A297494 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.

513, 20708, 584874, 4714408, 72449100, 200562418, 1012788198, 1953009460, 6172747128, 24788658690, 37242612640, 107770200778, 198936710910, 265200653548, 449592659568, 931777815258, 1775665528380, 2155635964450, 3812897562148, 5368106367720, 6351988507678

Offset: 1

Author

Seiichi Manyama, Dec 31 2017

Keywords

• nonn

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1000
• N. Lygeros, O. Rozier, A new solution to the equation tau(p) == 0 (mod p), J. Int. Seq. 13 (2010) # 10.7.4.
• Eric Weisstein's World of Mathematics, Tau Function.

Crossrefs

(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).

Cf. A000594, A259825, A295645, A297127.

Formula

Let b(n) = 42*n^6 - 90*n^4 - 75*n^3 - 35*n^2 - 9*n - 1.

a(n) = b(prime(n)) - tau(prime(n)) where tau(n)=A000594(n) is Ramanujan's tau function.

So tau(prime(n)) + 1 == -a(n) (mod prime(n)).