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 1-2 of 2 results.

A303428 Number of ways to write n as x*(3*x-2) + y*(3*y-2) + 3^u + 3^v, where x,y,u,v are integers with x <= y and 0 <= u <= v.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Apr 23 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1. Moreover, any integer n > 1 can be written as x*(3*x+2) + y*(3*y+2) + 3^z + 3^w, where x is an integer and y,z,w are nonnegative integers.
a(n) > 0 for all n = 2..3*10^8. Those x*(3*x-2) with x integral are called generalized octagonal numbers (A001082). 76683391 is the least integer n > 1 not representable as the sum of two generalized octagonal numbers and two powers of 2.
See also A303389, A303401 and A303432 for similar conjectures.

Examples

			a(2) = 1 with 2 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^0.
a(3) = 1 with 3 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^0.
a(4) = 2 with 4 = 1*(3*1-2) + 1*(3*1-2) + 3^0 + 3^0 = 0*(3*0-2) + 0*(3*0-2) + 3^0 + 3^1.
a(5) = 1 with 5 = 0*(3*0-2) + 1*(3*1-2) + 3^0 + 3^1.
a(9) = 1 with 9 = (-1)*(3*(-1)-2) + 0*(3*0-2) + 3^0 + 3^1.
a(4360) = 4 with 4360 = (-35)*(3*(-35)-2) + (-13)*(3*(-13)-2) + 3^0 + 3^4 = (-37)*(3*(-37)-2) + (-7)*(3*(-7)-2) + 3^2 + 3^2 = (-27)*(3*(-27)-2) + (-23)*(3*(-23)-2) + 3^5 + 3^5 = (-25)*(3*(-25)-2) + (-1)*(3*(-1)-2) + 3^5 + 3^7.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    f[n_]:=f[n]=FactorInteger[n];
    g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
    QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
    tab={};Do[r=0;Do[If[QQ[3(n-3^j-3^k)+2],Do[If[SQ[3(n-3^j-3^k-x(3x-2))+1],r=r+1],{x,-Floor[(Sqrt[3(n-3^j-3^k)/2+1]-1)/3],(Sqrt[3(n-3^j-3^k)/2+1]+1)/3}]],
    {j,0,Log[3,n/2]},{k,j,Log[3,n-3^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A280455 Number of ways to write n as x*(3x-1)/2 + y*(3y+1)/2 + p(z), where x,y,z are nonnegative integers with z > 0, and p(.) is the partition function given by A000041.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 03 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) lim_n a(n)/(log n)^2 = 1/Pi^2.
This is similar to the author's conjecture in A280386. At the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified part (i) of the above conjecture for n up to 10^9.
We also have some other similar conjectures. For example, we conjecture that any positive integer can be expressed as the sum of two triangular numbers and a partition number.
As the main term of log p(n) is Pi*sqrt(2n/3), the partition function p(n) eventually grows faster than any polynomial.
See also A280472 for a similar conjecture.

Examples

			a(1) = 1 since 1 = 0*(3*0-1)/2 + 0*(3*0+1)/2 + p(1).
a(2) = 2 since 2 = 1*(3*1-1)/2 + 0*(3*0+1)/2 + p(1) = 0*(3*0-1)/2 + 0*(3*0+1)/2 + p(2).
a(2771) = 1 since 2771 = 35*(3*35-1)/2 + 25*(3*25+1)/2 + p(1).
a(9426) = 1 since 9426 = 4*(3*4-1)/2 + 79*(3*79+1)/2 + p(3).
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    p[n_]:=p[n]=PartitionsP[n];
    Pen[n_]:=Pen[n]=SQ[24n+1]&&Mod[Sqrt[24n+1],6]==1;
    Do[r=0;m=1;Label[bb];If[p[m]>n,Goto[cc]];Do[If[Pen[n-p[m]-x(3x-1)/2],r=r+1],{x,0,(Sqrt[24(n-p[m])+1]+1)/6}];m=m+1;Goto[bb];Label[cc];Print[n," ",r];Label[aa];Continue,{n,1,80}]
Showing 1-2 of 2 results.