# Asymptotics related to the Erdos--Moser diophantine equation

I share the authorship of this question with Pieter Moree. In our recent joint work with Y. Gallot (arXiv:0907.1356 [math.NT]) we attack the Erdős--Moser diophantine equation $$1^k+2^k+\dots+(m-2)^k+(m-1)^k=m^k,$$ which conjecturally has no solutions in integers $k>1$ and $m$, using the theory of continued fractions (an approach initiated by M. Best and H. te Riele) together with certain divisibility criteria. The main result of the paper is that the existence of a solution $k,m$ implies quite strong (and unlikely happening) conditions on the continued fraction expansion of $\log2$.

The main analytic argument rests upon the asymptotic behaviour of real (rather than integer) $k$ as function of $m$. Based on ideas of Best and te Riele we prove that $$\begin{eqnarray*} k &=cm-\frac{3}{2}c-\biggl(\frac{25}{12}c-3c^2\biggr)m^{-1} +\biggl(-\frac{73}{8}c+\frac{61}2c^2-25c^3\biggr)m^{-2} \\\ &\qquad +\biggl(-\frac{41299}{720}c+\frac{657}2c^2-598c^3+\frac{1405}4c^4\biggr)m^{-3}+O(m^{-4}) \\\ &\approx 0.69314718m-1.03972077-0.00269758m^{-1}+0.00323260m^{-2} \\\ &\qquad +0.00217182m^{-3}+O(m^{-4}), \end{eqnarray*} \qquad{(1)}$$ where $c=\log2$. Moreover, we present a simple recipe to get as many terms of the asymptotic series (1) as required, although no closed form for its coefficients is (expected to be) obtained. Our approach works for the more general equation $$1^k+2^k+\dots+(m-1)^k=tm^k, \qquad{(2)}$$ with $t\in\mathbb Z_{>0}$ (or even $t\in\mathbb R_{>0}$) fixed, as well. What we can show in this way is the existence of the asymptotic series $$\frac{k_t(m)}m =c_0+\frac{c_1}m+\frac{c_2}{m^2}+\dots+\frac{c_n}{m^n}\dots, \qquad{(3)}$$ where $$c_0=c(t)=\log\biggl(1+\frac1t\biggr)=\log\frac{t+1}t, \qquad c_1=-\biggl(t+\frac12\biggr)c,$$ $$c_2=\biggl(t+\frac12\biggr)^3c^2-\biggl(t+\frac12\biggr)^2c -\frac14\biggl(t+\frac12\biggr)c^2+\frac c6$$ and, in general, $c_n$ is a polynomial in $t$ and $c$ of degree $2n-1$ and $n$, respectively, for each $n=1,2,\dots$\,. Moreover, $c_n(-(t+1))=(-1)^{n+1}c_n(t)$ for $n=0,1,2,\dots$; this reflects the equivalence of equation (2) and $$1^k+2^k+\dots+(m-1)^k+m^k=(t+1)m^k.$$

Although the desirable thing is to prove the irrationality of the quotient $k_t(m)/m$ for positive integers $t$ (which would clearly imply the absence of integer solutions of the corresponding equation (2)), we already do not know how to treat a `simpler' but analytical problem.

Problem. Determine the convergence domain of the asymptotic series on the right-hand side of (3). Specify the asymptotic behaviour of $c_n=c_n(t)$ as $n\to\infty$ and provide sharp (closed-form) estimates for $|c_n(t)|$.

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