

A291635


Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that p = x + 2*y + 5*z, p  2, p + 4 and p + 10 are all prime.


2



0, 1, 1, 0, 1, 3, 1, 0, 1, 2, 2, 0, 3, 7, 3, 0, 4, 4, 1, 0, 4, 7, 3, 0, 3, 5, 2, 0, 4, 6, 2, 0, 2, 3, 3, 0, 4, 8, 3, 0, 5, 8, 2, 0, 2, 5, 2, 0, 5, 8, 4, 0, 4, 5, 2, 0, 5, 6, 4, 0, 1, 8, 5, 0, 3, 9, 3, 0, 6, 8, 3, 0, 5, 13, 5, 0, 9, 9, 2, 0, 4, 6, 6, 0, 7, 11, 4, 0, 8, 10, 5, 0, 2, 11, 5, 0, 3, 10, 4, 0
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OFFSET

1,6


COMMENTS

Conjecture: a(n) > 0 for all n > 1 not divisible by 4.
This is stronger than the conjecture in A291624. Obviously, it implies that there are infinitely many prime quadruples (p2, p, p+4, p+10).
We have verified that a(n) > 0 for any integer 1 < n < 10^7 not divisible by 4.


LINKS



EXAMPLE

a(61) = 1 since 61 = 4^2 + 0^2 + 3^2 + 6^2 with 4 + 2*0 + 5*3 = 19, 19  2 = 17, 19 + 4 = 23 and 19 + 10 = 29 all prime.
a(253) = 1 since 253 = 12^2 + 8^2 + 3^2 + 6^2 with 12 + 2*8 + 5*3 = 43, 43  2 = 41, 43 + 4 = 47 and 43 + 10 = 53 all prime.
a(725) = 1 since 725 = 7^2 + 0^2 + 0^2 + 26^2 with 7 + 2*0 + 5*0 = 7, 7  2 = 5, 7 + 4 = 11 and 7 + 10 = 17 all prime.
a(1511) = 1 since 1511 = 18^2 + 15^2 + 11^2 + 29^2 with 18 + 2*15 + 5*11 = 103, 103  2 = 101, 103 + 4 = 107 and 103 + 10 = 113 all prime.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[p_]:=PQ[p]=PrimeQ[p]&&PrimeQ[p2]&&PrimeQ[p+4]&&PrimeQ[p+10]
Do[r=0; Do[If[SQ[nx^2y^2z^2]&&PQ[x+2y+5z], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[nx^2]}, {z, 0, Sqrt[nx^2y^2]}]; Print[n, " ", r], {n, 1, 100}]


CROSSREFS

Cf. A000040, A000118, A000290, A022004, A172454, A271518, A281976, A290935, A291150, A291191, A291455, A291624.


KEYWORD



AUTHOR



STATUS

approved



