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%I #11 Oct 26 2014 17:29:54
%S 0,7,2439,2439111,5358727111,21949346247111,150550565908935111,
%T 1603062425798341063111,25047850403099079111111111,
%U 549850412048830984647111111111,16380593625346723863622087111111111
%N (A007559(n+1)^2-1)/9, where A007559(n) = 1*4*7*...*(3n-2).
%C 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).
%H B. Sahu in reply to S. Klein, <a href="http://lnkd.in/bKzkYcS">A neat infinite sum ...</a>, Number Theory group on LinkedIn, Oct. 2014.
%o (PARI) a(n)=(prod(k=1,n,3*k+1)^3-1)/9
%K nonn
%O 0,2
%A _M. F. Hasler_, Oct 26 2014