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A279612 Number of ways to write n = x^2 + y^2 + z^2 + w^2 with x + 2*y - 2*z a power of 4 (including 4^0 = 1), where x,y,z,w are nonnegative integers.

%I A279612 #12 Jan 24 2017 17:31:09
%S A279612 1,1,1,2,3,1,1,1,3,5,2,1,3,4,1,1,3,5,5,4,3,2,3,2,4,5,1,3,4,4,1,1,5,7,
%T A279612 7,2,3,7,3,2,4,3,4,2,8,5,1,1,6,8,3,6,7,8,2,3,3,6,8,4,6,5,2,2,9,7,7,7,
%U A279612 7,12,3,1,9,10,7,1,10,10,2,3
%N A279612 Number of ways to write n = x^2 + y^2 + z^2 + w^2 with x + 2*y - 2*z a power of 4 (including 4^0 = 1), where x,y,z,w are nonnegative integers.
%C A279612 Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*q (k = 0,1,2,... and q = 1, 2, 3, 6, 7, 8, 12, 15, 27, 31, 47, 72, 76, 92, 111, 127).
%C A279612 (ii) Let a and b be positive integers with gcd(a,b) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y a power of two (including 2^0 = 1) if and only if (a,b) = (1,1), (2,1), (2,3).
%C A279612 (iii) Let a,b,c be positive integers with a <= b and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z a power of two if and only if (a,b,c) is among the triples (1,1,1), (1,1,2), (1,2,1), (1,2,2), (1,3,1), (1,3,2), (1,3,3), (1,3,4), (1,3,5), (1,4,1), (1,4,2), (1,4,3), (1,4,4), (1,5,1), (1,5,2), (1,5,4), (1,5,5,), (1,6,3), (1,7,4), (1,7,7), (1,8,1), (1,9,2), (2,3,1), (2,3,3), (2,3,4), (2,5,1), (2,5,3), (2,5,4), (2,5,5), (2,7,1), (2,7,3), (2,7,7), (2,9,3), (2,11,5), (3,4,3), (7,8,7).
%C A279612 (iv) Let a,b,c be positive integers with b <= c and gcd(a,b,c) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z a power of two if and only if (a,b,c) is among the triples (2,2,1), (4,2,1), (4,3,1), (4,4,3).
%C A279612 (v) Let a,b,c be positive integers with a <= b, c <= d, and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y - c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,2,1,1), (1,2,1,2), (1,2,1,3), (1,3,1,2), (1,3,2,3), (1,3,2,4), (1,4,1,2), (1,7,2,6), (1,9,1,4), (2,2,2,3), (2,3,1,2), (2,3,1,3), (2,3,2,3), (2,3,6,1), (2,4,1,2), (2,5,1,2), (2,5,2,3), (2,5,3,4), (3,4,1,2), (3,4,1,3), (3,4,1,5), (3,4,2,5), (3,4,3,4), (3,8,1,10), (3,8,2,3), (4,5,1,5).
%C A279612 (vi) Let a,b,c be positive integers with a <= b <= c and gcd(a,b,c,d) odd. Then any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z -d*w a power of two if and only if (a,b,c,d) is among the quadruples (1,1,2,2), (1,1,2,3), (1,1,2,4), (1,2,2,3), (1,2,3,4), (1,2,4,3), (1,2,6,7), (1,3,4,4), (1,4,6,5), (2,3,5,4).
%C A279612 (vii) For any positive integers a,b,c,d, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x - b*y - c*z -d*w a power of two.
%C A279612 (viii) Let a and b be positive integers, and c and d be nonnegative integers. Then, not all positive integers can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and a*x + b*y + c*z + d*w a power of two.
%C A279612 We have verified a(n) > 0 for all n = 1..2*10^7. The conjecture that a(n) > 0 for all n > 0 appeared in arXiv:1701.05868.
%H A279612 Zhi-Wei Sun, <a href="/A279612/b279612.txt">Table of n, a(n) for n = 1..10000</a>
%H A279612 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. Also available from <a href="http://arxiv.org/abs/1604.06723">arXiv:1604.06723 [math.NT]</a>.
%H A279612 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%e A279612 a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 2*1 - 2*1 = 4^0.
%e A279612 a(15) = 1 since 15 = 3^2 + 1^2 + 2^2 + 1^2 with 3 + 2*1 - 2*2 = 4^0.
%e A279612 a(27) = 1 since 27 = 4^2 + 1^2 + 1^2 + 3^2 with 4 + 2*1 - 2*1 = 4.
%e A279612 a(31) = 1 since 31 = 3^2 + 2^2 + 3^2 + 3^2 with 3 + 2*2 - 2*3 = 4^0.
%e A279612 a(47) = 1 since 47 = 3^2 + 2^2 + 3^2 + 5^2 with 3 + 2*2 - 2*3 = 4^0.
%e A279612 a(72) = 1 since 72 = 8^2 + 0^2 + 2^2 + 2^2 with 8 + 2*0 - 2*2 = 4.
%e A279612 a(76) = 1 since 76 = 1^2 + 5^2 + 5^2 + 5^2 with 1 + 2*5 - 2*5 = 4^0.
%e A279612 a(92) = 1 since 92 = 4^2 + 6^2 + 6^2 + 2^2 with 4 + 2*6 - 2*6 = 4.
%e A279612 a(111) = 1 since 111 = 9^2 + 1^2 + 5^2 + 2^2 with 9 + 2*1 - 2*5 = 4^0.
%e A279612 a(127) = 1 since 127 = 7^2 + 2^2 + 5^2 + 7^2 with 7 + 2*2 - 2*5 = 4^0.
%t A279612 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t A279612 FP[n_]:=FP[n]=n>0&&IntegerQ[Log[4,n]];
%t A279612 Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&FP[x+2y-2z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]
%Y A279612 Cf. A000079, A000118, A000290, A271518, A275656, A275675, A275676, A275738, A278560, A279616.
%K A279612 nonn
%O A279612 1,4
%A A279612 _Zhi-Wei Sun_, Dec 15 2016