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.

A299924 Number of ways to write n^2 as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x + 2*y + 3*z is a power of 4 (including 4^0 = 1).

Original entry on oeis.org

1, 1, 1, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 3, 1, 1, 11, 4, 6, 7, 7, 8, 4, 4, 6, 14, 4, 6, 17, 10, 1, 10, 6, 10, 7, 4, 4, 16, 2, 3, 10, 2, 1, 9, 6, 3, 2, 1, 5, 2, 3, 7, 9, 3, 1, 6, 2, 3, 7, 1, 4, 4, 1, 3, 4, 3, 1, 13, 20
Offset: 1

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Author

Zhi-Wei Sun, Feb 21 2018

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 3, 7, 13, 49, 61, 4^k*m (k = 0,1,2,... and m = 1, 2, 11, 14, 17).
(ii) Let a,b,c,d be nonnegative integers with a >= b >= c >= d, b positive, and gcd(a,b,c,d) not divisible by 4. Then, any positive square can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*x + b*y + c*z + d*w = 4^k for some k = 0,1,2,..., if and only if d = 0 and (a,b,c) is among the following ordered triples: (3,2,1), (2,1,0), (3,1,0), (4,2,0), (8,1,0), (15,1,0) and (16,2,0).
By Theorem 1.1(i) of arXiv:1701.05868, any positive square can be written as (4^k)^2 + x^2 + y^2 + z^2 with k,x,y,z nonnegative integers.
We have verified that a(n) > 0 for all n = 1..50000.

Examples

			a(1) = 1 since 1^2 = 1^2 + 0^2 + 0^2 + 0^2 with 1 + 2*0 + 3*0 = 4^0.
a(2) = 1 since 2^2 = 0^2 + 2^2 + 0^2 + 0^2 with 0 + 2*2 + 3*0 = 4.
a(3) = 1 since 3^2 = 2^2 + 1^2 + 0^2 + 2^2 with 2 + 2*1 + 3*0 = 4.
a(7) = 1 since 7^2 = 2^2 + 4^2 + 2^2 + 5^2 with 2 + 2*4 + 3*2 = 4^2.
a(11) = 1 since 11^2 = 2^2 + 1^2 + 4^2 + 10^2 with 2 + 2*1 + 4*3 = 4^2.
a(13) = 1 since 13^2 = 8^2 + 1^2 + 2^2 + 10^2 with 8 + 2*1 + 3*2 = 4^2.
a(14) = 1 since 14^2 = 4^2 + 6^2 + 0^2 + 12^2 with 4 + 2*6 + 3*0 = 4^2.
a(17) = 1 since 17^2 = 0^2 + 8^2 + 0^2 + 15^2 with 0 + 2*8 + 4*0 = 4^2.
a(49) = 1 since 49^2 = 22^2 + 3^2 + 12^2 + 42^2 with 22 + 2*3 + 3*12 = 4^3.
a(61) = 1 since 61^2 = 6^2 + 20^2 + 6^2 + 57^2 with 6 + 2*20 + 3*6 = 4^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
    Do[r=0;Do[If[SQ[n^2-x^2-y^2-z^2]&&Pow[x+2y+3z],r=r+1],{x,0,n},{y,0,Sqrt[n^2-x^2]},{z,0,Sqrt[n^2-x^2-y^2]}];Print[n," ",r],{n,1,70}]