A306418 Regular triangle read by rows where T(n, k) is the number of set partitions of {1, ..., n} requiring k steps of removing singletons and cyclical adjacency initiators until reaching a fixed point, n >= 0, 0 <= k <= n.
1, 0, 1, 0, 2, 0, 0, 2, 3, 0, 1, 2, 12, 0, 0, 0, 12, 35, 5, 0, 0, 5, 56, 100, 42, 0, 0, 0, 14, 282, 343, 231, 7, 0, 0, 0, 66, 1406, 1476, 1088, 104, 0, 0, 0, 0, 307, 7592, 7383, 4929, 909, 27, 0, 0, 0, 0, 1554, 44227, 40514, 22950, 6240, 470, 20, 0, 0, 0, 0
Offset: 0
Examples
Triangle begins: 1 0 1 0 2 0 0 2 3 0 1 2 12 0 0 0 12 35 5 0 0 5 56 100 42 0 0 0 14 282 343 231 7 0 0 0 66 1406 1476 1088 104 0 0 0 0 307 7592 7383 4929 909 27 0 0 0 0
Links
- David Callan, On conjugates for set partitions and integer compositions, arXiv:math/0508052 [math.CO], 2005.
Crossrefs
Programs
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Mathematica
sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}]; qbj[stn_]:=With[{ini=Join@@Table[Select[s,If[#==Max@@Max@@@stn,MemberQ[s,First[Union@@stn]],MemberQ[s,(Union@@stn)[[Position[Union@@stn,#][[1,1]]+1]]]]&],{s,stn}],sng=Join@@Select[stn,Length[#]==1&]},DeleteCases[Table[Complement[s,Union[sng,ini]],{s,stn}],{}]]; Table[Length[Select[sps[Range[n]],Length[FixedPointList[qbj,#]]-2==k&]],{n,0,8},{k,0,n}]
Comments