%I #8 Mar 19 2018 04:20:49

%S 1,2,2,2,2,3,1,2,3,4,3,4,2,3,2,2,4,5,3,4,4,2,1,3,2,6,5,4,3,4,2,2,6,5,

%T 4,6,3,3,3,4,6,8,2,5,5,3,2,4,4,5,6,5,5,4,3,3,6,5,2,6,3,4,3,2,6,10,3,5,

%U 8,1,2,5,5,6,5,6,5,3,1,4

%N Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0 <= z <= w such that x^2 + 7*y^2 = 2^k*m for some k = 0,1,2 and m = 1,2,3,....

%C Conjecture: a(n) > 0 for all n > 0.

%C We have verified this for n up to 10^8.

%C See also A301303 and A301314 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A301304/b301304.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], 2017-2018.

%e a(79) = 1 since 79 = 5^2 + 1^2 + 2^2 + 7^2 with 5^2 + 7*1^2 = 2^2*2^3.

%e a(323) = 1 since 323 = 3^2 + 1^2 + 12^2 + 13^2 with 3^2 + 7*1^2 = 2*2^3.

%e a(646) = 1 since 646 = 22^2 + 11^2 + 4^2 + 5^2 with 22^2 + 7*11^2 = 11^3.

%e a(815) = 1 since 815 = 9^2 + 5^2 + 15^2 + 22^2 with 9^2 + 7*5^2 = 2^2*4^3.

%e a(1111) = 1 since 1111 = 1^2 + 1^2 + 22^2 + 25^2 with 1^2 + 7*1^2 = 2^3.

%e a(2822) = 1 since 2822 = 2^2 + 0^2 + 3^2 + 53^2 with 2^2 + 7*0^2 = 2^2*1^3.

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

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];

%t QQ[n_]:=n>0&&(CQ[n]||CQ[n/2]||CQ[n/4]);

%t tab={};Do[r=0;Do[If[QQ[x^2+7y^2],Do[If[SQ[n-x^2-y^2-z^2],r=r+1],{z,0,Sqrt[(n-x^2-y^2)/2]}]],{y,0,Sqrt[n/2]},{x,y,Sqrt[n-y^2]}];tab=Append[tab,r],{n,1,80}];Print[tab]

%Y Cf. A000118, A000290, A271510, A271513, A271518, A281976, A300666, A300667, A300708, A300712, A300751, A300752, A300791, A300792, A300844, A300908, A301303, A301314.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Mar 17 2018