A124323 Triangle read by rows: T(n,k) is the number of partitions of an n-set having k singleton blocks (0<=k<=n).

1, 0, 1, 1, 0, 1, 1, 3, 0, 1, 4, 4, 6, 0, 1, 11, 20, 10, 10, 0, 1, 41, 66, 60, 20, 15, 0, 1, 162, 287, 231, 140, 35, 21, 0, 1, 715, 1296, 1148, 616, 280, 56, 28, 0, 1, 3425, 6435, 5832, 3444, 1386, 504, 84, 36, 0, 1, 17722, 34250, 32175, 19440, 8610, 2772, 840, 120, 45, 0, 1

Offset: 0

Author

Emeric Deutsch, Oct 28 2006

Keywords

• nonn
• tabl

Comments

Row sums are the Bell numbers (A000110). T(n,0)=A000296(n). T(n,k) = binomial(n,k)*T(n-k,0). Sum(k*T(n,k),k=0..n) = A052889(n) = n*B(n-1), where B(q) are the Bell numbers (A000110).

Exponential Riordan array [exp(exp(x)-1-x),x]. - Paul Barry, Apr 23 2009

Sum_{k=0..n} T(n,k)*2^k = A000110(n+1) is the number of binary relations on an n-set that are both symmetric and transitive. - Geoffrey Critzer, Jul 25 2014

Also the number of set partitions of {1, ..., n} with k cyclical adjacencies (successive elements in the same block, where 1 is a successor of n). Unlike A250104, we count {{1}} as having 1 cyclical adjacency. - Gus Wiseman, Feb 13 2019

Examples

			T(4,2)=6 because we have 12|3|4, 13|2|4, 14|2|3, 1|23|4, 1|24|3 and 1|2|34 (if we take {1,2,3,4} as our 4-set).
Triangle starts:
     1
     0    1
     1    0    1
     1    3    0    1
     4    4    6    0    1
    11   20   10   10    0    1
    41   66   60   20   15    0    1
   162  287  231  140   35   21    0    1
   715 1296 1148  616  280   56   28    0    1
  3425 6435 5832 3444 1386  504   84   36    0    1
From _Gus Wiseman_, Feb 13 2019: (Start)
Row n = 5 counts the following set partitions by number of singletons:
  {{1234}}    {{1}{234}}  {{1}{2}{34}}  {{1}{2}{3}{4}}
  {{12}{34}}  {{123}{4}}  {{1}{23}{4}}
  {{13}{24}}  {{124}{3}}  {{12}{3}{4}}
  {{14}{23}}  {{134}{2}}  {{1}{24}{3}}
                          {{13}{2}{4}}
                          {{14}{2}{3}}
... and the following set partitions by number of cyclical adjacencies:
  {{13}{24}}      {{1}{2}{34}}  {{1}{234}}  {{1234}}
  {{1}{24}{3}}    {{1}{23}{4}}  {{12}{34}}
  {{13}{2}{4}}    {{12}{3}{4}}  {{123}{4}}
  {{1}{2}{3}{4}}  {{14}{2}{3}}  {{124}{3}}
                                {{134}{2}}
                                {{14}{23}}
(End)
From _Paul Barry_, Apr 23 2009: (Start)
Production matrix is
0, 1,
1, 0, 1,
1, 2, 0, 1,
1, 3, 3, 0, 1,
1, 4, 6, 4, 0, 1,
1, 5, 10, 10, 5, 0, 1,
1, 6, 15, 20, 15, 6, 0, 1,
1, 7, 21, 35, 35, 21, 7, 0, 1,
1, 8, 28, 56, 70, 56, 28, 8, 0, 1 (End)
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened
• David Callan, On conjugates for set partitions and integer compositions, arXiv:math/0508052 [math.CO], 2005.
• T. Mansour, A. O. Munagi, Set partitions with circular successions, European Journal of Combinatorics, 42 (2014), 207-216.

Crossrefs

Cf. A000110, A052889, A124324.

A250104 is an essentially identical triangle, differing only in row 1.

For columns see A000296, A250105, A250106, A250107.

Cf. A000126, A001610, A032032, A052841, A066982, A080107, A169985, A187784, A324011, A324014, A324015.

Programs

Maple

G:=exp(exp(z)-1+(t-1)*z): Gser:=simplify(series(G,z=0,14)): for n from 0 to 11 do P[n]:=sort(n!*coeff(Gser,z,n)) od: for n from 0 to 11 do seq(coeff(P[n],t,k),k=0..n) od; # yields sequence in triangular form
# Program from R. J. Mathar, Jan 22 2015:
A124323 := proc(n,k)
    binomial(n,k)*A000296(n-k) ;
end proc:

Mathematica

Flatten[CoefficientList[Range[0,10]! CoefficientList[Series[Exp[x y] Exp[Exp[x] - x - 1], {x, 0,10}], x], y]] (* Geoffrey Critzer, Nov 24 2011 *)
sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
Table[Length[Select[sps[Range[n]],Count[#,{}]==k&]],{n,0,9},{k,0,n}] (* _Gus Wiseman, Feb 13 2019 *)

Formula

T(n,k) = binomial(n,k)*[(-1)^(n-k)+sum((-1)^(j-1)*B(n-k-j), j=1..n-k)], where B(q) are the Bell numbers (A000110).

E.g.f.: G(t,z) = exp(exp(z)-1+(t-1)*z).

G.f.: 1/(1-xy-x^2/(1-xy-x-2x^2/(1-xy-2x-3x^2/(1-xy-3x-4x^2/(1-... (continued fraction). - Paul Barry, Apr 23 2009