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.

A325191 Number of integer partitions of n such that the difference between the length of the minimal triangular partition containing and the maximal triangular partition contained in the Young diagram is 1.

Original entry on oeis.org

0, 0, 2, 0, 3, 3, 0, 4, 6, 4, 0, 5, 10, 10, 5, 0, 6, 15, 20, 15, 6, 0, 7, 21, 35, 35, 21, 7, 0, 8, 28, 56, 70, 56, 28, 8, 0, 9, 36, 84, 126, 126, 84, 36, 9, 0, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0, 11, 55, 165, 330, 462
Offset: 0

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Author

Gus Wiseman, Apr 11 2019

Keywords

Comments

The Heinz numbers of these partitions are given by A325196.
Under the Bulgarian solitaire step, these partitions form cycles of length >= 2. Length >= 2 means not the length=1 self-loop which occurs from the triangular partition when n is a triangular number. See A074909 for self-loops included. - Kevin Ryde, Sep 27 2019

Examples

			The a(2) = 2 through a(12) = 10 partitions (empty columns not shown):
  (2)   (22)   (32)   (322)   (332)   (432)   (4322)   (4332)
  (11)  (31)   (221)  (331)   (422)   (3321)  (4331)   (4422)
        (211)  (311)  (421)   (431)   (4221)  (4421)   (4431)
                      (3211)  (3221)  (4311)  (5321)   (5322)
                              (3311)          (43211)  (5331)
                              (4211)                   (5421)
                                                       (43221)
                                                       (43311)
                                                       (44211)
                                                       (53211)
		

Crossrefs

Programs

  • Mathematica
    otb[ptn_]:=Min@@MapIndexed[#1+#2[[1]]-1&,Append[ptn,0]];
    otbmax[ptn_]:=Max@@MapIndexed[#1+#2[[1]]-1&,Append[ptn,0]];
    Table[Length[Select[IntegerPartitions[n],otb[#]+1==otbmax[#]&]],{n,0,30}]
  • PARI
    a(n) = my(t=ceil(sqrtint(8*n+1)/2), r=n-t*(t-1)/2); if(r==0,0, binomial(t,r)); \\ Kevin Ryde, Sep 27 2019

Formula

Positions of zeros are A000217 = n * (n + 1) / 2.
a(n) = A074909(n) - A010054(n). - Kevin Ryde, Sep 27 2019