%I
%S 10,61,119,348,543,1166,2947,3848,7809,11774,14245,20332,32877,50489,
%T 57691,83963,105882,118326,162292,197743,261426,368872,433585,468962,
%U 546165,588159,679364,1083936,1227083,1467814,1555421,2053685,2166190
%N a(n) = (prime(n)^4  1) / 240.
%C 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.
%C From _JeanClaude Babois_, Jan 13 2012: (Start)
%C 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^41)/240.
%C Victor Miller replied on Jan 29 2012 via email, with an explanation of this observation. The following is an abridged version of his reply:
%C Let q = exp(2*Pi*i*z). Define the Eisenstein series E_4(z) = 1 + 240*sum_{n >= 1} n^3*q^n/(1q^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.
%C 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)
%H Charles R Greathouse IV, <a href="/A089034/b089034.txt">Table of n, a(n) for n = 4..10000</a>
%t Select[(Prime[Range[50]]^41)/240,IntegerQ] (* _Harvey P. Dale_, Nov 28 2015 *)
%o (PARI) a(n)=(prime(n)^4  1)/240 \\ _Charles R Greathouse IV_, May 31 2013
%Y Cf. A024702.
%K nonn
%O 4,1
%A _Lekraj Beedassy_, Nov 12 2003
%E More terms from _Ray Chandler_, Nov 12 2003
