A130681 Sum[ 1/k^(2p-1), {k,1,p-1}] divided by p^3, for prime p>3.

41361119, 126941659254799099843, 201945187495172518712395211386399925751676163316330287629003467281801, 534565103485593943310791656810688803242468895931876288948761507813750601446840308490623197040810555162527973

Offset: 3

Author

Alexander Adamchuk, Jun 29 2007

Keywords

• frac
• nonn

Comments

The generalized harmonic number is H(n,m) = Sum[ 1/k^m, {k,1,n} ]. The numerator of H(p-1,2p-1) is divisible by p^3 for prime p>3. Also the numerator of H(p-1,p) is divisible by p^3 for prime p>3. See A119722(n).

Examples

			Prime[3] = 5.
a(3) = numerator[ 1 + 1/2^9 + 1/3^9 + 1/4^9 ] / 5^3 = 5170139875/125 = 41361119.
		

Links

• Alexander Adamchuk, Table of n, a(n) for n = 3..10
• Eric Weisstein's World of Mathematics, Wolstenholme's Theorem
• Eric Weisstein's World of Mathematics, Harmonic Number

Crossrefs

Cf. A119722.

Programs

Mathematica

Table[ Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k,1,Prime[n]-1} ] ] / Prime[n]^3, {n,3,10} ]

PARI

a(n)=p=prime(n);numerator(sum(i=1,p-1,1/i^(2*p-1)))/p^3 \\ Ralf Stephan, Nov 10 2013

Formula

a(n) = Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k,1,Prime[n]-1} ] ] / Prime[n]^3 for n>2.

a(n) = A228426(A000040(n))/A000040(n)^3.

Extensions

Edited by Ralf Stephan, Nov 10 2013