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.

A309513 Number of even parts in the partitions of n into 3 parts.

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 5, 4, 7, 8, 12, 12, 18, 18, 24, 24, 31, 32, 41, 40, 49, 50, 60, 60, 72, 72, 84, 84, 97, 98, 113, 112, 127, 128, 144, 144, 162, 162, 180, 180, 199, 200, 221, 220, 241, 242, 264, 264, 288, 288, 312, 312, 337, 338, 365, 364, 391, 392, 420, 420
Offset: 0

Views

Author

Wesley Ivan Hurt, Aug 05 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) |     0      1      2      5      4      7      8     12      ...
-----------------------------------------------------------------------
		

Crossrefs

Programs

  • Mathematica
    Table[Sum[Sum[Mod[i - 1, 2] + Mod[j - 1, 2] + Mod[n - i - j - 1, 2], {i, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 80}]

Formula

a(n) = Sum_{j=1..floor(n/3)} Sum_{i=j..floor((n-j)/2)} (((i-1) mod 2) + ((j-1) mod 2) + ((n-i-j-1) mod 2)). [Corrected by Georg Fischer, Mar 11 2025]
From Colin Barker, Aug 06 2019: (Start)
G.f.: x^4*(1 + x + 3*x^2 - x^3 + 2*x^4) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) + a(n-6) - a(n-7) - a(n-10) + a(n-11) for n>10. (End)
a(n) = 3*A069905(n) - A309511(n). - Ray Chandler, Mar 13 2025