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A089034
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a(n) = (prime(n)^4 - 1) / 240.
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3
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10, 61, 119, 348, 543, 1166, 2947, 3848, 7809, 11774, 14245, 20332, 32877, 50489, 57691, 83963, 105882, 118326, 162292, 197743, 261426, 368872, 433585, 468962, 546165, 588159, 679364, 1083936, 1227083, 1467814, 1555421, 2053685, 2166190
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OFFSET
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4,1
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COMMENTS
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Mod 2, odd primes p are 1 and mod 4 or mod 6, p=-+1, so that p^2==p^4==1 (mod 2*4*6). Moreover, mod 5, p==-+1, -+2 for p>5, implying p^2==-+1 or p^4==1, so that finally p^4==1 (mod 2*4*6*5), i.e., 240 divides (p^4 - 1) for p>5.
From Simon Plouffe's web site we know that sum_{n >= 1} n^3/(exp(2*n*Pi / 7) - 1) = 10.0000000000000001901617..., very close to a(1). Extensive calculations suggest that more generally, for any prime p >= 7, Sum_{n >=1} n^3/(exp(2*n*Pi / p) - 1) is similarly very close to (p^4-1)/240.
Victor Miller replied on Jan 29 2012 via email, with an explanation of this observation. The following is an abridged version of his reply:
Let q = exp(2*Pi*i*z). Define the Eisenstein series E_4(z) = 1 + 240*sum_{n >= 1} n^3*q^n/(1-q^n). For your observation we take z = i/p, so that q = exp(-2*Pi / p). So what you've evaluated numerically is (E_4(i/p) - 1)/240.
The Eisenstein series obeys the transformation law E_4(-1/z) = z^4*E_4(z), or E_4(i/p) = p^4*E_4(i*p). Your observation reduces to showing that E_4(i*p) is very close to 1. In this case q = exp(-2*Pi*p), so E_4(i*p) - 1 is bounded by a geometric series in q. In your first case, when p = 7, q is around exp(-44), which is already quite small. (End)
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LINKS
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MATHEMATICA
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Select[(Prime[Range[50]]^4-1)/240, IntegerQ] (* Harvey P. Dale, Nov 28 2015 *)
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PROG
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CROSSREFS
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KEYWORD
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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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