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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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).

Links

Crossrefs

Cf. A000041, A000217, A000326, A005449, A280386, A280444, A280472.

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}]

A280472 Number of ways to write n as the sum of an octagonal number (A000567), a second octagonal number (A045944), and a strict partition number (A000009).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 04 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) lim_n a(n)/(log n)^2 = 1/Pi^2.
On 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.
See also A280455 for a similar conjecture of the author involving the partition function.

Examples

			a(1) = 1 since 1 = 0*(3*0-2) + 0*(3*0+2) + A000009(2).
a(50) = 2 since 50 = 4*(3*4-2) + 1*(3*1+2) + A000009(7) = 4*(3*4-2) + 0*(3*0+2) + A000009(10).
a(1399) = 1 since 1399 = 1*(3*1-2) + 18*(3*18+2) + A000009(32).

Links

Crossrefs

Cf. A000009, A000567, A045944, A280386, A280455.

Programs

  • Mathematica
    Oct[n_]:=Oct[n]=IntegerQ[Sqrt[3n+1]]&&Mod[Sqrt[3n+1],3]==1;
q[n_]:=q[n]=PartitionsQ[n];
Do[r=0;m=2;Label[bb];If[q[m]>n,Goto[cc]];Do[If[Oct[n-q[m]-x(3x-2)],r=r+1],{x,0,(Sqrt[3(n-q[m])+1]+1)/3}];m=m+If[m<3,2,1];Goto[bb];Label[cc];Print[n," ",r];Continue,{n,1,80}]
Showing 1-2 of 2 results.

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