A282542 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 3*y + 5*z and (at least) one of y,z,w are squares.

1, 2, 2, 2, 2, 1, 1, 1, 1, 3, 3, 2, 1, 2, 4, 2, 2, 4, 5, 3, 2, 2, 2, 2, 1, 5, 5, 2, 1, 5, 8, 1, 2, 3, 3, 3, 2, 3, 5, 5, 2, 8, 5, 1, 1, 6, 6, 1, 2, 5, 9, 5, 4, 2, 5, 5, 2, 5, 4, 5, 2, 1, 5, 3, 2, 7, 9, 5, 2, 3, 6, 2, 2, 8, 9, 5, 3, 5, 9, 2, 1

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....

This is stronger than the 1-3-5 conjecture (cf. A271518).

By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.

Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.

Example

a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 3*1 + 5*1 = 3^2 and 1 = 1^2.
a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 + 3*1 + 5*1 = 3^2 and 1 = 1^2.
a(47) = 1 since 47 = 3^2 + 1^2 + 6^2 + 1^2 with 3 + 3*1 + 5*6 = 6^2 and 1 = 1^2.
a(92) = 1 since 92 = 1^2 + 1^2 + 9^2 + 3^2 with 1 + 3*1 + 5*1 = 3^2 and 9 = 3^2.
a(188) = 1 since 188 = 7^2 + 9^2 + 3^2 + 7^2 with 7 + 3*9 + 5*3 = 7^2 and 9 = 3^2.
a(248) = 1 since 248 = 10^2 + 2^2 + 0^2 + 12^2 with 10 + 3*2 + 5*0 = 4^2 and 0 = 0^2.
a(388) = 1 since 388 = 13^2 + 1^2 + 13^2 + 7^2 with 13 + 3*1 + 5*13 = 9^2 and 1 = 1^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&(SQ[y]||SQ[z]||SQ[Sqrt[n-x^2-y^2-z^2]])&&SQ[x+3y+5z], r=r+1], {x, 0, n^(1/2)}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000118, A000290, A270969, A271518, A281976.

Sequence in context: A367912 A368109 A275301 * A271518 A352629 A106825

Adjacent sequences: A282539 A282540 A282541 * A282543 A282544 A282545

Keyword

nonn

Author

Zhi-Wei Sun, Feb 17 2017

Status

approved