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.

A350714 Least positive integer m such that m^12*n = x^4 + y^3 + z^2 for some nonnegative integers x,y,z.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 3, 1, 1, 1, 2, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 4, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 1
Offset: 0

Views

Author

Zhi-Wei Sun, Feb 02 2022

Keywords

Comments

4-3-2 Conjecture: a(n) exists for any nonnegative integer n. Equivalently, each nonnegative rational number can be written x^4 + y^3 + z^2 with x,y,z nonnegative rational numbers.
Note that m/n = (m*n^11)/n^12 for any positive integers m and n.
a(n) <= 4 for n <= 40000 with the only exception a(23710) = 5.
a(n) <= 4 for n = 77000..100000, and a(n) = 4 for n = 78367, 79479, 83494, 84694, 85979, 86822, 87395, 87814, 90047, 90278, 92891, 93715.
Qing-Hu Hou verified a(n) <= 4 for 40000 < n < 77000. - Zhi-Wei Sun Feb 04 2022
a(n) <= 5 for 10^5 < n <= 2*10^5, and a(n)=5 for n=107206, 117615, and 148079. - Qing-Hu Hou, Feb 05 2022

Examples

			a(6) = 1 with 1^12*6 = 1^4 + 1^3 + 2^2.
a(7) = 2 with 2^12*7 = 2^4 + 15^3 + 159^2.
a(75) = 4 with 4^12*75 = 122^4 + 1007^3 + 3951^2.
a(1140) = 3 with 3^12*1140 = 0^4 + 531^3 + 21357^2.
a(23710) = 5 with 5^12*23710 = 217^4 + 17897^3 + 232166^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[m=1;Label[bb];k=m^12;Do[If[SQ[k*n-x^4-y^3],tab=Append[tab,m];Goto[aa]],{x,0,(k*n)^(1/4)},{y,0,(k*n-x^4)^(1/3)}];m=m+1;Goto[bb];Label[aa],{n,0,100}];Print[tab]

A280153 Number of ways to write n as x^3 + 2*y^3 + z^2 + 4^k, where x is a positive integer and y,z,k are nonnegative integers.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Dec 27 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 3, 4, 7, 10, 17, 24, 36, 38, 52, 62, 115, 136, 185, 990.
(ii) Each positive integer n can be written as x^2 + 2*y^2 + z^3 + 8^k with x,y,z,k nonnegative integers.
(iii) Let a,b,c be positive integers with b <= c. Then any positive integer can be written as a*x^3 + b*y^2 + c*z^2 + 4^k with x,y,z,k nonnegative integers, if and only if (a,b,c) is among the following triples: (1,1,1), (1,1,2), (1,1,3), (1,1,5), (1,1,6), (1,2,3), (1,2,5), (2,1,1), (2,1,2), (2,1,3), (2,1,6), (4,1,2), (5,1,2), (8,1,2), (9,1,2).
We have verified that a(n) > 0 for all n = 2..10^6, and that part (ii) of the conjecture holds for all n = 1..10^6.
For any positive integer n, it is easy to see that at least one of n-1, n-8, n-64 is not of the form 4^k*(8m+7) with k and m nonnegative integers, thus, by the Gauss-Legendre theorem on sums of three squares, n = x^2 + y^2 + z^2 + 8^k for some nonnegative integers x,y,z and k < 3.
See also A280356 for a similar conjecture involving powers of two.

Examples

			a(7) = 1 since 7 = 1^3 + 2*1^3 + 0^2 + 4^1.
a(10) = 1 since 10 = 2^3 + 2*0^3 + 1^2 + 4^0.
a(17) = 1 since 17 = 1^3 + 2*0^3 + 0^2 + 4^2.
a(24) = 1 since 24 = 2^3 + 2*0^3 + 0^2 + 4^2.
a(36) = 1 since 36 = 2^3 + 2*1^3 + 5^2 + 4^0.
a(38) = 1 since 38 = 1^3 + 2*0^3 + 6^2 + 4^0.
a(52) = 1 since 52 = 3^3 + 2*0^3 + 3^2 + 4^2.
a(62) = 1 since 62 = 2^3 + 2*1^3 + 6^2 + 4^2.
a(115) = 1 since 115 = 2^3 + 2*3^3 + 7^2 + 4^1.
a(136) = 1 since 136 = 2^3 + 2*0^3 + 8^2 + 4^3.
a(185) = 1 since 185 = 3^3 + 2*3^3 + 10^2 + 4^1.
a(990) = 1 since 990 = 7^3 + 2*3^3 + 23^2 + 4^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-4^k-x^3-2y^3],r=r+1],{k,0,Log[4,n]},{x,1,(n-4^k)^(1/3)},{y,0,((n-4^k-x^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,1,80}]

