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.

A300844 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x or 2*y or z is a square and (12*x)^2 + (21*y)^2 + (28*z)^2 is also a square.

Original entry on oeis.org

1, 4, 4, 2, 5, 6, 2, 2, 4, 5, 7, 1, 3, 7, 2, 3, 5, 7, 7, 2, 6, 1, 2, 2, 2, 11, 7, 3, 3, 8, 5, 1, 4, 5, 9, 4, 6, 8, 6, 4, 7, 9, 3, 3, 2, 9, 2, 1, 3, 6, 16, 5, 9, 7, 6, 5, 1, 5, 9, 4, 4, 7, 5, 5, 5, 17, 6, 4, 7, 3, 6, 3, 6, 11, 11, 4, 3, 1, 8, 2, 6
Offset: 0

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Author

Zhi-Wei Sun, Mar 13 2018

Keywords

Comments

Conjecture 1: a(n) > 0 for all n = 0,1,2,....
Conjecture 2: Any positive integer can be written as x^2 + y^2 + z^2 + w^2 with w a positive integer and x,y,z nonnegative integers such that x or y or z is a square and 144*x^2 + 505*y^2 + 720*z^2 is also a square.
By the author's 2017 JNT paper, each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w integers such that x (or 2*x) is a square.
In 2016, the author conjectured in A271510 that each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and y >= z such that (3*x)^2 + (4*y)^2 + (12*z)^2 is a square.
See also A300791 and A300792 for similar conjectures.

Examples

			a(11) = 1 since 11 = 0^2 + 1^2 + 1^2 + 3^2 with 0 = 0^2 and (12*0)^2 + (21*1)^2 + (28*1)^2 = 35^2.
a(56) = 1 since 56 = 4^2 + 6^2 + 2^2 + 0^2 with 4 = 2^2 and (12*4)^2 + (21*6)^2 + (28*2)^2 = 146^2.
a(77) = 1 since 77 = 4^2 + 0^2 + 5^2 + 6^2 with 4 = 2^2 and (12*4)^2 + (21*0)^2 + (28*5)^2 = 148^2.
a(184) = 1 since 184 = 12^2 + 2^2 + 0^2 + 6^2 with 0 = 0^2 and (12*12)^2 + (21*2)^2 + (28*0)^2 = 150^2.
a(599) = 1 since 599 = 21^2 + 11^2 + 1^2 + 6^2 with 1 = 1^2 and (12*21)^2 + (21*11)^2 + (28*1)^2 = 343^2.
a(7836) = 1 since 7836 = 38^2 + 18^2 + 68^2 + 38^2 with 2*18 = 6^2 and (12*38)^2 + (21*18)^2 + (28*68)^2 = 1994^2.
a(15096) = 1 since 15096 = 16^2 + 6^2 + 52^2 + 110^2 with 16 = 4^2 and (12*16)^2 + (21*6)^2 + (28*52)^2 = 1474^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[(SQ[x]||SQ[2y]||SQ[z])&&SQ[(12x)^2+(21y)^2+(28z)^2]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];tab=Append[tab,r],{n,0,80}]