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%I #23 Mar 01 2018 11:39:43
%S 1,1,1,1,3,4,1,1,2,2,1,2,1,1,3,1,1,11,4,6,7,7,8,4,4,6,14,4,6,17,10,1,
%T 10,6,10,7,4,4,16,2,3,10,2,1,9,6,3,2,1,5,2,3,7,9,3,1,6,2,3,7,1,4,4,1,
%U 3,4,3,1,13,20
%N Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 2*y + 3*z is a power of 4 (including 4^0 = 1).
%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 13, 49, 61, 4^k*m (k = 0,1,2,... and m = 1, 2, 11, 14, 17).
%C (ii) Let a,b,c,d be nonnegative integers with a >= b >= c >= d, b positive, and gcd(a,b,c,d) not divisible by 4. Then, any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*x + b*y + c*z + d*w = 4^k for some k = 0,1,2,..., if and only if d = 0 and (a,b,c) is among the following ordered triples: (3,2,1), (2,1,0), (3,1,0), (4,2,0), (8,1,0), (15,1,0) and (16,2,0).
%C By Theorem 1.1(i) of arXiv:1701.05868, any positive square can be written as (4^k)^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.
%C We have verified that a(n) > 0 for all n = 1..50000.
%H Zhi-Wei Sun, <a href="/A299924/b299924.txt">Table of n, a(n) for n = 1..500</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(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1 + 2*0 + 3*0 = 4^0.
%e a(2) = 1 since 2^2 = 0^2 + 2^2 + 0^2 + 0^2 with 0 + 2*2 + 3*0 = 4.
%e a(3) = 1 since 3^2 = 2^2 + 1^2 + 0^2 + 2^2 with 2 + 2*1 + 3*0 = 4.
%e a(7) = 1 since 7^2 = 2^2 + 4^2 + 2^2 + 5^2 with 2 + 2*4 + 3*2 = 4^2.
%e a(11) = 1 since 11^2 = 2^2 + 1^2 + 4^2 + 10^2 with 2 + 2*1 + 4*3 = 4^2.
%e a(13) = 1 since 13^2 = 8^2 + 1^2 + 2^2 + 10^2 with 8 + 2*1 + 3*2 = 4^2.
%e a(14) = 1 since 14^2 = 4^2 + 6^2 + 0^2 + 12^2 with 4 + 2*6 + 3*0 = 4^2.
%e a(17) = 1 since 17^2 = 0^2 + 8^2 + 0^2 + 15^2 with 0 + 2*8 + 4*0 = 4^2.
%e a(49) = 1 since 49^2 = 22^2 + 3^2 + 12^2 + 42^2 with 22 + 2*3 + 3*12 = 4^3.
%e a(61) = 1 since 61^2 = 6^2 + 20^2 + 6^2 + 57^2 with 6 + 2*20 + 3*6 = 4^3.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
%t Do[r=0;Do[If[SQ[n^2-x^2-y^2-z^2]&&Pow[x+2y+3z],r=r+1],{x,0,n},{y,0,Sqrt[n^2-x^2]},{z,0,Sqrt[n^2-x^2-y^2]}];Print[n," ",r],{n,1,70}]
%Y Cf. A000118, A000302, A271518, A279612, A281976, A299825, A300219.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Feb 21 2018
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