A268197 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w*(25*w + 24*x + 48*y + 96*z) a square, where w is a positive integer and x,y,z are nonnegative integers.

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

Offset

1,2

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 15, 23, 43, 55, 463, 4^k*m (k = 0,1,2,... and m = 1, 31, 34).

(ii) For each triple (a,b,c) = (1,3,4), (2,3,4), (2,4,6), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w*(25*w + 24*(a*x+b*y+c*z)) a square, where w is a positive integer and x,y,z are nonnegative integers.

For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

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.

Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.

Example

a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 1*(25*1 + 24*0 + 48*0 + 96*0) = 5^2.
a(2) = 2 since 2 = 1^2 + 0^2 + 0^2 + 1^2 with 1 > 0 and 1*(25*1 + 24*0 + 48*0 + 96*1) = 11^2, and also 2 = 1^2 + 1^2 + 0^2 + 0^2 with 1 > 0 and 1*(25*1 + 24*1 + 48*0 + 96*0) = 7^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 1^2 + 1^2 with 1 > 0 and 1*(25*1 + 24*0 + 48*1 + 96*1) = 13^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 > 0 and 1*(25*1 + 24*1 + 48*1 + 96*2) = 17^2.
a(15) = 1 since 15 = 1^2 + 3^2 + 2^2 + 1^2 with 1 > 0 and 1*(25*1 + 24*3 + 48*2 + 96*1) = 17^2.
a(23) = 1 since 23 = 3^2 + 2^2 + 3^2 + 1^2 with 3 > 0 and 3*(25*3 + 24*2 + 48*3 + 96*1) = 33^2.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 > 0 and 1*(25*1 + 24*1 + 48*2 + 96*5) = 25^2.
a(34) = 1 since 34 = 1^2 + 1^2 + 4^2 + 4^2 with 1 > 0 and 1*(25*1 + 24*1 + 48*4 + 96*4) = 25^2.
a(43) = 1 since 43 = 3^2 + 3^2 + 3^2 + 4^2 with 3 > 0 and 3*(25*3 + 24*3 + 48*3 + 96*4) = 45^2.
a(55) = 1 since 55 = 3^2 + 1^2 + 6^2 + 3^2 with 3 > 0 and 3*(25*3 + 24*1 + 48*6 + 96*3) = 45^2.
a(463) = 1 since 463 = 3^2 + 18^2 + 11^2 + 3^2 with 3 > 0 and 3*(25*3 + 24*18 + 48*11 + 96*3) = 63^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[25x^2+24x(y+2z+4*Sqrt[n-x^2-y^2-z^2])], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 70}]

Crossrefs

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268507, A269400, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620.

Sequence in context: A276520 A242062 A025848 * A065382 A066888 A029313

Adjacent sequences: A268194 A268195 A268196 * A268198 A268199 A268200

Keyword

nonn

Author

Zhi-Wei Sun, May 04 2016

Status

approved