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A297492 a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^6*H(4*p-k^2) where H() is the Hurwitz class number and p is the n-th prime.

%I A297492 #15 Jan 01 2018 04:55:31
%S A297492 33,308,2874,11528,72060,141218,414918,648260,1394328,3528690,4608800,
%T A297492 9358298,14113470,17077148,24378288,39426858,60555180,69195410,
%U A297492 100714868,127012680,141942878,194693840,237229188,313639470,442561238,520209690,562658408,655294428
%N A297492 a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^6*H(4*p-k^2) where H() is the Hurwitz class number and p is the n-th prime.
%H A297492 Seiichi Manyama, <a href="/A297492/b297492.txt">Table of n, a(n) for n = 1..1000</a>
%H A297492 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.
%F A297492 Let b(n) = 5*n^4 - 9*n^2 - 5*n - 1.
%F A297492 a(n) = b(prime(n)).
%o A297492 (PARI) b(n) = 5*n^4 - 9*n^2 - 5*n - 1;
%o A297492 a(n) = b(prime(n)); \\ _Michel Marcus_, Jan 01 2018
%Y A297492 (1/2) * Sum_{|k|<=2*sqrt(p)} k^m*H(4*p-k^2): A000040 (m=0), A084920 (m=2), A297491 (m=4), this sequence (m=6), A297493 (m=8), A297494 (m=10).
%Y A297492 Cf. A259825.
%K A297492 nonn
%O A297492 1,1
%A A297492 _Seiichi Manyama_, Dec 31 2017