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.

Showing 1-2 of 2 results.

A347827 Number of ways to write n as w^4 + x^4 + (y^2 + 23*z^2)/16, where w is zero or a power of two (including 2^0 = 1), and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 4, 4, 4, 3, 2, 2, 2, 4, 5, 2, 2, 5, 4, 1, 4, 8, 8, 8, 6, 2, 2, 3, 6, 12, 9, 5, 9, 9, 4, 2, 8, 8, 6, 5, 4, 6, 4, 4, 11, 11, 6, 7, 6, 3, 3, 5, 11, 8, 7, 3, 9, 10, 5, 11, 9, 3, 4, 5, 3, 2, 3, 7, 10, 10, 6, 2, 7, 5, 8, 10, 5, 9, 7, 6, 4, 1, 6, 9, 9, 9, 10, 7, 5, 4, 6, 5, 13, 11, 6, 5, 3, 6, 16, 11, 6, 15, 15, 7, 5
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 23 2022

Keywords

Comments

23-Conjecture: a(n) > 0 for all n = 0,1,2,....
This is stronger than the conjecture in A347824, and it has been verified for n up to 3*10^6. See also a similar conjecture in A347562.
It seems that a(n) = 1 only for n = 0, 15, 77, 231, 291, 437, 471, 1161, 1402, 4692, 7107, 9727.

Examples

			a(231) = 1 with 231 = 0^4 + 3^4 + (10^2 + 23*10^2)/16.
a(437) = 1 with 437 = 3^4 + 4^4 + (40^2 + 23*0^2)/16.
a(1402) = 1 with 1402 = 2^4 + 5^4 + (3^2 + 23*23^2)/16.
a(9727) = 1 with 9727 = 0^4 + 6^4 + (367^2 + 23*3^2)/16.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];tab={};Do[r=0;Do[If[w==0||IntegerQ[Log[2,w]],Do[If[SQ[16(n-w^4-x^4)-23z^2],r=r+1],{x,0,(n-w^4)^(1/4)},{z,0,Sqrt[16(n-w^4-x^4)/23]}]],{w,0,n^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A347824 Number of ways to write n as x^4 + y^4 + (z^2 + 23*w^2)/16, where x,y,z,w are nonnegative integers with x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 23 2022

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....
This has been verified for n up to 2*10^6. See also A347827 for a further refinement.
It seems that a(n) = 1 only for n = 0, 7, 15, 21, 22, 67, 77, 137, 252, 291, 437, 471, 477, 597, 1161, 4692, 7107.
For m = 32, 48, we also conjecture that every n = 0,1,2,... can be written as x^4 + y^4 + (z^2 + 23*w^2)/m, where x,y,z,w are nonnegative integers.

Examples

			a(7) = 1 with 7 = 0^4 + 1^4 + (2^2 + 23*2^2)/16.
a(15) = 1 with 15 = 1^4 + 1^4 + (1^2 + 23*3^2)/16.
a(67) = 1 with 67 = 1^4 + 2^4 + (15^2 + 23*5^2)/16.
a(477) = 1 with 477 = 0^4 + 2^4 + (27^2 + 23*17^2)/16.
a(597) = 1 with 597 = 2^4 + 4^4 + (5^2 + 23*15^2)/16.
a(1161) = 1 with 1161 = 2^4 + 2^4 + (89^2 + 23*21^2)/16.
a(4692) = 1 with 4692 = 2^4 + 5^4 + (248^2 + 23*12^2)/16.
a(7107) = 1 with 7107 = 1^4 + 5^4 + (239^2 + 23*45^2)/16.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[16(n-x^4-y^4)-23z^2],r=r+1],{x,0,(n/2)^(1/4)},{y,x,(n-x^4)^(1/4)},{z,0,Sqrt[16(n-x^4-y^4)/23]}];tab=Append[tab,r],{n,0,100}];Print[tab]
Showing 1-2 of 2 results.