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%I #34 May 10 2020 13:42:23
%S 1,11,341,2953,388853,403553,142339079,1163882707,31983746689,
%T 32452469713,43725835522403,44184852180503,97954699428176291,
%U 98731028315167091,99421162547987123,800313205356878959,3953829021224881128767,3973669953994085875967
%N Numerator of Sum_{k=1..n} H(k)/k^2, where H(k) is the k-th harmonic number.
%C Lim_{n-> infinity} (a(n)/A195506(n)) = 2*Zeta(3) [L. Euler].
%C Sum_{k = 1..n} H(k)/k^2 is an example of a multiple harmonic (star) sum. Euler's result Sum_{k = 1..inf} H(k)/k^2 = 2*zeta(3) was the first evaluation of a multiple zeta star value. - _Peter Bala_, Jan 31 2019
%H Seiichi Manyama, <a href="/A195505/b195505.txt">Table of n, a(n) for n = 1..768</a>
%H Leonhard Euler, <a href="http://eulerarchive.maa.org/pages/E477.html">Meditationes circa singulare serierum genus</a>, Novi. Comm. Acad. Sci. Petropolitanae, 20 (1775), 140-186.
%F From _Peter Bala_, Jan 31 2019: (Start)
%F Let S(n) = Sum_{k = 1..n} H(k)/k^2. Then
%F S(n) = 1 + (1 + 1/2^3)*(n-1)/(n+1) + (1/2^3 + 1/3^3)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/3^3 + 1/4^3)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...
%F S(n)/n = 1 + (1/2^4 - 1)*(n-1)/(n+1) + (1/3^4 - 1/2^4)*(n-1)*(n-2)/((n+1)*(n+2)) + (1/4^4 - 1/3^4)*(n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ...
%F For odd n >= 3, 1/2*S((n-1)/2) = (n-1)/(n+1) + 1/2^3*(n-1)*(n-3)/((n+1)*(n+3)) + 1/3^3*(n-1)*(n-3)*(n-5)/((n+1)*(n+3)*(n+5)) + ....
%F Cf. A001008. See the Bala link in A036970. (End)
%e a(2) = 11 because 1 + (1 + 1/2)/2^2 = 11/8.
%e The first few fractions are 1, 11/8, 341/216, 2953/1728, 388853/216000, 403553/216000, 142339079/74088000, 1163882707/592704000, ... = A195505/A195506. - _Petros Hadjicostas_, May 06 2020
%t s = 0; Table[s = s + HarmonicNumber[n]/n^2; Numerator[s], {n, 20}] (* _T. D. Noe_, Sep 20 2011 *)
%o (PARI) H(n) = sum(k=1, n, 1/k);
%o a(n) = numerator(sum(k=1, n, H(k)/k^2)); \\ _Michel Marcus_, May 07 2020
%Y Cf. A001008, A002117, A036970, A195506 (denominators).
%K nonn,frac,easy
%O 1,2
%A _Franz Vrabec_, Sep 19 2011
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