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%I #11 Apr 08 2019 11:08:26
%S 0,3,13,197,1105,9211,130277,82987349,331950131,16929464521,
%T 29241805241,3538258509761,6259995854281,1057939300471201,
%U 1057939300716589,51133732870640471,372975463296151087,107789908892879155343,51058377896658637853,681986753565766904623961
%N Numerator of 3*Sum_{i=1..n} 1/(i^2*C(2*i,i)).
%H Robert Israel, <a href="/A112093/b112093.txt">Table of n, a(n) for n = 0..772</a>
%H C. Elsner, <a href="http://www.fq.math.ca/Papers1/43-1/paper43-1-5.pdf">On recurrence formulas for sums involving binomial coefficients</a>, Fib. Q., 43,1 (2005), 31-45.
%F 3*Sum_{i >= 1} 1/(i^2*C(2*i, i)) = zeta(2) = Pi^2/6.
%p 0, 3/2, 13/8, 197/120, 1105/672, 9211/5600, 130277/79200, 82987349/50450400, ... -> Pi^2/6.
%p X:= [0,seq(3/(i^2*binomial(2*i,i)),i=1..20)]:
%p S:= ListTools:-PartialSums(X):
%p map(numer,S); # _Robert Israel_, Apr 08 2019
%o (PARI) a(n) = numerator(3*sum(i=1, n, 1/(i^2*binomial(2*i, i)))); \\ _Michel Marcus_, Mar 10 2016
%Y Cf. A112094.
%K nonn,frac
%O 0,2
%A _N. J. A. Sloane_, Nov 30 2005