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%I #45 Feb 17 2024 10:26:31
%S 1,13,197,1105,9211,130277,82987349,331950131,16929464521,29241805241,
%T 3538258509761,6259995854281,1057939300471201,1057939300716589,
%U 51133732870640471,372975463296151087,107789908892879155343
%N Numerators of partial sums for a series for 2*Zeta(2)/3 = (Pi^2)/9.
%C Denominators are given in A130550.
%C The r(n) = 2*Sum_{j = 1..n} 1/(j^2*binomial(2*j,j) tend, for n -> infinity, to 2*Zeta(2)/3 = (Pi^2)/9, which is approximately 1.096622711.
%C A related result is zeta(2) = 3*Sum_{j = 1..n} 1/(j^2*binomial(2*j,j)) + n!^4/(2*n)!*Sum_{j >= 1} 1/( Product_{i = 0..n} (j + i)^2 ) valid for n >= 0. See Wilf, equation 5, p. 191. - _Peter Bala_, Oct 30 2023
%D L. Berggren, T. Borwein and P. Borwein, Pi: A Source Book, Springer, New York, 1997, p. 687.
%D A. van der Poorten, A proof that Euler missed..., reprinted in Pi: A Source Book, pp. 439-447, eq. 2', with a hint for the proof in footnote 4.
%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.
%H Wolfdieter Lang, <a href="/A130549/a130549.txt">Rationals and limit.</a>
%H A. J. van der Poorten, <a href="http://dx.doi.org/10.1007/BF03028234">A proof that Euler missed ... Apery's proof of the irrationality of zeta(3)</a>, Math. Intelligencer 1 (1978/1979), 195-203.
%H Herbert S. Wilf, <a href="https://doi.org/10.46298/dmtcs.265">Accelerated series for universal constants, by the WZ method</a>, Discrete Mathematics & Theoretical Computer Science, Vol 3, No 4 (1999).
%F a(n) = numerator(r(n)), n>=1, with the rationals r(n) defined above.
%F Numerator of 2*Sum_{i=1..n} 1/(i^2*C(2*i,i)). - _Wolfdieter Lang_, Oct 07 2008; edited by _Michel Marcus_, Mar 10 2016
%F a(n) = A112093(n) for n >= 2. - _Georg Fischer_, Nov 03 2018
%F From _Peter Bala_, Feb 17 2024: (Start)
%F The sequences {(2*n)! : n >= 1} and {(2*n)!*r(n) : n >= 1} satisfy the same second-order recurrence u(n) = (5*n^2 - 4*n + 1)*u(n-1) - 2*(n - 1)^3*(2*n - 3)*u(n-2) leading to the continued fraction representations r(n) = 1/(1 - 1/(13 - 48/(34 - 270/(65 - ... - 2*(2*n - 3)*(n - 1)^3/(5*n^2 - 4*n + 1 ))))) and Pi^2/9 = 1/(1 - 1/(13 - 48/(34 - 270/(65 - ... - 2*(2*n - 3)*(n - 1)^3/((5*n^2 - 4*n + 1) - ... ))))). (End)
%e Rationals r(n): [1, 13/12, 197/180, 1105/1008, 9211/8400, 130277/118800, ...].
%e r(3) = 1/(1 - 1/(13 - 48/(34))) = 197/180. - _Peter Bala_, Feb 17 2024
%p seq(numer(add(2/(k^2*binomial(2*k, k)), k = 1 .. n)), n = 1 .. 17); # _Peter Bala_, Mar 03 2015
%t Table[2*Sum[1/(i^2*Binomial[2*i, i]), {i, 1, n}], {n, 1, 20}] // Numerator
%t Accumulate[Table[1/(n^2 Binomial[2n,n]),{n,20}]]//Numerator (* _Harvey P. Dale_, Jan 27 2019 *)
%o (PARI) a(n) = numerator(2*sum(i=1, n, 1/(i^2*binomial(2*i, i)))); \\ _Michel Marcus_, Mar 10 2016
%Y Cf. A112093, A112099, A112100, A112102, A112103, A130550.
%K nonn,frac,easy
%O 1,2
%A _Wolfdieter Lang_, Jul 13 2007
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