A275301 Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y a cube, where x,y,z,w are nonnegative integers.

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

Offset

0,2

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 5, 6, 7, 8, 11, 14, 15, 30, 78, 90, 93, 106, 111, 117, 125, 223, 335.

(ii) Any 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 + 2*y is a square.

See also A275344 for a similar conjecture.

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 = 0^3.
a(2) = 2 since 2 = 1^2 + 0^2 + 1^2 + 2*0^2 with 1 + 2*0 =1^3, and 2 = 0^2 + 0^2 + 0^2 + 2*1^2 with 0 + 2*0 = 0^3.
a(5) = 1 since 5 = 1^2 + 0^2 + 2^2 + 2*0^2 with 1 + 2*0 = 1^3.
a(6) = 1 since 6 = 0^2 + 0^2 + 2^2 + 2*1^2 with 0 + 2*0 = 0^3.
a(7) = 1 since 7 = 1^2 + 0^2 + 2^2 + 2*1^2 with 1 + 2*0 = 1^3.
a(8) = 1 since 8 = 0^2 + 0^2 + 0^2 + 2*2^2 with 0 + 2*0 = 0^3.
a(11) = 1 since 11 = 0^2 + 0^2 + 3^2 + 2*1^2 with 0 + 2*0 = 0^3.
a(14) = 1 since 14 = 2^2 + 3^2 + 1^2 + 2*0^2 with 2 + 2*3 = 2^3.
a(15) = 1 since 15 = 2^2 + 3^2 + 0^2 + 2*1^2 with 2 + 2*3 = 2^3.
a(30) = 1 since 30 = 2^2 + 3^2 + 3^2 + 2*2^2 with 2 + 2*3 = 2^3.
a(78) = 1 since 78 = 6^2 + 1^2 + 3^2 + 2*4^2 with 6 + 2*1 = 2^3.
a(90) = 1 since 90 = 1^2 + 0^2 + 9^2 + 2*2^2 with 1 + 2*0 = 1^3.
a(93) = 1 since 93 = 4^2 + 2^2 + 1^2 + 2*6^2 with 4 + 2*2 = 2^3.
a(106) = 1 since 106 = 4^2 + 2^2 + 6^2 + 2*5^2 with 4 + 2*2 = 2^3.
a(111) = 1 since 111 = 2^2 + 3^2 + 0^2 + 2*7^2 with 2 + 2*3 = 2^3.
a(117) = 1 since 117 = 4^2 + 2^2 + 5^2 + 2*6^2 with 4 + 2*2 = 2^3.
a(125) = 1 since 125 = 6^2 + 1^2 + 4^2 + 2*6^2 with 6 + 2*1 = 2^3.
a(223) = 1 since 223 = 11^2 + 8^2 + 6^2 + 2*1^2 with 11 + 2*8 = 3^3.
a(335) = 1 since 335 = 11^2 + 8^2 + 10^2 + 2*5^2 with 11 + 2*8 = 3^3.

Mathematica

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

Crossrefs

Cf. A000290, A000578, A271518, A275297, A275300, A275344.

Sequence in context: A327051 A367912 A368109 * A282542 A271518 A352629

Adjacent sequences: A275298 A275299 A275300 * A275302 A275303 A275304

Keyword

nonn

Author

Zhi-Wei Sun, Jul 22 2016

Status

approved