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Primes p such that denominator of Sum_{k=1..p-1} 1/k^3} is a cube.
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Chai Wah Wu, <a href="/A127046/b127046.txt">Table of n, a(n) for n = 1..10000</a>
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d[n_] := Module[{}, su = 0; a = {}; For[i = 1, i <= n, i++, su = su + 1/ i^3; If[PrimeQ[i + 1], If[IntegerQ[(Denominator[su])^(1/3)], AppendTo[a, i + 1]]]]; a] d[10000]
d[n_] := Module[{}, su = 0; a = {}; For[i = 1, i <= n, i++, su = su + 1/ i^3; If[PrimeQ[i + 1], If[IntegerQ[(Denominator[su])^(1/3)], AppendTo[a, i + 1]]]]; a]; d[2000]
Select[Prime[Range[200]], IntegerQ[Surd[Denominator[Sum[1/k^3, {k, #-1}]], 3]]&] (* Harvey P. Dale, Mar 13 2013 *)
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Select[Prime[Range[200]], IntegerQ[Surd[Denominator[Sum[1/k^3, {k, #-1}]], 3]]&] (* Harvey P. Dale, Mar 13 2013 *)
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