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 1-4 of 4 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]

A349945 Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/5 with a,b,c,d nonnegative integers.

Original entry on oeis.org

1, 3, 4, 3, 3, 6, 7, 3, 1, 4, 5, 4, 2, 3, 8, 5, 3, 9, 10, 6, 7, 11, 10, 3, 2, 6, 8, 9, 3, 9, 16, 5, 4, 11, 9, 7, 9, 9, 12, 7, 2, 8, 11, 7, 2, 11, 14, 4, 3, 10, 10, 9, 8, 9, 21, 9, 3, 9, 5, 7, 4, 10, 17, 8, 3, 15, 15, 9, 9, 16, 20, 5, 3, 5, 7, 11, 3, 11, 18, 4, 6, 22, 18, 11, 14, 15, 19, 10, 2, 9, 16, 10, 3, 9, 16, 11, 7, 19, 16, 13, 12
Offset: 0

Views

Author

Zhi-Wei Sun, Dec 06 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 2^(4*k+3) (k = 0,1,2,...).
This has been verified for all n <= 10^5.
Conjecture 2: Each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + w^2)/5 with x,y,z,w nonnegative integers, provided that (a,b,c) is among the four triples (1,2,4), (2,1,1), (6,1,1), (6,1,6).
See also A349942 for a similar conjecture.
Via a computer search, we have found many tuples (a,b,c,d,m) of positive integers (such as (1,1,4,2,3), (4,1,1,2,3) and (1,1,19,1,4900)) for which we guess that each n = 0,1,2,... can be written as a*x^4 + b*y^2 + (c*z^4 + d*w^2)/m with x,y,z,w nonnegative integers.

Examples

			a(8) = 1 with 8 = 0^4 + 2^2 + (2^4 + 2^2)/5.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[5(n-x^4-y^2)-z^4],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,(5(n-x^4-y^2))^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A349992 Number of ways to write n as x^4 + y^2 + (z^2 + 2*4^w)/3, where x, y, z are nonnegative integers, and w is 0 or 1.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 08 2021

Keywords

Comments

Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 30, 64, 80, 302, 350, 472, 480, 847, 3497, 13582, 25630, 38064.
This has been verified for n up to 10^6.
Conjecture 2: If (a,b,c,m) is one of the ordered tuples (1,1,11,12), (1,1,11,60), (1,1,14,15), (1,1,23,24), (1,1,23,32), (1,1,23,48), (1,2,23,96), (2,1,11,60), (2,1,23,24), (2,1,23,48), (4,1,23,48), then each n = 1 2,3,... can be written as a*x^4 + b*y^2 + (z^2 + c*4^w)/m, where x,y,z are nonnegative integers, and w is 0 or 1.
We have verified Conjecture 2 for n up to 2*10^5.

Examples

			a(30) = 1 with 30 = 1^4 + 5^2 + (2^2 + 2*4)/3.
a(480) = 1 with 480 = 1^4 + 14^2 + (29^2 + 2*4)/3.
a(847) = 1 with 847 = 0^4 + 29^2 + (4^2 + 2*4^0)/3.
a(3497) = 1 with 3497 = 4^4 + 48^2 + (53^2 + 2*4^0)/3.
a(13582) = 1 with 13582 = 9^4 + 28^2 + (53^2 + 2*4^0)/3.
a(25630) = 1 with 25630 = 5^4 + 158^2 + (11^2 + 2*4^0)/3.
a(38064) = 1 with 38064 = 3^4 + 157^2 + (200^2 + 2*4^0)/3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[3(n-x^4-y^2)-2*4^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]

A350012 Number of ways to write n as 4*x^4 + y^2 + (z^2 + 4^w)/2 with x,y,z,w nonnegative integers.

Original entry on oeis.org

1, 2, 1, 1, 4, 4, 1, 3, 5, 5, 3, 3, 4, 7, 3, 2, 6, 5, 2, 4, 6, 2, 2, 5, 4, 6, 2, 2, 6, 7, 2, 2, 6, 5, 5, 4, 3, 7, 5, 5, 8, 6, 2, 6, 9, 4, 2, 4, 5, 8, 3, 3, 5, 8, 3, 6, 5, 3, 6, 4, 6, 5, 6, 1, 10, 9, 2, 6, 11, 8, 1, 7, 5, 11, 6, 4, 7, 10, 3, 6, 10, 4, 8, 8, 6, 8, 6, 5, 11, 13, 5, 1, 11, 8, 3, 4, 4, 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 since (x^2 + y^2)/2 = ((x+y)/2)^2 + ((x-y)/2)^2. We have verified the conjecture for n up to 10^6.
See also A349661 for a similar conjecture.
We also have some other conjectures of such a type.

Examples

			a(1) = 4*0^4 + 0^2 + (1^2 + 4^0)/2.
a(3) = 1 with 3 = 4*0^4 + 1^2 + (0^2 + 4)/2.
a(4) = 1 with 4 = 4*0^4 + 0^2 + (2^2 + 4)/2.
a(7) = 1 with 7 = 4*1^4 + 1^2 + (0^2 + 4)/2.
a(71) = 1 with 71 = 4*1^4 + 3^2 + (10^2 + 4^2)/2.
a(92) = 1 with 92 = 4*1^4 + 6^2 + (10^2 + 4)/2.
a(167) = 1 with 167 = 4*1^4 + 9^2 + (10^2 + 4^3)/2.
a(271) = 1 with 271 = 4*1^4 + 11^2 + (6^2 + 4^4)/2.
a(316) = 1 with 316 = 4*1^4 + 4^2 + (24^2 + 4^2)/2.
a(4796) = 1 with 4796 = 4*5^4 + 36^2 + (44^2 + 4^3)/2.
a(14716) = 1 with 14716 = 4*5^4 + 4^2 + (156^2 + 4^3)/2.
a(24316) = 1 with 24316 = 4*3^4 + 84^2 + (184^2 + 4^2)/2.
		

Crossrefs

Programs

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