A123489 - OEIS

A123489

a(n) = Sum_{k=0..p-1} Kronecker(4k^3+1, p) where p is the n-th prime of the form 6k+1.

1, -5, 7, 4, -11, -8, 1, -5, 7, -17, 19, 13, -2, -20, -23, 19, -14, 25, 7, -23, -11, 13, 28, 22, -17, -29, -26, -32, 16, -35, 1, -5, 37, -35, 13, -29, 34, 31, 19, -2, 28, 10, -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

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OFFSET

1,2

COMMENTS

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.

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

REFERENCES

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.

LINKS

Table of n, a(n) for n=1..74.

EXAMPLE

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.

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

PROG

(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)) )};

CROSSREFS

Cf. A002338 is the unsigned version, A002339 is the y.

Cf. A002476 (primes of the form 6k+1).

Sequence in context: A198998 A096437 A096458 * A002338 A324791 A226021

Adjacent sequences: A123486 A123487 A123488 * A123490 A123491 A123492

KEYWORD

sign

AUTHOR

Michael Somos, Sep 30 2006

STATUS

approved