cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A300752 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 3*x or y or z is a square.

Original entry on oeis.org

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, 5, 4, 2, 8, 5, 1, 3, 5, 6, 1, 2, 4, 8, 5, 4, 2, 4, 4, 2, 6, 5, 5, 2, 1, 6, 4, 1, 8, 9, 6, 2, 3, 4, 1, 2, 6, 8, 5, 4, 5, 8, 2, 1
Offset: 1

Views

Author

Zhi-Wei Sun, Mar 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n = 1,2,3,....
This is stronger than the author's 1-3-5 conjecture in A271518. See also A300751 for a similar conjecture stronger than the 1-3-5 conjecture.
In 2020, A. Machiavelo, R. Reis and N. Tsopanidis verified a(n) > 0 for n up to 1.05*10^11. - Zhi-Wei Sun, Oct 06 2020

Examples

			a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 1 = 1^2 and 3 + 3*1 + 5*6 = 6^2.
a(248) = 1 since 248 = 10^2 + 2^2 + 0^2 + 12^2 with 0 = 0^2 and 10 + 3*2 + 5*0 = 4^2.
a(263) = 1 since 263 = 13^2 + 2^2 + 9^2 + 3^2 with 9 = 3^2 and 13 + 3*2 + 5*9 = 8^2.
a(808) = 1 since 808 = 12^2 + 14^2 + 18^2 + 12^2 with 3*12 = 6^2 and 12 + 3*14 + 5*18 = 12^2.
a(1288) = 1 since 1288 = 12^2 + 18^2 + 26^2 + 12^2 with 3*12 = 6^2 and 12 + 3*18 + 5*26 = 14^2.
a(3544) = 1 since 3544 = 14^2 + 34^2 + 16^2 + 44^2 with 16 = 4^2 and 14 + 3*34 + 5*16 = 14^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[3(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]