A323955 Regular triangle read by rows where T(n, k) is the number of set partitions of {1, ..., n} with no block containing k cyclically successive vertices, n >= 1, 2 <= k <= n + 1.

1, 1, 2, 1, 4, 5, 4, 10, 14, 15, 11, 36, 46, 51, 52, 41, 145, 184, 196, 202, 203, 162, 631, 806, 855, 869, 876, 877, 715, 3015, 3847, 4059, 4115, 4131, 4139, 4140, 3425, 15563, 19805, 20813, 21056, 21119, 21137, 21146, 21147, 17722, 86144, 109339, 114469

Offset: 1

Author

Gus Wiseman, Feb 10 2019

Keywords

• nonn
• tabl

Comments

Cyclically successive means 1 is a successor of n.

Examples

			Triangle begins:
    1
    1    2
    1    4    5
    4   10   14   15
   11   36   46   51   52
   41  145  184  196  202  203
  162  631  806  855  869  876  877
  715 3015 3847 4059 4115 4131 4139 4140
Row 4 counts the following partitions:
  {{13}{24}}      {{12}{34}}      {{1}{234}}      {{1234}}
  {{1}{24}{3}}    {{13}{24}}      {{12}{34}}      {{1}{234}}
  {{13}{2}{4}}    {{14}{23}}      {{123}{4}}      {{12}{34}}
  {{1}{2}{3}{4}}  {{1}{2}{34}}    {{124}{3}}      {{123}{4}}
                  {{1}{23}{4}}    {{13}{24}}      {{124}{3}}
                  {{12}{3}{4}}    {{134}{2}}      {{13}{24}}
                  {{1}{24}{3}}    {{14}{23}}      {{134}{2}}
                  {{13}{2}{4}}    {{1}{2}{34}}    {{14}{23}}
                  {{14}{2}{3}}    {{1}{23}{4}}    {{1}{2}{34}}
                  {{1}{2}{3}{4}}  {{12}{3}{4}}    {{1}{23}{4}}
                                  {{1}{24}{3}}    {{12}{3}{4}}
                                  {{13}{2}{4}}    {{1}{24}{3}}
                                  {{14}{2}{3}}    {{13}{2}{4}}
                                  {{1}{2}{3}{4}}  {{14}{2}{3}}
                                                  {{1}{2}{3}{4}}
		

Crossrefs

First column (k = 2) is A000296. Second column (k = 3) is A323949. Rightmost terms are A000110. Second to rightmost terms are A058692.

Cf. A000126, A000325, A001610, A001644, A169985, A306351, A306357, A323950, A323954.

Programs

Mathematica

spsu[,{}]:={{}};spsu[foo,set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,_}];
Table[Length[spsu[Select[Subsets[Range[n]],Select[Partition[Range[n],k,1,1],Function[ed,UnsameQ@@ed&&Complement[ed,#]=={}]]=={}&],Range[n]]],{n,7},{k,2,n+1}]