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A130681
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Sum[ 1/k^(2p-1), {k,1,p-1}] divided by p^3, for prime p>3.
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2
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41361119, 126941659254799099843, 201945187495172518712395211386399925751676163316330287629003467281801, 534565103485593943310791656810688803242468895931876288948761507813750601446840308490623197040810555162527973
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OFFSET
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3,1
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COMMENTS
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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).
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LINKS
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FORMULA
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a(n) = Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k,1,Prime[n]-1} ] ] / Prime[n]^3 for n>2.
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EXAMPLE
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Prime[3] = 5.
a(3) = numerator[ 1 + 1/2^9 + 1/3^9 + 1/4^9 ] / 5^3 = 5170139875/125 = 41361119.
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MATHEMATICA
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Table[ Numerator[ Sum[ 1/k^(2*Prime[n]-1), {k, 1, Prime[n]-1} ] ] / Prime[n]^3, {n, 3, 10} ]
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PROG
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(PARI) a(n)=p=prime(n); numerator(sum(i=1, p-1, 1/i^(2*p-1)))/p^3 \\ Ralf Stephan, Nov 10 2013
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CROSSREFS
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KEYWORD
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frac,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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