A275656 Number of ordered ways to write n as 4^k*(1+x^2+y^2)+z^2, where k,x,y,z are nonnegative integers with x <= y <= z.

1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 3, 1, 1, 2, 3, 2, 2, 3, 2, 1, 1, 2, 2, 2, 2, 4, 2, 2, 1, 1, 4, 2, 2, 3, 3, 1, 2, 3, 3, 3, 1, 2, 3, 2, 1, 2, 4, 2, 3, 5, 3, 2, 3, 2, 4, 2, 1, 4, 2, 2, 2, 2, 5, 3, 3, 4, 4, 2, 1, 3, 5, 2, 3, 4, 1, 2, 1, 3

Offset

1,4

Comments

Conjecture: a(n) > 0 for all n > 0.

This is stronger than Lagrange's four-square theorem and the conjecture in A275648.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

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

Example

a(22) = 1 since 22 = 4^0*(1+1^2+2^2) + 4^2 with 1 < 2 < 4.
a(31) = 1 since 31 = 4^0*(1+1^2+2^2) + 5^2 with 1 < 2 < 5.
a(59) = 1 since 59 = 4^0*(1+0^2+3^2) + 7^2 with 0 < 3 < 7.
a(79) = 1 since 79 = 4^0*(1+2^2+5^2) + 7^2 with 2 < 5 < 7.
a(94) = 1 since 94 = 4^0*(1+2^2+5^2) + 8^2 with 2 < 5 < 8.
a(128) = 1 since 128 = 4^3*(1+0^2+0^2) + 8^2 with 0 = 0 < 8.
a(134) = 1 since 134 = 4^0*(1+4^2+6^2) + 9^2 with 4 < 6 < 9.
a(221) = 1 since 221 = 4*(1+3^2+5^2) + 9^2 with 3 < 5 < 9.
a(254) = 1 since 254 = 4^0*(1+3^2+10^2) + 12^2 with 3 < 10 < 12.
a(349) = 1 since 349 = 4*(1+5^2+7^2) + 7^2  with 5 < 7 = 7.
a(608) = 1 since 608 = 4^2*(1+0^2+1^2) + 24^2 with 0 < 1 < 24.
a(797) = 1 since 797 = 4*(1+0^2+4^2) + 27^2 with 0 < 4 < 27.
a(1181) = 1 since 1181 = 4*(1+9^2+9^2) + 23^2 with 9 = 9 < 23.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-4^k*(1+x^2+y^2)], r=r+1], {k, 0, Log[4, n]}, {x, 0, Sqrt[(n-4^k)/(2*4^k+1)]}, {y, x, Sqrt[(n-4^k*(1+x^2))/(4^k+1)]}]; Print[n, " ", r]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000118, A000290, A271518, A275648.

Sequence in context: A184257 A365721 A352219 * A228825 A324381 A331383

Adjacent sequences: A275653 A275654 A275655 * A275657 A275658 A275659

Keyword

nonn

Author

Zhi-Wei Sun, Aug 04 2016

Status

approved