Thanks very much for your kind words, and I’m glad that you found the information interesting. I read your page and am happy to hear that you are teaching your son mathematics and experimental mathematics. The confidence that one can discover things on one’s own is a fantastic gift to give to a child.

Peter

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Thanks for your clear analysis and explanations.

]]>Thanks for the clarification — I had looked at the wikipedia page for the polylogarithm… they listed dozens of properties of it, and I looked through them but apparently missed that one somehow. This formulation probably provides an easier avenue for a combinatorial understanding of the series. The coefficients are still somewhat difficult to compute (by hand anyway) but certainly represent a drastic improvement over iteration (something like k^2 computations versus k!).

]]>As I stated previously, you are studying one of the special functions, the polylogarithm, about which much is known,

Note that you are studying the case for and a negative integer.

For

where

are the so-called Eulerian numbers. Note that this is a finite sum! I believe this is the formula you are looking for. I suppose one could derive it from

with a little work.

Take care, and keep up the good blogging,

…

]]>As for the OEIS entry, I don’t really get what they mean by “sets of labeled beads” — it seems somewhat ambiguous. For n=1, the sum is 2, so there are two necklaces that can be made from a single bead. I guess the empty necklace counts? But then for n=2, I seem to get more than 6 if the empty necklace counts. Maybe I’m not understanding what they mean.

As for a more heuristic argument, I’m not sure. One can certainly argue from a more elementary point of view. For instance, the summand hits its maximum at . So, for , one can replace a single in the numerator of the summand with so that . It’s not hard to see the same inequality holds for larger values of n by a more direct comparison. Then by iterating this estimate, one gets the correct upper bound. The opposite direction intuitively then comes from the fact that the only terms in the sum which contribute are ones near the maximum and that the summand is approximately symmetric around that maximum. Then one should be able to get a similar lower bound, possibly losing a constant factor (or a factor which tends to 1 as k gets large).

]]>Have fun! ]]>

are essentially the Euler polynomials (which means yes, there is probably a formula in terms of Bernoulli numbers); then you just need to plug in x = 1/2.

Anyway, two questions:

– In the words of the OEIS entry, what do necklaces of sets of labeled beads have to do with the cumulants of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses?

– Can you think of a heuristic argument that gives that estimate?

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