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.

A309683 Number of odd parts appearing among the smallest parts of the partitions of n into 3 parts.

Original entry on oeis.org

0, 0, 0, 1, 1, 2, 2, 3, 3, 5, 5, 7, 7, 9, 9, 12, 12, 15, 15, 18, 18, 22, 22, 26, 26, 30, 30, 35, 35, 40, 40, 45, 45, 51, 51, 57, 57, 63, 63, 70, 70, 77, 77, 84, 84, 92, 92, 100, 100, 108, 108, 117, 117, 126, 126, 135, 135, 145, 145, 155, 155, 165, 165, 176
Offset: 0

Views

Author

Wesley Ivan Hurt, Aug 12 2019

Keywords

Examples

			Figure 1: The partitions of n into 3 parts for n = 3, 4, ...
                                                          1+1+8
                                                   1+1+7  1+2+7
                                                   1+2+6  1+3+6
                                            1+1+6  1+3+5  1+4+5
                                     1+1+5  1+2+5  1+4+4  2+2+6
                              1+1+4  1+2+4  1+3+4  2+2+5  2+3+5
                       1+1+3  1+2+3  1+3+3  2+2+4  2+3+4  2+4+4
         1+1+1  1+1+2  1+2+2  2+2+2  2+2+3  2+3+3  3+3+3  3+3+4    ...
-----------------------------------------------------------------------
  n  |     3      4      5      6      7      8      9     10      ...
-----------------------------------------------------------------------
a(n) |     1      1      2      2      3      3      5      5      ...
-----------------------------------------------------------------------
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[Mod[j, 2], {i, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 80}]
    LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 0, 0, 1, 1, 2, 2, 3, 3}, 50] (* Wesley Ivan Hurt, Aug 28 2019 *)

Formula

a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} (j mod 2).
From Colin Barker, Aug 22 2019: (Start)
G.f.: x^3 / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) for n>8.
(End)