

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.


3



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
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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

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, Refining Lagrange's foursquare 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[n2w^2x^2y^2]&&CQ[x+2*y], r=r+1], {w, 0, (n/2)^(1/2)}, {x, 0, Sqrt[n2w^2]}, {y, 0, Sqrt[n2w^2x^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000290, A000578, A271518, A275297, A275300, A275344.
Sequence in context: A037805 A327144 A327051 * A282542 A271518 A106825
Adjacent sequences: A275298 A275299 A275300 * A275302 A275303 A275304


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jul 22 2016


STATUS

approved



