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%I #33 Feb 28 2022 12:14:31
%S 1,11,79,479,5297,69071,69203,471181,8960447,44831407,1031626241,
%T 5160071143,15484789693,64166447971,1989542332021,3979714828967,
%U 27861681000449,1030996803010973,1031094241305773,42278288849598913,1818093633186492859,1818204269645957299,85460151199040573933
%N Numerator of partial sums of a certain series. First member (m = 2) of a family.
%C The denominators are given in A101029.
%C The limit s = lim_{n -> infinity} s(n) with s(n) defined below equals 3*Sum_{k >= 1} zeta(2*k+1)/2^(2*k) with Euler's (or Riemann's) zeta function. This limit is 3*(2*log(2)-1) = 1.158883083...; see the Abramowitz-Stegun reference p. 259, eq. 6.3.15 with z = 1/2 together with p. 258, eqs. 6.3.5 and 6.3.3.
%C This is the first member (m = 2) of a family of rational partial sum sequences s(n,m) = (m-1)*m*(m+1)*Sum_{k = 1..n} 1/((m*k-1)*(m*k)*(m*k+1)) which have limit s(m) = lim_{n -> infinity} s(n,m) = -(gamma + Psi(1/m) + m/2 + Pi*cot(Pi*x)/2), with the Euler-Mascheroni constant gamma and the digamma function Psi. The same limit is reached by (m^2-1)*Sum_{k >= 0} zeta(2*k+1)/m^(2*k).
%C From _Peter Bala_, Feb 17 2022: (Start)
%C Let F(n) = (6*n/(2*n-1))*( 1/(1*2)*(n-1)/n - 1/(2*3)*(n-1)*(n-2)/(n*(n+1)) + 1/(3*4)*(n-1)*(n-2)*(n-3)/(n*(n+1)*(n+2)) - 1/(4*5)*(n-1)*(n-2)*(n-3)*(n-4)/(n*(n+1)*(n+2)*(n+3)) + ...). Then F(n+1) = 6*Sum_{k = 1..n} 1/((2*k-1)*(2*k)*(2*k+1)). Cf. A082687.
%C This identity allows us to extend the definition of Sum_{k = 1..n} 1/((2*k-1)*(2*k)*(2*k+1)) to non-integral values of n. (End)
%D M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, pp. 258-259.
%D Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
%H M. Abramowitz and I. A. Stegun, eds., <a href="http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP">Handbook of Mathematical Functions</a>, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
%H Wolfdieter Lang, <a href="/A101028/a101028.txt">Rationals s(n) and more.</a>
%F a(n) = numerator(s(n)) with s(n) = 6*Sum_{k = 1..n} (1/((2*k-1)*(2*k)*(2*k+1)).
%e s(3)= 6*(1/(1*2*3)+ 1/(3*4*5) + 1/(5*6*7)) = 79/70, hence a(3)=79 and A101029(3)=70.
%o (PARI) a(n) = numerator(6*sum(k=1, n, 1/((2*k-1)*(2*k)*(2*k+1)))); \\ _Michel Marcus_, Feb 28 2022
%Y Cf. A101627 (m=3), A101629 (m=4), A101631 (m=5).
%Y Cf. A082687, A101029 (denominators).
%K nonn,frac,easy
%O 1,2
%A _Wolfdieter Lang_, Dec 17 2004
%E More terms from _Michel Marcus_, Feb 28 2022
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