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Reading a paper regarding Bernoulli numbers, and I stumbled onto a definition. First let $$\frac{x}{e^x-1}=\sum_{k=0}^{\infty}B_k\frac{x^k}{k!}$$ The author then goes on to define new terms.
Let $$\beta_n=(-1)^n\frac{B_n}{n}, n\ge 1$$ noting that $\beta_n=B_n/n$ except for $\beta_1=-B_1=1/2$ since $B_n=0$ when $n$ is odd and greater than 1. Now he defines $$\beta_n^{(j)}=\frac1{j!}\sum_{i_1+i_2+...+i_j=n}\binom{n}{i_1,i_2,...,i_j}\beta_{i_1}...\beta_{i_1}$$ I'm not even sure where to start with this, but, I suppose I'm confused about the role of the bound $i_1+i_2+...+i_j=n.$ I would have to say that the sum of the $i_k$'s are integer partitions of a number $n$. Should I take this as unique integers? Or is this simply all possible integer partitions of $n$? So for $n=4$, so i take $$\{4, 3+1, 2+2, 2+1+1, 1+1+1+1\}$$ or $$\{4, 3+1, 1+3, 2+2, 2+1+1, 1+2+1, 1+1+2, 1+1+1+1\}$$ And what of the $j$? So this tells me, for example if $j=2$ that my summation is $$\beta_4^{(2)}=\frac{1}{2!}\left[\binom{4}{2,2}\frac{B_2}{2}\frac{B_2}{2}+\binom{4}{3,1}\frac{B_3}{3}\frac{B_1}{1}\right]$$ Or do i have to include the case of $\binom{4}{1,3}$. Or am i completely off to begin with?

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    $\begingroup$ It seems Gessel does not mention that $\beta_n$ are the cumulants of the uniform distribution on the interval $[-1,0]$. ${}\qquad{}$ $\endgroup$ – Michael Hardy Oct 13 '14 at 17:08
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Looking at formula (4) at the top of page 3 of the paper makes it clear that Gessel uses the usual convention, in which summation over $i_1+\cdots+i_j = n$ is a summation over all non-negative integers $i_1,\ldots,i_j$ which sum to $n$, rather than all partitions. In the case of (4) we also have a lower bound on $i_1,\ldots,i_j$. If he wanted to sum over all partitions he would have added the condition $i_1\geq\cdots\geq i_j$, so in (4), instead of $i,j,k \geq 2$ he would have had $i\geq j\geq k\geq 2$.

Reading further down the paper, the definition you quote, formula (5), appears on the top of page (4), followed by the description "where the sum is over positive integers $i_1,\ldots,i_j$". Again, it is not mentioned that $i_1,\ldots,i_j$ are ordered.

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  • $\begingroup$ Okay, so in the example I gave, $\beta_n^{(j)}$, since $j=2$, then the integer sums would just be the set of ordered pairs that sum to $4$, i.e., $\{(4,0),(3,1),(2,2),(3,1),(0,4)\}$? $\endgroup$ – Eleven-Eleven Oct 17 '14 at 9:17
  • $\begingroup$ Actually, the accompanying text restricts all these integers to be strictly positive, so you should remove $(4,0)$ and $(0,4)$. $\endgroup$ – Yuval Filmus Oct 17 '14 at 14:12
  • $\begingroup$ Thank you! +50 for the help...when it allows me... $\endgroup$ – Eleven-Eleven Oct 17 '14 at 17:00

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