A271518 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x + 3*y + 5*z a square, where w, x, y and z are nonnegative integers.

1, 2, 2, 2, 2, 1, 1, 1, 1, 3, 3, 2, 2, 2, 4, 2, 2, 5, 5, 3, 2, 2, 2, 3, 1, 5, 5, 2, 2, 5, 8, 1, 2, 6, 3, 3, 2, 3, 7, 5, 2, 8, 6, 1, 4, 6, 6, 2, 2, 6, 9, 5, 4, 3, 7, 6, 2, 6, 7, 5, 2, 1, 6, 6, 2, 10, 9, 6, 3, 3, 6, 2, 3, 8, 12, 5, 5, 7, 11, 5, 1

Offset: 0

Author

Zhi-Wei Sun, Apr 09 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 4^k*6 (k = 0,1,2,...), 16^k*m (k = 0,1,2,... and m = 5, 7, 8, 31, 43, 61, 116).

(ii) Any integer n > 15 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 6*x + 10*y + 12*z a square.

(iii) Each nonnegative integer n not among 7, 15, 23, 71, 97 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 2*x + 6*y + 10*z a square. Also, any nonnegative integer n not among 7, 43, 79 can be written as w^2 + x^2 + y^2 + z^2 with w, x, y, z nonnegative integers and 3*x + 5*y + 6*z a square.

See also A271510 and A271513 for related conjectures.

a(n) > 0 verified for all n <= 3*10^7. - Zhi-Wei Sun, Nov 28 2016

Qing-Hu Hou at Tianjin Univ. has verified a(n) > 0 and parts (ii) and (iii) of the above conjecture for n up to 10^9. - Zhi-Wei Sun, Dec 04 2016

The conjecture that a(n) > 0 for all n = 0,1,2,... is called the 1-3-5-Conjecture and the author has announced a prize of 1350 US dollars for its solution. - Zhi-Wei Sun, Jan 17 2017

Qing-Hu Hou has finished his verification of a(n) > 0 for n up to 10^10. - Zhi-Wei Sun, Feb 17 2017

The 1-3-5 conjecture was finally proved by António Machiavelo and Nikolaos Tsopanidis in a JNT paper published in 2021. This is a great achivement! - Zhi-Wei Sun, Mar 31 2021

Examples

			a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 0^2 with 1 + 3*0 + 5*0 = 1^2.
a(6) = 1 since 6 = 2^2 + 1^2 + 1^2 + 0^2 with 1 + 3*1 + 5*0 = 2^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 1^2 + 1^2 with 1 + 3*1 + 5*1 = 3^2.
a(8) = 1 since 8 = 0^2 + 0^2 + 2^2 + 2^2 with 0 + 3*2 + 5*2 = 4^2.
a(24) = 1 since 24 = 4^2 + 0^2 + 2^2 + 2^2 with 0 + 3*2 + 5*2 = 4^2.
a(31) = 1 since 31 = 1^2 + 5^2 + 2^2 + 1^2 with 5 + 3*2 + 5*1 = 4^2.
a(43) = 1 since 43 = 1^2 + 1^2 + 5^2 + 4^2 with 1 + 3*5 + 5*4 = 6^2.
a(61) = 1 since 61 = 6^2 + 0^2 + 0^2 + 5^2 with 0 + 3*0 + 5*5 = 5^2.
a(116) = 1 since 116 = 10^2 + 4^2 + 0^2 + 0^2 with 4 + 3*0 + 5*0 = 2^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• António Machiavelo and Nikolaos Tsopanidis, Zhi-Wei Sun's 1-3-5 Conjecture and Variations, arXiv:2003.02592 [math.NT], 2020.
• António Machiavelo and Nikolaos Tsopanidis, Zhi-Wei Sun's 1-3-5 Conjecture and Variations, J. Number Theory 222 (2021), 1-20.
• António Machiavelo, Rogério Reis, and Nikolaos Tsopanidis, Report on Zhi-Wei Sun's "1-3-5 conjecture" and some of its refinements, arXiv:2005.13526 [math.NT], 2020.
• António Machiavelo, Rogério Reis, and Nikolaos Tsopanidis, Report on Zhi-Wei Sun's 1-3-5 conjecture and some of its refinements, J. Number Theory 222 (2021), 21-29.
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.NT], 2016.
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190. (See Conjecture 4.3(i) and Remark 4.3.)

Crossrefs

Cf. A000118, A000290, A270969, A271510, A271513, A273294, A273302, A276533, A278560.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x+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}]