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

1,3

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.

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.

Example

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.

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}]

Crossrefs

Cf. A000040, A000290.

Sequence in context: A304302 A305516 A305182 * A033178 A029418 A185736

Adjacent sequences: A291044 A291045 A291046 * A291048 A291049 A291050

Keyword

nonn

Author

Zhi-Wei Sun, Aug 16 2017

Status

approved