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.

A338096 Number of ways to write 2*n+1 as x^2 + y^2 + z^2 + w^2 with x + 2*y + 3*z a positive power of two, where x, y, z, w are nonnegative integers.

Original entry on oeis.org

1, 1, 5, 1, 3, 2, 3, 2, 5, 1, 5, 2, 4, 4, 7, 2, 5, 5, 3, 3, 6, 1, 5, 3, 2, 6, 6, 2, 4, 2, 2, 2, 8, 2, 7, 3, 5, 6, 6, 1, 5, 6, 7, 7, 8, 4, 6, 5, 5, 7, 11, 3, 13, 5, 3, 6, 11, 4, 7, 6, 3, 7, 9, 5, 8, 6, 3, 8, 9, 5, 10, 3, 9, 8, 7, 2, 7, 6, 5, 4, 4, 3, 12, 7, 3, 9, 9, 5, 11, 8, 2, 5, 10, 3, 5, 5, 2, 9, 9, 4, 13
Offset: 0

Views

Author

Zhi-Wei Sun, Oct 09 2020

Keywords

Comments

Conjecture 1 (1-2-3 Conjecture): a(n) > 0 for all n >= 0. In other words, any positive odd integer m can be written as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers such that x + 2*y + 3*z = 2^k for some positive integer k.
Conjecture 2 (Strong Version of the 1-2-3 Conjecture): For any integer m > 4627 not congruent to 0 or 2 modulo 8, we can write m as x^2 + y^2 + z^2 + w^2 with x, y, z, w nonnegative integers such that x + 2*y + 3*z = 4^k for some positive integer k.
We have verified Conjectures 1 and 2 for m up to 5*10^6. Conjecture 2 implies that A299924(n) > 0 for all n > 0.
By Theorem 1.2(v) of the author's 2017 JNT paper, any positive integer n can be written as x^2 + y^2 + z^2 + 4^k with k, x, y, z nonnegative integers.
See also A338094 and A338095 for similar conjectures.

Examples

			a(1) = 1, and 2*1 + 1 = 1^2 + 0^2 + 1^2 + 1^2 with 1 + 2*0 + 3*1 = 2^2.
a(3) = 1, and 2*3 + 1 = 1^2 + 2^2 + 1^2 + 1^2 with 1 + 2*2 + 3*1 = 2^3.
a(9) = 1, and 2*9 + 1 = 1^2 + 6^2 + 1^2 + 1^2 with 1 + 2*6 + 3*1 = 2^4.
a(21) = 1, and 2*21 + 1 = 5^2 + 4^2 + 1^2 + 1^2 with 5 + 2*4 + 3*1 = 2^4.
a(39) = 1, and 2*39 + 1 = 1^2 + 5^2 + 7^2 + 2^2 with 1 + 2*5 + 3*7 = 2^5.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    PQ[n_]:=PQ[n]=n>1&&IntegerQ[Log[2,n]];
    tab={};Do[r=0;Do[If[SQ[2n+1-x^2-y^2-z^2]&&PQ[x+2y+3z],r=r+1],{x,0,Sqrt[2n+1]},{y,Boole[x==0],Sqrt[2n+1-x^2]},{z,0,Sqrt[2n+1-x^2-y^2]}]; tab=Append[tab,r],{n,0,100}];Print[tab]