A282426 Number of ways to write n as x^4 + 4*y^2 + z^2 + 3^k, where x,y,z are nonnegative integers and k is among 0,1,2,3,4.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Feb 14 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Each n = 1,2,3,... can be written as x^4 + y^2 + z^2 + 2^k with x,y,z nonnegative integers and k among 0,1,2,3,4, 5.
(iii) Any positive integer n can be written as 4*x^4 + y^3 + z^2 + 3^k with k,x,y,z nonnegative integers.
We have verified parts (i) and (ii)-(iii) for n up to 2*10^7 and 10^7 respectively.
See arXiv:1701.05868 for more such conjectures.

Examples

			a(1) = 1 since 1 = 0^4 + 4*0^2 + 0^2 + 3^0.
a(16) = 1 since 16 = 0^4 + 4*1^2 + 3^2 + 3^1.
a(475) = 1 since 475 = 3^4 + 4*6^2 + 13^2 + 3^4.
a(556) = 1 since 556 = 0^4 + 4*0^2 + 23^2 + 3^3.
a(8641) = 1 since 8641 = 9^4 + 4*21^2 + 17^2 + 3^3.
a(52696) = 1 since 52696 = 12^4 + 4*87^2 + 41^2 + 3^1.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-x^4-4y^2-3^k],r=r+1],{k,0,Min[4,Log[3,n]]},{x,0,(n-3^k)^(1/4)},{y,0,Sqrt[(n-3^k-x^4)/4]}];Print[n," ",r];Continue,{n,1,80}]

A297995 Number of ways to write n as 4*u + v^2 + x^3 + y^4 + 2*z^8, where u is 0 or 1, v is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 4, 4, 4, 5, 6, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 6, 5, 5, 5, 7, 6, 6, 5, 4, 5, 5, 6, 3, 5, 4, 4, 5, 5, 5, 5, 6, 5, 5, 5, 4, 4, 4, 2, 2, 2, 3, 4, 6, 6, 5, 7, 6, 7, 5, 6, 4, 4, 4, 3, 4, 3, 5, 4
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 10 2018

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 398, 496, 498, 500, 1507, 6419, 7843, 7983, 8688, 8947, 9175, 9251, 12923, 12976, 48381.
(ii) For any positive integers a,b,c,d,e and integers g,h,i,j,k greater than one, if each n = 0,1,2,... can be written as a*u^g + b*v^h + c*x^i + d*y^j + e*z^k with u,v,x,y,z nonnegative integers, then 1/g + 1/h + 1/i + 1/j + 1/k is greater than 1/2 + 1/3 + 1/4 + 1/8 = 29/24.

Examples

			a(500) = 1 since 500 = 4*0 + 22^2 + 0^3 + 2^4 + 2*0^8.
a(1507) = 1 since 1507 = 4*1 + 13^2 + 11^3 + 1^4 + 2*1^8.
a(9251) = 1 since 9251 = 4*0 + 91^2 + 7^3 + 5^4 + 2*1^8.
a(12923) = 1 since 12923 = 4*1 + 54^2 + 1^3 + 10^4 + 2*1^8.
a(12976) = 1 since 12976 = 4*0 + 92^2 + 15^3 + 5^4 + 2*2^8.
a(48381) = 1 since 48381 = 4*1 + 140^2 + 6^3 + 13^4 + 2*0^8.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]];
    Do[r=0;Do[If[SQ[n-4u-2z^8-y^4-x^3],r=r+1],{u,0,Boole[n>3]},{z,0,((n-4u)/2)^(1/8)},{y,0,(n-4u-2z^8)^(1/4)},{x,0,(n-4u-2z^8-y^4)^(1/3)}];Print[n," ",r],{n,1,80}]
Showing 1-4 of 4 results.