<k0\leq h<k, gcd(h,k)=1\gcd(h,k)=1, and l1l\geq1, the limit limNCh,k,l(N)\lim_{N\to\infty}C_{h,k,l}(N) exists and equals

Rh,k,l:=2π(π12)3/2eπi(s(h,k)+2hl/k)k5/2Δαl1L3/2(π26k2(α+1)),R_{h,k,l}:=-2\pi\left(\frac{\pi}{12}\right)^{3/2}\frac{e^{\pi i(s(h,k)+2hl/k)}}{k^{5/2}}\Delta^{l-1}_{\alpha}L_{3/2}\left(-\frac{\pi^2}{6k^2}(\alpha+1)\right),

evaluated at α=124\alpha=\frac{1}{24}, where

s(h,k)=μ=1k1(μkμk12)(hμkhμk12)s(h,k)=\sum_{\mu=1}^{k-1}\left(\frac{\mu}{k}-\left\lfloor\frac{\mu}{k}\right\rfloor-\frac12\right)\left(\frac{h\mu}{k}-\left\lfloor\frac{h\mu}{k}\right\rfloor-\frac12\right)

and

Δαjf(α)=h=0j(1)h(jh)f(α+jh).\Delta_\alpha^j f(\alpha)=\sum_{h=0}^j(-1)^h\binom{j}{h}f(\alpha+j-h).

The conjecture was open for nearly four decades, but the paper presents overwhelming evidence that these limits do not exist: the sequences oscillate and attain arbitrarily large positive and negative values. Thus the conjecture is almost certainly false.

Sources & referencesView supporting material

Primary source

Andrew V. Sills and Doron Zeilberger, “Rademacher's infinite partial fraction conjecture is (almost certainly) false”, arXiv:1110.4932 (2011).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.