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A195505
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Numerator of Sum_{k=1..n} H(k)/k^2, where H(k) is the k-th harmonic number.
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3
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1, 11, 341, 2953, 388853, 403553, 142339079, 1163882707, 31983746689, 32452469713, 43725835522403, 44184852180503, 97954699428176291, 98731028315167091, 99421162547987123, 800313205356878959, 3953829021224881128767, 3973669953994085875967
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OFFSET
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1,2
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COMMENTS
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Lim_{n-> infinity} (a(n)/A195506(n)) = 2*Zeta(3) [L. Euler].
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
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LINKS
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FORMULA
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Let S(n) = Sum_{k = 1..n} H(k)/k^2. Then
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)) + ...
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)) + ...
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)) + ....
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EXAMPLE
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a(2) = 11 because 1 + (1 + 1/2)/2^2 = 11/8.
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
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MATHEMATICA
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s = 0; Table[s = s + HarmonicNumber[n]/n^2; Numerator[s], {n, 20}] (* T. D. Noe, Sep 20 2011 *)
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PROG
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(PARI) H(n) = sum(k=1, n, 1/k);
a(n) = numerator(sum(k=1, n, H(k)/k^2)); \\ Michel Marcus, May 07 2020
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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STATUS
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approved
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