A291047 Number of ways to write 4*n+1 as p^2 + q^2 + 8*r^2, where p is prime, and q and r are nonnegative integers.

1, 1, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 5, 1, 1, 3, 2, 3, 6, 1, 3, 4, 2, 1, 1, 2, 4, 4, 5, 1, 5, 3, 3, 3, 1, 4, 9, 2, 1, 4, 3, 3, 6, 4, 2, 5, 3, 4, 5, 2, 6, 3, 3, 4, 6, 1, 4, 5, 4, 2, 8, 2, 2, 6, 1, 3, 5, 2, 3, 3, 5, 6, 8, 3, 1, 9, 4, 4

Offset: 1

Author

Zhi-Wei Sun, Aug 16 2017

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and the only values of n >= 1000 with a(n) = 1 are 1000, 1052, 1472, 1675, 1967, 4787, 4822, 11962.

(ii) For any positive integer n, we can write 6*n+3 as p^2 + 2*q^2 + 3*r^2, where p is prime, and q and r are integers.

Examples

			a(1) = 1 since 4*1+1 = 2^2 + 1^2 + 8*0^2 with 2 prime.
a(2) = 1 since 4*2+1 = 3^2 + 0^2 + 8*0^2 with 3 prime.
a(4) = 1 since 4*4+1 = 3^2 + 0^2 + 8*1^2 with 3 prime.
a(16) = 1 since 4*16+1 = 7^2 + 4^2 + 8*0^2 with 7 prime.
a(17) = 1 since 4*17+1 = 5^2 + 6^2 + 8*1^2 with 5 prime.
a(4787) = 1 since 4*4787+1 = 31^2 + 126^2 + 8*17^2 with 31 prime.
a(4822) = 1 since 4*4822+1 = 29^2 + 4^2 + 8*48^2 with 29 prime.
a(11962) = 1 since 4*11962+1 = 109^2 + 160^2 + 8*36^2 with 109 prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Crossrefs

Cf. A000040, A000290.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[PrimeQ[p]&&SQ[4n+1-p^2-8q^2],r=r+1],{p,2,Sqrt[4n+1]},{q,0,Sqrt[(4n+1-p^2)/8]}];
tab=Append[tab,r];Continue,{n,1,80}]