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OEIS sequence A210247 gives the signs of $\text{li}(-n,-1/3) = \sum_{k=1}^\infty (-1)^k k^n/3^k$, also the signs of the Maclaurin coefficients of $4/(3 + \exp(4x))$.

Mikhail Kurkov noticed that it appeared that $a(n+28) = -a(n)$ for this sequence. It's not quite true: the first $n$'s for which this is not true are $578, 1143,$ and $1736$. But still it's remarkably close to true.

A nice illustration of the Strong Law of Small Numbers? Is there any explanation for the almost-identity, or is it just coincidence? Is it still true for large $n$ that $a(n+28)$ is usually $-a(n)$?

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1 Answer 1

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Consider $F(z) = 4/(3+\exp(4z))$ as a function of the complex variable $z$. It is meromorphic and has simple poles where the denominator vanishes. Namely when $4z = \log 3 + (2k +1)\pi i$ for integers $k$. The poles with the smallest magnitude of $z$ occur when $4z = \log 3 \pm \pi i$. We can compute the Taylor series coefficients of $F$ by looking at $$ \frac{1}{2\pi i} \int_{|z|= r} F(z) z^{-n} \frac{dz}{z}, $$ starting with $r$ suitably small. Now we can estimate the integral asymptotically by taking larger values of $r$, and accounting for poles that are encountered. As noted above, the smallest poles are at $(\log 3 \pm \pi i)/4$ and these will account for the leading asymptotics of these coefficients. Now, the argument of $(\log 3 \pm \pi i)/4$ is $\pm 1.23438\ldots $ which is very nearly $11 \pi/28=1.23419\ldots$. This accounts for the observed phenomenon.

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  • $\begingroup$ Out of curiosity: does it answer the question that for large $n$ that $a(n+28)$ is usually $-a(n)$? $\endgroup$
    – Andrew
    Commented Sep 11, 2018 at 13:32
  • $\begingroup$ @Andrew: Yes, for large $n$ most of the time the signs will flip, but a small proportion of the time it won't. The answer roughly says that the signs look like the signs of $\cos (1.23438 n)$ (not being too careful here), which will have the similar feature of flipping very often when $n$ is increased by $28$. $\endgroup$
    – Lucia
    Commented Sep 11, 2018 at 14:29
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    $\begingroup$ I just asked WolframAlpha. It's surely no big deal to compute $\arctan (\pi/(4\log 3))$. $\endgroup$
    – Lucia
    Commented Sep 17, 2018 at 19:59

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