A249352 (A007559(n+1)^2-1)/9, where A007559(n) = 1*4*7*...*(3n-2).

0, 7, 2439, 2439111, 5358727111, 21949346247111, 150550565908935111, 1603062425798341063111, 25047850403099079111111111, 549850412048830984647111111111, 16380593625346723863622087111111111

Offset: 0

Author

M. F. Hasler, Oct 26 2014

Keywords

• nonn

Comments

These are the numerators of the partial sums S(n) = Sum_{k=1..n} (3k^3+3k^2+k)/A007559(k+1)^2 before simplification, i.e., a(n) = S(n)*A007559(n+1)^2. The series S(n) has sum 1/9, actually S(n) = 1/9 - 1/(9*A007559(n+1)^2).

Links

• B. Sahu in reply to S. Klein, A neat infinite sum ..., Number Theory group on LinkedIn, Oct. 2014.

Programs

PARI

a(n)=(prod(k=1,n,3*k+1)^3-1)/9