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%I #12 Feb 12 2019 08:45:32
%S 41361119,126941659254799099843,
%T 201945187495172518712395211386399925751676163316330287629003467281801,
%U 534565103485593943310791656810688803242468895931876288948761507813750601446840308490623197040810555162527973
%N Sum[ 1/k^(2p-1), {k,1,p-1}] divided by p^3, for prime p>3.
%C 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).
%H Alexander Adamchuk, <a href="/A130681/b130681.txt">Table of n, a(n) for n = 3..10</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/WolstenholmesTheorem.html">Wolstenholme's Theorem</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HarmonicNumber.html">Harmonic Number</a>
%F a(n) = Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k,1,Prime[n]-1} ] ] / Prime[n]^3 for n>2.
%F a(n) = A228426(A000040(n))/A000040(n)^3.
%e Prime[3] = 5.
%e a(3) = numerator[ 1 + 1/2^9 + 1/3^9 + 1/4^9 ] / 5^3 = 5170139875/125 = 41361119.
%t Table[ Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k,1,Prime[n]-1} ] ] / Prime[n]^3, {n,3,10} ]
%o (PARI) a(n)=p=prime(n);numerator(sum(i=1,p-1,1/i^(2*p-1)))/p^3 \\ _Ralf Stephan_, Nov 10 2013
%Y Cf. A119722.
%K frac,nonn
%O 3,1
%A _Alexander Adamchuk_, Jun 29 2007
%E Edited by _Ralf Stephan_, Nov 10 2013
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