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 1-3 of 3 results.

A351617 Number of ways to write n as 11^w + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 10 2022

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Let c be among 3, 4, 5, 7, 8. Then each positive integer n can be written as c^w + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.
This has been verified for all n = 1..3*10^5.

Examples

			a(6) = 1 with 6 = 11^0 + 0^2 + 2*1^2 + 3*1^2 + 0*1*1.
a(24) = 1 with 24 = 11^1 + 1^2 + 2*0^2 + 3*2^2 + 1*0*2.
a(71) = 1 with 71 = 11^0 + 4^2 + 2*3^2 + 3*2^2 + 4*3*2.
a(89) = 1 with 89 = 11^0 + 4^2 + 2*6^2 + 3*0^2 + 4*6*0.
a(107) = 1 with 107 = 11^1 + 8^2 + 2*4^2 + 3*0^2 + 8*4*0.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[4(n-11^w-2y^2-3z^2)+y^2*z^2],r=r+1],{w,0,Log[11,n]},{z,0,Sqrt[(n-11^w)/3]},{y,0,Sqrt[(n-11^w-3z^2)/2]}];tab=Append[tab,r],{n,1,100}];Print[tab]

A352259 Number of ways to write n as w^6 + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Mar 10 2022

Keywords

Comments

Conjecture 1: (i) a(n) > 0 for every n = 0,1,2,.... Moreover, 106, 744, 5469 and 331269 are the only nonnegative integers not in the set {w + x^2 + 2*y^2 + 3*z^2 + x*y*z: w = 0,1; x,y,z = 0,1,2,...}.
(ii) Let k be one of 4, 5, 6, 7. Then each n = 0,1,2,... can be written as 10*w^k + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.
(iii) Let c be among 1, 3, 4, 6, 7, and let k be 4 or 5. Then every n = 0,1,2,... can be written as c*w^k + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.
(iv) Each n = 0,1,2,... can be written as 9*w^4 + x^2 + 2*y^2 + 3*z^2 + x*y*z, where w,x,y,z are nonnegative integers.
Conjecture 2: Every n = 0,1,2,... can be written as 2*w^4 + 3*x^2 + y^2 + z^2 + x*y*z, where w,x,y,z are nonnegative integers.
We have verified Conjectures 1 and 2 for all n <= 10^5.

Examples

			a(24) = 1 with 24 = 0^6 + 4^2 + 2*2^2 + 3*0^2 + 4*2*0.
a(106) = 1 with 106 = 2^6 + 1^2 + 2*2^2 + 3*3^2 + 1*2*3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[4(n-w^6-2y^2-3z^2)+y^2*z^2],r=r+1],{w,0,n^(1/6)},{z,0,Sqrt[(n-w^6)/3]},{y,0,Sqrt[(n-w^6-3z^2)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]

A352356 Number of ways to write 12*n + 5 as 2*x^2 + 5*y^2 + 9*z^2 + x*y*z, where x, y and z are nonnegative integers.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 2, 4, 3, 3, 2, 5, 1, 3, 4, 3, 3, 7, 4, 2, 5, 3, 4, 3, 5, 5, 5, 6, 4, 4, 4, 2, 4, 5, 6, 3, 6, 5, 6, 5, 4, 5, 6, 7, 4, 4, 6, 4, 7, 6, 5, 3, 3, 8, 3, 7, 7, 4, 5, 7, 5, 6, 6, 8, 4, 1, 4, 7, 4, 8, 6, 5, 8, 9, 8, 4, 8, 3, 7, 4, 4, 12, 3, 4, 11, 8, 1, 6, 7, 5, 5, 8, 9, 5, 8, 12, 5, 6, 6, 6, 6
Offset: 0

Views

Author

Zhi-Wei Sun, Mar 13 2022

Keywords

Comments

Conjecture: For each n = 0,1,2,... we can write 12*n + 5 as 2*x^2 + 5*y^2 + 9*z^2 + x*y*z with x,y,z nonnegative integers.
This has been verified for all n = 0..10^6.
It seems that a(n) = 1 only for n = 0, 2, 5, 12, 65, 86, 155, 338, 21030.

Examples

			a(0) = 1 with 12*0 + 5 = 5 = 2*0^2 + 5*1^2 + 9*0^2 + 0*1*0.
a(2) = 1 with 12*2 + 5 = 29 = 2*0^2 + 5*2^2 + 9*1^2 + 0*2*1.
a(5) = 1 with 12*5 + 5 = 65 = 2*3^2 + 5*1^2 + 9*2^2 + 3*1*2.
a(12) = 1 with 12*12 + 5 = 149 = 2*0^2 + 5*1^2 + 9*4^2 + 0*1*4.
a(65) = 1 with 12*65 + 5 = 785 = 2*1^2 + 5*9^2 + 9*6^2 + 1*9*6.
a(86) = 1 with 12*86 + 5 = 1037 = 2*6^2 + 5*1^2 + 9*10^2 + 6*1*10.
a(155) = 1 with 12*155 + 5 = 1865 = 2*2^2 + 5*6^2 + 9*13^2 + 2*6*13.
a(338) = 1 with 12*338 + 5 = 4061 = 2*20^2 + 5*6^2 + 9*13^2 + 20*6*13.
a(21030) = 1 with 12*21030 + 5 = 252365 = 2*32^2 + 5*126^2 + 9*39^2 + 32*126*39.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[8(12n+5-5y^2-9z^2)+y^2*z^2]&&Mod[Sqrt[8(12n+5-5y^2-9z^2)+y^2*z^2]-y*z,4]==0,r=r+1],{y,0,Sqrt[(12n+5)/5]},{z,0,Sqrt[(12n+5-5y^2)/9]}];tab=Append[tab,r],{n,0,100}];Print[tab]
Showing 1-3 of 3 results.