A338094 Number of ways to write 2*n + 1 as x^2 + y^2 + z^2 + w^2 with x + y a positive power of two, where x, y, z, w are nonnegative integers with x <= y and z <= w.

1, 1, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 3, 3, 1, 5, 3, 2, 3, 5, 2, 5, 3, 4, 4, 4, 3, 6, 4, 3, 4, 5, 3, 7, 2, 4, 6, 5, 2, 6, 3, 3, 4, 7, 3, 6, 4, 4, 5, 5, 2, 7, 2, 2, 3, 5, 4, 6, 4, 4, 4, 6, 3, 9, 4, 5, 6, 5, 3, 7, 2, 5, 7, 7, 4, 10, 7, 6, 7, 9, 3, 8, 3, 4, 7, 7, 5, 10, 6, 5, 6, 10, 6, 11, 5, 5, 9, 5, 3, 12

Offset: 1

Author

Zhi-Wei Sun, Oct 09 2020

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0. Moreover, any integer m > 1987 not congruent to 0 or 6 modulo 8 can be written as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers and x + y a positive power of 4.

We have verified the latter version of the conjecture for m up to 3*10^7.

By Theorem 1.1(ii) of the author's IJNT paper, any positive integer can be written as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers and x - y a power of two (including 2^0 = 1).

See also A338121 for related information, and A338095 and A338096 for similar conjectures.

Examples

			a(1) = 1, and 2*1 + 1 = 1^2 + 1^2 + 0^2 + 1^2 with 1 + 1 = 2^1.
a(2) = 1, and 2*2 + 1 = 0^2 + 2^2 + 0^2 + 1^2 with 0 + 2 = 2^1.
a(3) = 1, and 2*3 + 1 = 1^2 + 1^2 + 1^2 + 2^2 with 1 + 1 = 2^1.
a(11) = 1, and 2*11 + 1 = 1^2 + 3^2 + 2^2 + 3^2 with 1 + 3 = 2^2.
a(15) = 1, and 2*15 + 1 = 1^2 + 1^2 + 2^2 + 5^2 with 1 + 1 = 2^1.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. See also arXiv:1604.06723 [math.NT].
• Zhi-Wei Sun, Restricted sums of four squares, Int. J. Number Theory 15(2019), 1863-1893. See also arXiv:1701.05868 [math.NT].
• Zhi-Wei Sun, Sums of four squares with certain restrictions, arXiv:2010.05775 [math.NT], 2020.

Crossrefs

Cf. A000079, A000118, A000290, A000302, A279612, A338095, A338096, A338103, A338103, A338119, A338121.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
PQ[n_]:=PQ[n]=n>1&&IntegerQ[Log[2,n]];
tab={};Do[r=0;Do[If[SQ[2n+1-x^2-y^2-z^2]&&PQ[x+y],r=r+1],{x,0,Sqrt[(2n+1)/2]},{y,x,Sqrt[2n+1-x^2]},{z,Boole[x+y==0],Sqrt[(2n+1-x^2-y^2)/2]}];
tab=Append[tab,r],{n,1,100}];Print[tab]