A291635 - OEIS

%I #19 Aug 29 2017 14:19:41

%S 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,

%T 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,

%U 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

%N 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.

%C Conjecture: a(n) > 0 for all n > 1 not divisible by 4.

%C This is stronger than the conjecture in A291624. Obviously, it implies that there are infinitely many prime quadruples (p-2, p, p+4, p+10).

%C We have verified that a(n) > 0 for any integer 1 < n < 10^7 not divisible by 4.

%H Zhi-Wei Sun, <a href="/A291635/b291635.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.

%e 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.

%e 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.

%e 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.

%e 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.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t PQ[p_]:=PQ[p]=PrimeQ[p]&&PrimeQ[p-2]&&PrimeQ[p+4]&&PrimeQ[p+10]

%t Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&PQ[x+2y+5z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r],{n,1,100}]

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

%K nonn,look

%O 1,6

%A _Zhi-Wei Sun_, Aug 28 2017