closed form for regular superfunction expressed as a periodic function

[mike3]

Doing some tests, it appears that

\( B_n(1, 2! a_2, ..., n! a_n) \)

has only one occurrence of \( n! a_n \), and no higher powers of it, and it never seems to be multiplied by any sort of n-dependent coefficient. This means that \( B_n(1, a_2, ..., a_n) - n! a_n = B_n(1, 2! a_2, ..., (n-1)! a_{n-1}, 0) \). I don't have a proof at this point, ...

Now I've got that proof. Here it goes:

We have

\( B_{n,k}(x_1, x_2, ..., x_{n-k+1}) = \sum \frac{n!}{j_1! j_2! ... j_{n-k+1}!} \left(\frac{x_1}{1!}\right)^{j_1} \left(\frac{x_2}{2!}\right)^{j_2} ... \left(\frac{x_{n-k+1}}{(n-k+1)!}\right)^{j_{n-k+1}} \)

(note how the connection to the Faà di Bruno's formula is clear)

with the sum taken over all sequences of non-negative integers \( j \) such that \( j_1 + j_2 + ... + j_{n-k+1} = k \) and \( j_1 + 2j_2 + ... + (n-k+1)j_{n-k+1} = n \), and \( B_n \) is the sum of the \( B_{n, k} \). As should be obvious from the formula, we see that the value \( x_n \) in \( B_n \) only occurs in the component \( B_{n, 1} \). What kind of occurrences of \( x_n \) are possible there? Using the formulas for the \( j \), we see that \( j_1 + j_2 + ... + j_n = 1 \), which means that all but one \( j_i \) must be zero, and that one that is must be 1. The second constraint, \( j_1 + 2j_2 + ... + nj_n = n \), would imply that if all \( j_i \) but one are zero, the nonzero one must equal \( \frac{n}{m} \) where \( m \) is its position in the sequence. This would mean \( m \) must divide \( n \). The first constraint, though, said it must be 1, which means \( m = n \) and so the only possible sequence of \( j \) is \( 0, 0, ..., 1 \) (\( n \) terms). This means \( B_{n, 1}(x_1, x_2, ..., x_n) = \frac{n!}{0!0!...1!} \left(\frac{x_1}{1!}\right)^0 \left(\frac{x_2}{2!}\right)^0 ... \left(\frac{x_n}{n!}\right)^1 \), which is obviously just \( x_n \). So \( B_n(x_1, x_2, ..., x_n) \) contains only one term with \( x_n \), which is just \( x_n \) itself. And then \( B_n(x_1, x_2, ..., x_n) - x_n = B_n(x_1, x_2, ..., x_{n-1}, 0) \) follows trivially.