|
%I #19 Jan 01 2018 04:54:58
%S 513,20708,584874,4714408,72449100,200562418,1012788198,1953009460,
%T 6172747128,24788658690,37242612640,107770200778,198936710910,
%U 265200653548,449592659568,931777815258,1775665528380,2155635964450,3812897562148,5368106367720,6351988507678
%N a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^10*H(4*p-k^2) where H() is the Hurwitz class number and p is n-th prime.
%H Seiichi Manyama, <a href="/A297494/b297494.txt">Table of n, a(n) for n = 1..1000</a>
%H N. Lygeros, O. Rozier, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Lygeros/lygeros5.html">A new solution to the equation tau(p) == 0 (mod p)</a>, J. Int. Seq. 13 (2010) # 10.7.4.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TauFunction.html">Tau Function.</a>
%F Let b(n) = 42*n^6 - 90*n^4 - 75*n^3 - 35*n^2 - 9*n - 1.
%F a(n) = b(prime(n)) - tau(prime(n)) where tau(n)=A000594(n) is Ramanujan's tau function.
%F So tau(prime(n)) + 1 == -a(n) (mod prime(n)).
%Y (1/2) * Sum_{|k|<=2*sqrt(p)} k^m*H(4*p-k^2): A000040 (m=0), A084920 (m=2), A297491 (m=4), A297492 (m=6), A297493 (m=8), this sequence (m=10).
%Y Cf. A000594, A259825, A295645, A297127.
%K nonn
%O 1,1
%A _Seiichi Manyama_, Dec 31 2017
|
