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%I #13 Jan 01 2018 04:55:01
%S 9,44,234,664,2628,4354,9774,13660,24264,48690,59488,101194,137718,
%T 158884,207504,297594,410580,453778,601324,715608,777814,985840,
%U 1143324,1409670,1825054,2060298,2185144,2449764,2589730,2885454,4096384,4495788,5142294
%N a(n) = (1/2) * Sum_{|k|<=2*sqrt(p)} k^4*H(4*p-k^2) where H() is the Hurwitz class number and p is n-th prime.
%H Seiichi Manyama, <a href="/A297491/b297491.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.
%F Let b(n) = 2*n^3 - 3*n - 1.
%F a(n) = b(prime(n)).
%Y (1/2) * Sum_{|k|<=2*sqrt(p)} k^m*H(4*p-k^2): A000040 (m=0), A084920 (m=2), this sequence (m=4), A297492 (m=6), A297493 (m=8), A297494 (m=10).
%Y Cf. A259825.
%K nonn
%O 1,1
%A _Seiichi Manyama_, Dec 31 2017