A300751 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x + 3*y + 5*z a positive square, where x,y,z,w are nonnegative integers such that 2*x or y or z is a square.

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

Offset: 1

Author

Zhi-Wei Sun, Mar 11 2018

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n = 1,2,3,..., and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 1, 4, 5, 6, 7, 8, 20, 24, 28, 31, 36, 43, 61, 71, 79, 100, 116, 157, 188, 200, 344, 351, 388, 632.

This is stronger than the author's 1-3-5 conjecture in A271518. See also A300752 for a similar conjecture stronger than the 1-3-5 conjecture.

a(n) > 0 for all n = 1..3*10^6. - Zhi-Wei Sun, Oct 06 2020

Examples

			a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 2*0 = 0^2 and 0 + 3*2 + 5*2 = 4^2.
a(61) = 1 since 61 = 0^2 + 0^2 + 5^2 + 6^2 with 0 = 0^2 and 0 + 3*0 + 5*5 = 5^2.
a(79) = 1 since 79 = 5^2 + 2^2 + 1^2 + 7^2 with 1 = 1^2 and 5 + 3*2 + 5*1 = 4^2.
a(188) = 1 since 188 = 7^2 + 9^2 + 3^2 + 7^2 with 9 = 3^2 and 7 + 3*9 + 5*3 = 7^2.
a(200) = 0 since 200 = 6^2 + 10^2 + 0^2 + 8^2 with 0 = 0^2 and 6 + 3*10 + 5*0 = 6^2.
a(632) = 1 since 632 = 6^2 + 16^2 + 18^2 + 4^2 with 16 = 4^2 and 6 + 3*16 + 5*18 = 12^2.
a(808) = 3 since 808 = 8^2 + 2^2 + 26^2 + 8^2 = 8^2 + 22^2 + 14^2 + 8^2 = 18^2 + 12^2 + 18^2 + 4^2  with 2*8 = 4^2, 2*18 = 6^2 and 8 + 3*2 + 5*26 = 8 + 3*22 + 5*14 = 18 + 3*12 + 5*18 = 12^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
• Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Crossrefs

Cf. A000118, A000290, A271518, A281976, A282542, A300666, A300667, A300708, A300712, A300752.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[(SQ[2(m^2-3y-5z)]||SQ[y]||SQ[z])&&SQ[n-(m^2-3y-5z)^2-y^2-z^2],r=r+1],{m,1,(35n)^(1/4)},{y,0,Min[m^2/3,Sqrt[n]]},{z,0,Min[(m^2-3y)/5,Sqrt[n-y^2]]}];tab=Append[tab,r],{n,1,80}];Print[tab]