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.

A337483 Number of ordered triples of positive integers summing to n that are either weakly increasing or weakly decreasing.

Original entry on oeis.org

0, 0, 0, 1, 2, 4, 5, 8, 10, 13, 16, 20, 23, 28, 32, 37, 42, 48, 53, 60, 66, 73, 80, 88, 95, 104, 112, 121, 130, 140, 149, 160, 170, 181, 192, 204, 215, 228, 240, 253, 266, 280, 293, 308, 322, 337, 352, 368, 383, 400, 416, 433, 450, 468, 485, 504, 522, 541, 560
Offset: 0

Views

Author

Gus Wiseman, Sep 07 2020

Keywords

Examples

			The a(3) = 1 through a(8) = 10 triples:
  (1,1,1)  (1,1,2)  (1,1,3)  (1,1,4)  (1,1,5)  (1,1,6)
           (2,1,1)  (1,2,2)  (1,2,3)  (1,2,4)  (1,2,5)
                    (2,2,1)  (2,2,2)  (1,3,3)  (1,3,4)
                    (3,1,1)  (3,2,1)  (2,2,3)  (2,2,4)
                             (4,1,1)  (3,2,2)  (2,3,3)
                                      (3,3,1)  (3,3,2)
                                      (4,2,1)  (4,2,2)
                                      (5,1,1)  (4,3,1)
                                               (5,2,1)
                                               (6,1,1)
		

Crossrefs

A001399(n - 3) = A069905(n) = A211540(n + 2) counts the unordered case.
2*A001399(n - 6) = 2*A069905(n - 3) = 2*A211540(n - 1) counts the strict case.
A001399(n - 6) = A069905(n - 3) = A211540(n - 1) counts the strict unordered case.
A329398 counts these compositions of any length.
A218004 counts strictly increasing or weakly decreasing compositions.
A337484 counts neither strictly increasing nor strictly decreasing compositions.

Programs

  • Mathematica
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{3}],LessEqual@@#||GreaterEqual@@#&]],{n,0,30}]

Formula

a(n > 0) = 2*A001399(n - 3) - A079978(n).
From Colin Barker, Sep 08 2020: (Start)
G.f.: x^3*(1 + x + x^2 - x^3) / ((1 - x)^3*(1 + x)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6) for n>6. (End)
E.g.f.: (36 - 9*exp(-x) + exp(x)*(6*x^2 + 6*x - 19) - 8*exp(-x/2)*cos(sqrt(3)*x/2))/36. - Stefano Spezia, Apr 05 2023