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%I #21 Jul 11 2022 17:12:19
%S 1,-5,7,4,-11,-8,1,-5,7,-17,19,13,-2,-20,-23,19,-14,25,7,-23,-11,13,
%T 28,22,-17,-29,-26,-32,16,-35,1,-5,37,-35,13,-29,34,31,19,-2,28,10,
%U -23,25,-32,43,-29,1,31,-11,-26,49,-47,-17,43,40,49,37,-8,-53,-44,-50,16,-41,-29,49,31,-56,-5,7,-35,13,-59,-47
%N a(n) = Sum_{k=0..p-1} Kronecker(4k^3+1, p) where p is the n-th prime of the form 6k+1.
%C Given a prime p == 1 (mod 6), the sum x is the unique solution to 4*p = x^2 + 27*y^2 where x == 1 (mod 3) and y is a positive integer.
%C Given a prime p == 1 (mod 6), the number of solutions to u^3 + v^3 == 1 (mod p) is p - 2 + x. - _Michael Somos_, Jul 11 2022
%D L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 3, p. 55.
%e If p = 37, then 4*37 = (-11)^2 + 27*(1)^2 where -11 = Sum_{k=0..36} Kronecker(4k^3+1, 37) and 37 is the 5th prime of the form 6k+1 so a(5) = -11.
%e If p = 13, then 4*p = x^2 + 26*y^2 where x = -5, y = 1. The solutions to u^3 + v^3 == 1 (mod p) is {(0,1), (1,0), (0,3), (3,0), (0,9), (9,0)} with cardinality = 6 = 13 - 2 + (-5). - _Michael Somos_, Jul 11 2022
%o (PARI) {a(n) = my(p, c); if(n<1, 0, c=0; p=0; while(c<n, p++; if(isprime(p)& p%6==1, c++)); sum(x=0, p-1, kronecker(4*x^3+1, p)) )};
%Y Cf. A002338 is the unsigned version, A002339 is the y.
%Y Cf. A002476 (primes of the form 6k+1).
%K sign
%O 1,2
%A _Michael Somos_, Sep 30 2006
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