For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(n).

A131657

For n >= 1, put A_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j and B_n(z) = Sum_{j>=0} (n*j)!/(j!^n) * z^j * (Sum_{k=1..j*n} (1/k)), and let b(n) be the largest integer for which exp(B_n(z)/(b(n)*A_n(z))) has integral coefficients. The sequence is b(n).