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Consider the function: $$S(n)=\sum_{j=1}^{\infty}\frac{j^n}{2^j}=\frac1 2+\frac{2^n}{4}+\frac{3^n}{8}+\frac{4^n}{16}+\frac{5^n}{32}+...$$

Euler found the sum of the first few of these as:

$S(0)=1$; (as per the usual geometric series.)

$S(1)=2$; etc. This creates a series of sums as follows:

1, 2, 6, 26, 150, 1082, 9366, 94586, … , which is the OEIS sequence A000629, “Number of necklaces of partitions of $n+1$ labelled beads”.

My question is - what connection is there between this infinite series and necklace combinatorics? Or is the connection accidental or illusory?

Hector Blandin
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Dottard
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    Both are given by $\operatorname{Li}{-n}(1/2)$. One way to view the connection is that we have $\operatorname{Li}{-n}(1/2) = 2^{n+1} \sum_{k=0}^{n-1} \left \langle !{n\atop k}!\right \rangle2^{k-n}$ and the Eulerian Numbers come up when studying increasing sequences in a partition (which is obviously related the necklace problem) – Brevan Ellefsen Oct 18 '19 at 00:25
  • By "Li", do you mean logarithmic integral? If not, what function? – Dottard Oct 18 '19 at 01:48
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    I am referring to the nth polylogarithm. – Brevan Ellefsen Oct 18 '19 at 01:54

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