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?

  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.