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A249352
(A007559(n+1)^2-1)/9, where A007559(n) = 1*4*7*...*(3n-2).
1
0, 7, 2439, 2439111, 5358727111, 21949346247111, 150550565908935111, 1603062425798341063111, 25047850403099079111111111, 549850412048830984647111111111, 16380593625346723863622087111111111
OFFSET
0,2
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.
PROG
(PARI) a(n)=(prod(k=1, n, 3*k+1)^3-1)/9
CROSSREFS
Sequence in context: A060759 A067504 A203382 * A264743 A210405 A203705
KEYWORD
nonn
AUTHOR
M. F. Hasler, Oct 26 2014
STATUS
approved