For the variable-base case,
\( a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!} \)
thus
\( a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n) \)
\( a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!} \).
\( a_n (L \log(B))^n = L \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n)}{n!} \)
thus
\( a_n L^{n-1} \log(B)^n n! = B_n(\log(B), \log(B) 2! a_2, ..., \log(B) n! a_n) \)
\( a_n = \frac{B_n(\log(B), \log(B) 2! a_2, ..., \log(B) (n-1)! a_{n-1}, 0)}{(L^{n-1} \log(B)^n - \log(B)) n!} \).