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.

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}]

A280444 Least positive integer m such that n - p(m) = x*(3x-1)/2 + y*(3y+1)/2 for some nonnegative integers x and y, or 0 if no such m exists, where p(.) is the partition function given by A000041.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 03 2017

Keywords

Comments

The conjecture in A280455 asserts that a(n) > 0 for all n > 0.

Examples

			a(12) = 4 since 12 - p(4) = 12 - 5 = 7 = 0*(3*0-1)/2 + 2*(3*2+1)/2.
a(35) = 6 since 35 - p(6) = 35 - 11 = 24 = 4*(3*4-1)/2 + 1*(3*1+1)/2.
a(4327) = 15 since 4327 - p(15) = 4327 - 176 = 4151 = 16*(3*16-1)/2 + 50*(3*50+1)/2.

Links

Crossrefs

Cf. A000041, A000326, A005449, A280455.

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[m=1;Label[bb];If[p[m]>n,Goto[cc]];Do[If[Pen[n-p[m]-x(3x-1)/2],Print[n," ",m];Goto[aa]],{x,0,(Sqrt[24(n-p[m])+1]+1)/6}];m=m+1;Goto[bb];Label[cc];Print[n," ",0];Label[aa];Continue,{n,1,80}]

A281826 Number of ways to write n as x^3 + 2*y^3 + 3*z^3 + p(k), where x,y,z are nonnegative integers, k is a positive integer, and p(.) is the partition function given by A000041.

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, Jan 31 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer n can be written as x^3 + 2*y^3 + 4*z^3 + p(k) with x,y,z nonnegative integers and k a positive integer.
(iii) For each c = 3, 4, any positive integer n can be written as x^3 + 2*y^3 + c*z^3 + A000009(k) with x,y,z nonnegative and k a positive integer.
We have verified the conjecture for n up to 1.3*10^6.
On the author's request, Prof. Qing-Hu Hou at Tianjin Univ. has verified all the three parts of the conjecture for n up to 10^9. - Zhi-Wei Sun, Feb 06 2017

Examples

			a(1) = 1 since 1 = 0^3 + 2*0^3 + 3*0^3 + p(1).
a(2) = 2 since 2 = 1^3 + 2*0^3 + 3*0^3 + p(1) = 0^3 + 2*0^3 + 3*0^3 + p(2).
a(75) = 3 since 75 = 4^3 + 2*0^3 + 3*0^3 + p(6) = 3^3 + 2*1^3 + 3*2^3 + p(8) = 0^3 + 2*2^3 + 3*1^3 + p(11).

Links

Crossrefs

Cf. A000009, A000041, A000578, A280455.

Programs

  • Mathematica
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
p[n_]:=p[n]=PartitionsP[n]
Do[r=0;Do[If[p[k]>n,Goto[bb]];Do[If[CQ[n-p[k]-3x^3-2y^3],r=r+1],{x,0,((n-p[k])/3)^(1/3)},{y,0,((n-p[k]-3x^3)/2)^(1/3)}];Continue,{k,1,n}];Label[bb];Print[n," ",r];Continue,{n,1,80}]
Showing 1-3 of 3 results.

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