Recently I asked this question as I was trying to see how to write a particular Identity. I asked about how to write the following sum:
$$\sum_{i_1+i_2+...+i_n=k}\binom{k}{i_1,i_2,...,i_n}\frac{f(i_1)f(i_2)...f(i_n)}{k!}$$
It is my "f" that is the topic of this question. My particular "f" are called Hypergeometric Bernoulli Numbers. The code I have to generate the numbers is below:
g[m_, x] := x^m/(m! (E^x - T[m - 1, x]))
where
T[m_, x_] := Sum[x^j/j!, {j, 0, m}]
The hypergeometric bernoulli numbers are extracted using the following
b[m, n, M] = b[m_, n_, M_] := Coefficient[ n! Normal[Series[g[m, x], {x, 0, M}]], x, n]
Now I want to sum over the $n$ term defined in the bernoulli number, so I basically want the "f" I wrote in the formula to be replaced by "b[m,n,M]". I had two helpers submit sequences on the linked page but the way I am implementing them seems incorrect. I was hoping to have someone help me integrate the "b" into the expression for the "f". Thank you.
b[m,n,M]
which is supposed to replace $f$ takes three arguments. Can you clarify the relationship betweenb
and $f$? $\endgroup$