A275344 Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y + 3*z a square, where x,y,z,w are nonnegative integers.

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

Offset

0,3

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 3, 5, 7, 14, 15, 16, 25, 30, 33, 84, 169, 225.

(ii) For each ordered pair (a,b) = (1,2), (1,3), (1,12), (1,23), (2,3), (2,4), (2,6), (2,7), (2,15), (2,16), any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that a*x + b*y is a square.

This is similar to the conjecture in A271518. It is known that any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers.

Links

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

Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Example

a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 2*0^2 with 0 + 2*0 + 3*0 = 0^2.
a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 2*0^2 with 1 + 2*0 + 3*0 = 1^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 0^2 + 2*1^2 with 1 + 2*0 + 3*0 = 1^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 2*0^2 with 2 + 2*1 + 3*0 = 2^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 0^2 + 2*1^2 with 2 + 2*1 + 3*0 = 2^2.
a(14) = 1 since 14 = 1^2 + 1^2 + 2^2 + 2*2^2 with 1 + 2*1 + 3*2 = 3^2.
a(15) = 1 since 15 = 3^2 + 0^2 + 2^2 + 2*1^2 with 3 + 2*0 + 3*2 = 3^2.
a(16) = 1 since 16 = 4^2 + 0^2 + 0^2 + 2*0^2 with 4 + 2*0 + 3*0 = 2^2.
a(25) = 1 since 25 = 1^2 + 4^2 + 0^2 + 2*2^2 with 1 + 2*4 + 3*0 = 3^2.
a(30) = 1 since 30 = 3^2 + 2^2 + 3^2 + 2*2^2 with 3 + 2*2 + 3*3 = 4^2.
a(33) = 1 since 33 = 1^2 + 0^2 + 0^2 + 2*4^2 with 1 + 2*0 + 3*0 = 1^2.
a(84) = 1 since 84 = 4^2 + 6^2 + 0^2 + 2*4^2 with 4 + 2*6 + 3*0 = 4^2.
a(169) = 1 since 169 = 10^2 + 6^2 + 1^2 + 2*4^2 with 10 + 2*6 + 3*1 = 5^2.
a(225) = 1 since 225 = 10^2 + 6^2 + 9^2 + 2*2^2 with 10 + 2*6 + 3*9 = 7^2.

Mathematica

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

Crossrefs

Cf. A000290, A271518, A275297, A275301.

Sequence in context: A112632 A254575 A355402 * A206826 A175835 A123738

Adjacent sequences: A275341 A275342 A275343 * A275345 A275346 A275347

Keyword

nonn

Author

Zhi-Wei Sun, Jul 24 2016

Status

approved