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.

A349942 Number of ways to write n as a^4 + b^2 + (c^4 + d^2)/25 with a,b,c,d nonnegative integers.

Original entry on oeis.org

1, 4, 6, 4, 3, 5, 4, 1, 1, 4, 8, 7, 2, 4, 6, 2, 4, 12, 13, 6, 7, 9, 4, 1, 2, 11, 19, 11, 2, 10, 10, 2, 6, 12, 12, 9, 11, 9, 8, 4, 3, 16, 18, 7, 1, 13, 10, 1, 4, 7, 17, 15, 11, 11, 10, 2, 4, 12, 11, 9, 4, 13, 12, 5, 3, 15, 25, 10, 10, 12, 8, 3, 4, 9, 17, 17, 4, 14, 16, 3, 5, 20, 20, 14, 13, 12, 14, 4, 3, 12, 30, 22, 3, 12, 13, 4, 4, 16, 24, 20, 11
Offset: 0

Views

Author

Zhi-Wei Sun, Dec 05 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 7, 8, 23, 44, 47).
We have verified this for n up to 3*10^5.
As m/n = (m*n^3)/n^4 for any nonnegative integers m and n > 0, the conjecture implies that each nonnegative rational number can be written as x^4 + 25*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
See also A349943 for similar conjectures.

Examples

			a(0) = 1 with 0 = 0^4 + 0^2 + (0^4 + 0^2)/25.
a(7) = 1 with 7 = 1^4 + 2^2 + (1^4 + 7^2)/25.
a(8) = 1 with 8 = 0^4 + 2^2 + (0^4 + 10^2)/25.
a(23) = 1 with 23 = 1^4 + 3^2 + (1^4 + 18^2)/25.
a(44) = 1 with 44 = 1^4 + 3^2 + (5^4 + 15^2)/25.
a(47) = 1 with 47 = 1^4 + 6^2 + (3^4 + 13^2)/25.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[25(n-x^4-y^2)-z^4],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,(25(n-x^4-y^2))^(1/4)}];tab=Append[tab,r],{n,0,100}];Print[tab]

A350860 Number of ways to write n as w^4 + (x^4 + y^2 + z^2)/81, where w,x,y,z are nonnegative integers with y <= z.

Original entry on oeis.org

1, 6, 8, 5, 4, 7, 8, 3, 2, 5, 10, 10, 3, 4, 6, 2, 6, 14, 12, 10, 8, 19, 18, 4, 4, 11, 23, 15, 2, 7, 8, 3, 8, 13, 19, 18, 15, 21, 13, 4, 4, 17, 24, 10, 3, 6, 13, 7, 5, 10, 21, 23, 14, 15, 13, 3, 5, 17, 15, 12, 4, 13, 21, 4, 4, 13, 36, 25, 14, 20, 14, 3, 6, 13, 19, 18, 5, 14, 11, 3, 7, 32, 45, 19, 17, 22, 21, 8, 4, 17, 31
Offset: 0

Views

Author

Zhi-Wei Sun, Jan 19 2022

Keywords

Comments

Conjecture 1: a(n) > 0 for any nonnegative integer n. Also, every n = 0,1,2,... can be written as 4*w^4 + (x^4 + y^2 + z^2)/81 with w,x,y,z integers.
This implies that each nonnegative rational number can be written as w^4 + x^4 + y^2 + z^2 (or 4*w^4 + x^4 + y^2 + z^2) with w,x,y,z rational numbers.
Conjecture 2: For any positive integer c, there is a positive integer m such that every n = 0,1,2,... can be written as w^4 + (c^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive integer c each nonnegative rational number can be written as w^4 + c^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.
Conjecture 3: For any positive odd number d, there is a positive integer m such that every n = 0,1,2,... can be written as 4*w^4 + (d^2*x^4 + y^2 + z^2)/m^4 with w,x,y,z integers.
This implies that for any positive odd number d each nonnegative rational number can be written as 4*w^4 + d^2*x^4 + y^2 + z^2 with w,x,y,z rational numbers.

Examples

			a(8) = 2 with 8 = 0^4 + (0^4 + 18^2 + 18^2)/81 = 0^4 + (4^4 + 14^2 + 14^2)/81.
a(15) = 2 with 15 = 1^4 + (3^4 + 18^2 + 27^2)/81 = 1^4 + (5^4 + 5^2 + 22^2)/81.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[81(n-w^4)-x^4-y^2],r=r+1],{w,0,n^(1/4)},{x,0,3(n-w^4)^(1/4)},{y,0,Sqrt[(81(n-w^4)-x^4)/2]}];tab=Append[tab,r],{n,0,90}];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]

A351206 Least positive integer m such that n = x^4 + (y^4 + z^4 + 7*w^2)/m^4 for some nonnegative integers x,y,z,w with y <= z.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 2, 4, 3, 1, 1, 1, 6, 2, 2, 10, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 3, 4, 2, 4, 10, 1, 1, 2, 4, 2, 1, 1, 1, 3, 1, 3, 2, 2, 2, 2, 2, 1, 3, 4, 2, 2, 1, 1, 3, 1, 1, 1, 1, 2, 4, 2, 2, 3, 8, 10, 2, 2, 1, 10, 5, 1, 1, 1, 1, 1, 4, 2, 2, 8, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 4, 4
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 04 2022

Keywords

Comments

Conjecture: a(n) exists for any nonnegative integer n.
This implies that each nonnegative rational number can be written as 7*w^2 + x^4 + y^4 + z^4 with w,x,y,z rational numbers.

Examples

			a(6) = 2 with 6 = 1^4 + (1^4 + 2^4 + 7*3^2)/2^4.
a(19) = 6 with 19 = 0^4 + (1^4 + 4^4 + 7*59^2)/6^4.
a(22) = 10 with 22 = 2^4 + (2^4 + 13^4 + 7*67^2)/10^4.
a(5797) = 20 with 5797 = 0^4 + (81^4 + 164^4 + 7*4797^2)/20^4.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[m=1; Label[bb]; k=m^4; Do[If[SQ[(k*(n-x^4)-y^4-z^4)/7], tab=Append[tab,m]; Goto[aa]],  {x, 0, n^(1/4)}, {y, 0, (k*(n-x^4)/2)^(1/4)},{z,y,(k*(n-x^4)-y^4)^(1/4)}]; m=m+1; Goto[bb]; Label[aa], {n,0,100}];Print[tab]
Showing 1-4 of 4 results.