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%I #28 Sep 08 2022 08:46:21
%S 1,1,2,2,1,2,2,5,6,2,6,3,6,5,14,8,6,5,6,15,10,6,14,24,14,6,12,12,6,16,
%T 19,18,18,36,18,10,16,20,20,12,28,18,8,24,38,27,40,20,17,30,52,18,22,
%U 26,29,21,42,31,26,26,18,44,38,40,46,26,30,44,38,36,52,28,27,38,103,22,38,78
%N a(n) is the number of partitions of (prime(n)^2 + 2)/3 into 3 squares.
%C If p >= 5 is a prime number it can be written as p = 6m-1 or p = 6m+1. The identities ((6m-1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m-1)^2 and ((6m+1)^2 + 2)/3 = (2m)^2 + (2m)^2 + (2m+1)^2 show that the number (p^2 + 2)/3 can be written as a sum of 3 squares of integers in at least one way.
%D Ion Cucurezeanu, Pătrate și cuburi perfecte de numere întregi (Squares and perfect cubes of integer numbers), Ed. Gil., Zalău, 2007, ch. 1, p. 21, pr. 166.
%D Laurențiu Panaitopol, Dinu Șerbănescu, Number theory and combinatorial problems for juniors, Ed. Gil, Zalău, (2003), ch. 1, p. 5, pr. 4. (in Romanian).
%e For n = 3, p = prime(3) = 5, (5^2+2)/3 = 9 = 2^2 + 2^2 + 1^2, so a(3) = 1.
%e For n = 9, p = prime(9) = 23, (23^2+2)/3 = 177 = 13^2 + 2^2 + 2^2 = 8^2 + 8^2 + 7^2, so a(9) = 2.
%e For n = 17, p = prime(17) = 59, (59^2+2)/3 = 1161 = 34^2 + 2^2 + 1^2 = 33^2 + 6^2 + 6^2 = 24^2 + 11^2 + 4^2 = 31^2 + 14^2 + 2^2 = 31^2 + 10^2 + 10^2 = 30^2 + 15^2 + 6^2 = 29^2 + 16^2 + 8^2 = 28^2 + 19^2 + 4^2 = 28^2 + 16^2 + 11^2 = 26^2 + 22^2 + 1^2 = 26^2 + 17^2 + 14^2 = 24^2 + 24^2 + 3^2 = 24^2 + 21^2 + 12^2 = 20^2 + 20^2 + 19^2, so a(17) = 14.
%o (Magma) [#RestrictedPartitions(Floor((p*p+2)/3),3,{d*d:d in [1..p]}): p in PrimesInInterval(5,500) ];
%o (PARI) a(n)={k=(prime(n+2)^2+2)/3;sum(a=1, k, sum(b=1, a, sum(c=1, b, a^2+b^2+c^2==k)));} \\ _Jinyuan Wang_, Mar 30 2019
%Y Cf. A000040, A000378, A024795, A025427.
%K nonn
%O 3,3
%A _Marius A. Burtea_, Mar 29 2019
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