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.
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
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)
Links
- FindStat, St000380: Half the perimeter of the largest rectangle that fits inside the diagram of an integer partition
- FindStat, St000384: The maximal part of the shifted composition of an integer partition
- FindStat, St000783: The maximal number of occurrences of a colour in a proper colouring of a Ferrers diagram
- Eric Weisstein's World of Mathematics, Graph Distance
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.
Comments