A195505 Numerator of Sum_{k=1..n} H(k)/k^2, where H(k) is the k-th harmonic number.

1, 11, 341, 2953, 388853, 403553, 142339079, 1163882707, 31983746689, 32452469713, 43725835522403, 44184852180503, 97954699428176291, 98731028315167091, 99421162547987123, 800313205356878959, 3953829021224881128767, 3973669953994085875967

Offset: 1

Author

Franz Vrabec, Sep 19 2011

Keywords

• nonn
• frac
• easy

Comments

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

Examples

			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
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..768
• Leonhard Euler, Meditationes circa singulare serierum genus, Novi. Comm. Acad. Sci. Petropolitanae, 20 (1775), 140-186.

Crossrefs

Cf. A001008, A002117, A036970, A195506 (denominators).

Programs

Mathematica

s = 0; Table[s = s + HarmonicNumber[n]/n^2; Numerator[s], {n, 20}] (* T. D. Noe, Sep 20 2011 *)

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

Formula

From Peter Bala, Jan 31 2019: (Start)

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)) + ....

Cf. A001008. See the Bala link in A036970. (End)