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.

A349661 Number of ways to write n as x^4 + y^2 + (z^2 + 4^w)/2 with x,y,z,w nonnegative integers.

Original entry on oeis.org

1, 3, 3, 2, 4, 5, 2, 2, 4, 5, 6, 4, 3, 6, 4, 1, 6, 7, 4, 6, 8, 5, 1, 4, 6, 10, 10, 3, 6, 10, 2, 3, 8, 6, 10, 10, 5, 6, 4, 5, 12, 14, 6, 5, 9, 6, 2, 3, 6, 12, 14, 7, 5, 8, 2, 7, 14, 6, 9, 9, 5, 9, 4, 2, 10, 15, 7, 7, 8, 7, 3, 5, 5, 7, 14, 5, 9, 9, 1, 4, 11, 8, 11, 13, 7, 13, 7, 2, 11, 17, 12, 8, 5, 6, 7, 5, 7, 11, 12, 8
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This has been verified for n up to 10^6.
As (x^2 + y^2)/2 = ((x+y)/2)^2 + ((x-y)/2)^2, the conjecture gives a new refinement of Lagrange's four-square theorem.
See also A350012 for a similar conjecture.

Examples

			a(1) = 1 with 1 = 0^4 + 0^2 + (1^2 + 4^0)/2.
a(23) = 1 with 23 = 1^4 + 3^2 + (5^2 + 4^0)/2.
a(79) = 1 with 79 = 1^4 + 2^2 + (12^2 + 4^1)/2.
a(1199) = 1 with 1199 = 5^4 + 18^2 + (22^2 + 4^2)/2.
a(3679) = 1 with 3679 = 5^4 + 2^2 + (78^2 + 4^2)/2.
a(6079) = 1 with 6079 = 3^4 + 42^2 + (92^2 + 4^1)/2.
a(33439) = 1 with 33439 = 1^4 + 175^2 + (75^2 + 4^0)/2.
a(50399) = 1 with 50399 = 13^4 + 135^2 + (85^2 + 4^0)/2.
a(207439) = 1 with 207439 = 1^4 + 142^2 + (612^2 + 4^1)/2.
		

Crossrefs

Programs

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

A350021 Number of ways to write n as w^4 + x^2 + y^2 + z^2 with x - y a power of two (including 2^0 = 1).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 08 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This is a new refinement of Lagrange's four-square theorem, and we have verified it for n up to 10^6.
If x - y = 2^k, then x^2 + y^2 = ((x+y)^2 + (2^k)^2)/2 and x + y >= 2^k. So the above conjecture implies the conjecture in A349661.
In his 2017 JNT paper, the author proved that each n = 0,1,2,... can be written as w^4 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers.
In his 2019 IJNT paper, the author proved that any positive integer can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z nonnegative integers such that x - y is a power of two (including 2^0 = 1).

Examples

			a(3) = 1 with 3 = 1^4 + 1^2 + 0^2 + 1^2 and 1 - 0 = 2^0.
a(4) = 1 with 4 = 0^4 + 2^2 + 0^2 + 0^2 and  2 - 0 = 2^1.
a(7) = 1 with 7 = 1^4 + 2^2 + 1^2 + 1^2 and 2 - 1 = 2^0.
a(8) = 1 with 8 = 0^4 + 2^2 + 0^2 + 2^2 and 2 - 0 = 2^1.
a(12) = 1 with 12 = 1^4 + 3^2 + 1^2 + 1^2 and 3 - 1 = 2^1.
a(19) = 1 with 19 = 0^4 + 3^2 + 1^2 + 3^2 and 3 - 1 = 2^1.
a(28) = 1 with 28 = 1^4 + 5^2 + 1^2 + 1^2 and 5 - 1 = 2^2.
a(47) = 1 with 47 = 1^4 + 3^2 + 1^2 + 6^2 and 3 - 1 = 2^1.
a(55) = 1 with 55 = 1^4 + 2^2 + 1^2 + 7^2 and 2 - 1 = 2^0.
a(88) = 1 with 88 = 0^4 + 6^2 + 4^2 + 6^2 and 6 - 4 = 2^1.
a(103) = 1 with 103 = 3^4 + 3^2 + 2^2 + 3^2 and 3 - 2 = 2^0.
a(193) = 1 with 193 = 2^4 + 8^2 + 7^2 + 8^2 and 8 - 7 = 2^0.
a(439) = 1 with 439 = 3^4 + 5^2 + 3^2 + 18^2 and 5 - 3 = 2^1.
		

Crossrefs

Programs

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