A279522 Number of ways to write n as w^2 + x^2 + y^2 + z^2 with w + 2*x + 3*y + 5*z a square, where w,x,y,z are nonnegative integers.

1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 3, 1, 1, 2, 2, 2, 1, 3, 5, 2, 2, 5, 4, 1, 1, 2, 4, 1, 0, 7, 3, 1, 1, 1, 4, 6, 4, 5, 4, 4, 3, 3, 5, 2, 3, 5, 4, 2, 1, 4, 8, 5, 1, 5, 12, 1, 1, 2, 6, 3, 4, 3, 3, 9, 1, 6, 4, 2, 3, 8, 8, 2, 3, 7, 7, 7, 3, 8, 14, 3, 2

Offset

0,6

Comments

Conjecture: (i) a(n) = 0 if and only if n = 16^k*28 for some k = 0,1,2,....

(ii) For any positive integers a,b,c,d, there are infinitely many positive integers which cannot be written as w^2 + x^2 + y^2 + z^2 with a*w + b*x + c*y + d*z a square, where w,x,y,z are nonnegative integers.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..5000

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

Example

a(27) = 1 since 27 = 1^2 + 5^2 + 0^2 + 1^2 with 1 + 2*5 + 3*0 + 5*1 = 4^2.
a(31) = 1 since 31 = 1^2 + 2^2 + 5^2 + 1^2 with 1 + 2*2 + 3*5 + 5*1 = 5^2.
a(33) = 1 since 33 = 0^2 + 4^2 + 4^2 + 1^2 with 0 + 2*4 + 3*4 + 5*1 = 5^2.
a(52) = 1 since 52 = 4^2 + 6^2 + 0^2 + 0^2 with 4 + 2*6 + 3*0 + 5*0 = 4^2.
a(55) = 1 since 55 = 1^2 + 5^2 + 5^2 + 2^2 with 1 + 2*5 + 3*5 + 5*2 = 6^2.
a(56) = 1 since 56 = 0^2 + 4^2 + 6^2 + 2^2 with 0 + 2*4 + 3*6 + 5*2 = 6^2.
a(88) = 1 since 88 = 4^2 + 8^2 + 2^2 + 2^2 with 4 + 2*8 + 3*2 + 5*2 = 6^2.
a(137) = 1 since 137 = 10^2 + 6^2 + 1^2 + 0^2 with 10 + 2*6 + 3*1 + 5*0 = 5^2.
a(164) = 1 since 164 = 12^2 + 2^2 + 0^2 + 4^2 with 12 + 2*2 + 3*0 + 5*4 = 6^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2]+2x+3y+5z], r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000118, A000290, A270969, A271518, A273404.

Sequence in context: A016533 A122915 A327193 * A182592 A030298 A370221

Adjacent sequences: A279519 A279520 A279521 * A279523 A279524 A279525

Keyword

nonn

Author

Zhi-Wei Sun, Dec 14 2016

Status

approved