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.

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A271365 Number of ordered ways to write n as u^2 + v^3 + x^4 + y^5 + z^6, where u is a positive integer, and v, x, y, z are nonnegative integers.

Original entry on oeis.org

1, 4, 6, 5, 5, 6, 4, 1, 2, 7, 9, 6, 4, 3, 1, 1, 6, 12, 10, 4, 3, 3, 1, 1, 6, 12, 11, 7, 6, 4, 2, 4, 9, 12, 8, 5, 10, 12, 6, 2, 5, 9, 8, 8, 10, 6, 2, 3, 8, 13, 10, 8, 11, 8, 2, 1, 6, 10, 8, 7, 6, 2, 2, 7, 15, 20, 14, 9, 13, 11
Offset: 1

Views

Author

Zhi-Wei Sun, Apr 05 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 8, 15, 16, 23, 24, 56. Moreover, the only positive integers not represented by u^2+v^3+x^4+y^5 (u > 0 and v,x,y >= 0) are 8, 15, 23, 55, 62, 71, 471, 478, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471.
Note that 1/2+1/3+1/4+1/5+1/6 = 29/20 < 3/2.

Examples

			a(1) = 1 since 1 = 1^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(8) = 1 since 8 = 2^2 + 1^3 + 1^4 + 1^5 + 1^6.
a(15) = 1 since 15 = 2^2 + 2^3 + 1^4 + 1^5 + 1^6.
a(16) = 1 since 16 = 4^2 + 0^3 + 0^4 + 0^5 + 0^6.
a(23) = 1 since 23 = 2^2 + 1^3 + 2^4 + 1^5 + 1^6.
a(24) = 1 since 24 = 4^2 + 2^3 + 0^4 + 0^5 + 0^6.
a(56) = 1 since 56 = 4^2 + 2^3 + 0^4 + 2^5 + 0^6.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=n>0&&IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x1^6-x2^5-x3^4-x4^3],r=r+1],{x1,0,n^(1/6)},{x2,0,(n-x1^6)^(1/5)},{x3,0,(n-x1^6-x2^5)^(1/4)},{x4,0,(n-x1^6-x2^5-x3^4)^(1/3)}];Print[n," ",r];Continue,{n,1,70}]

A271381 Number of ordered ways to write n as u^2 + v^2 + x^3 + y^3, where u, v, x, y are nonnegative integers with 2 | u*v, u <= v and x <= y.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 06 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 except for n = 23, and a(n) = 1 only for n = 0, 3, 4, 7, 11, 12, 15, 19, 23, 30, 78, 203, 219.
(ii) Any natural number can be written as the sum of two squares and three fourth powers.

Examples

			a(3) = 1 since 3 = 0^2 + 1^2 + 1^3 + 1^3 with 0 even.
a(4) = 1 since 4 = 0^2 + 2^2 + 0^3 + 0^3 with 0 and 2 even.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^3 + 1^3 with 2 even.
a(11) = 1 since 11 = 0^2 + 3^2 + 1^3 + 1^3 with 0 even.
a(12) = 1 since 12 = 0^2 + 2^2 + 0^3 + 2^3 with 0 and 2 even.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^3 + 1^3 with 2 even.
a(19) = 1 since 19 = 1^2 + 4^2 + 1^3 + 1^3 with 4 even.
a(30) = 1 since 30 = 2^2 + 5^2 + 0^3 + 1^3 with 2 even.
a(78) = 1 since 78 = 2^2 + 3^2 + 1^3 + 4^3 with 2 even.
a(203) = 1 since 203 = 7^2 + 10^2 + 3^3 + 3^3 with 10 even.
a(219) = 1 since 219 = 8^2 + 8^2 + 3^3 + 4^3 with 8 even.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^3-y^3-u^2]&&(Mod[u*Sqrt[n-x^3-y^3-u^2],2]==0),r=r+1],{x,0,(n/2)^(1/3)},{y,x,(n-x^3)^(1/3)},{u,0,((n-x^3-y^3)/2)^(1/2)}];Print[n," ",r];Continue,{n,0,80}]
Previous Showing 21-22 of 22 results.