A275409 Number of ordered ways to write n as 2*w^2 + x^2 + y^2 + z^2 with w + x + 2*y + 4*z a square, where w,x,y,z are nonnegative integers.

1, 2, 1, 0, 2, 2, 2, 1, 1, 1, 0, 3, 1, 2, 1, 1, 3, 2, 5, 3, 4, 3, 1, 1, 1, 1, 2, 2, 2, 4, 2, 2, 4, 2, 7, 3, 1, 6, 2, 1, 2, 3, 4, 5, 1, 1, 3, 5, 3, 3, 4, 3, 7, 3, 2, 4, 3, 4, 4, 3, 1, 4, 5, 3, 6, 4, 4, 4, 5, 7, 7, 3, 6, 5, 5, 4, 3, 11, 2, 2, 4

Offset

0,2

Comments

Conjecture: (i) a(n) > 0 except for n = 3, 10, and a(n) = 1 only for n = 0, 2, 7, 8, 9, 12, 14, 15, 22, 23, 24, 25, 36, 39, 44, 45, 60, 87, 98, 106, 110, 111, 183.

(ii) 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 x + 2*y + 3*z - 3*w is a square.

(iii) For each triple (a,b,c) = (1,2,1), (1,2,3), (1,3,1), (2,4,1), (2,4,2), (2,4,3), (2,4,4), (2,4,8), (8,9,5), 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 - c*z is a square.

(iv) 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 x + 2*y - 2*z is twice a nonnegative cube. Also, each natural number can be written as x^2 + y^2 + z^2 + 2*w^3 with x,y,z,w nonnegative integers such that x + 3*y - z is a square.

See also A275344 and A275301 for related conjectures. We are able to show that each natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w integers such that x + y + z = t^2 for some t = 0, 1, 2.

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(2) = 1 since 2 = 2*1^2 + 0^2 + 0^2 + 0^2 with 1 + 0 + 2*0 + 4*0 = 1^2.
a(7) = 1 since 7 = 2*1^2 + 0^2 + 2^2 + 1^2 with 1 + 0 + 2*2 + 4*1 = 3^2.
a(8) = 1 since 8 = 2*1^2 + 2^2 + 1^2 + 1^2 with 1 + 2 + 2*1 + 4*1 = 3^2.
a(9) = 1 since 9 = 2*2^2 + 0^2 + 1^2 + 0^2 with 2 + 0 + 2*1 + 4*0 = 2^2.
a(12) = 1 since 12 = 2*2^2 + 2^2 + 0^2 + 0^2 with 2 + 2 + 2*0 + 4*0 = 2^2.
a(14) = 1 since 14 = 2*0^2 + 2^2 + 1^2 + 3^2 with 0 + 2 + 2*1 + 4*3 = 4^2.
a(15) = 1 since 15 = 2*1^2 + 2^2 + 3^2 + 0^2 with 1 + 2 + 2*3 + 4*0 = 3^2.
a(22) = 1 since 22 = 2*1^2 + 4^2 + 2^2 + 0^2 with 1 + 4 + 2*2 + 4*0 = 3^2.
a(23) = 1 since 23 = 2*3^2 + 2^2 + 0^2 + 1^2 with 3 + 2 + 2*0 + 4*1 = 3^2.
a(24) = 1 since 24 = 2*0^2 + 4^2 + 2^2 + 2^2 with 0 + 4 + 2*2 + 4*2 = 4^2.
a(25) = 1 since 25 = 2*0^2 + 4^2 + 0^2 + 3^2 with 0 + 4 + 2*0 + 4*3 = 4^2.
a(36) = 1 since 36 = 2*3^2 + 1^2 + 4^2 + 1^2 with 3 + 1 + 2*4 + 4*1 = 4^2.
a(39) = 1 since 39 = 2*1^2 + 6^2 + 1^2 + 0^2 with 1 + 6 + 2*1 + 4*0 = 3^2.
a(44) = 1 since 44 = 2*3^2 + 0^2 + 1^2 + 5^2 with 3 + 0 + 2*1 + 4*5 = 5^2.
a(45) = 1 since 45 = 2*0^2 + 5^2 + 2^2 + 4^2 with 0 + 5 + 2*2 + 4*4 = 5^2.
a(60) = 1 since 60 = 2*2^2 + 6^2 + 4^2 + 0^2 with 2 + 6 + 2*4 + 4*0 = 4^2.
a(87) = 1 since 87 = 2*3^2 + 2^2 + 8^2 + 1^2 with 3 + 2 + 2*8 + 4*1 = 5^2.
a(98) = 1 since 98 = 2*4^2 + 1^2 + 8^2 + 1^2 with 4 + 1 + 2*8 + 4*1 = 5^2.
a(106) = 1 since 106 = 2*2^2 + 8^2 + 3^2 + 5^2 with 2 + 8 + 2*3 + 4*5 = 6^2.
a(110) = 1 since 110 = 2*6^2 + 5^2 + 3^2 + 2^2 with 6 + 5 + 2*3 + 4*2 = 5^2.
a(111) = 1 since 111 = 2*5^2 + 3^2 + 6^2 + 4^2 with 5 + 3 + 2*6 + 4*4 = 6^2.
a(183) = 1 since 183 = 2*3^2 + 10^2 + 4^2 + 7^2 with 3 + 10 + 2*4 + 4*7 = 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[w+x+2y+4*Sqrt[n-2*w^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, A275344.

Sequence in context: A282459 A016154 A307332 * A029343 A137992 A047654

Adjacent sequences: A275406 A275407 A275408 * A275410 A275411 A275412

Keyword

nonn

Author

Zhi-Wei Sun, Jul 26 2016

Status

approved