OFFSET
0,8
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
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.
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
EXAMPLE
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)
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 *)
CROSSREFS
KEYWORD
AUTHOR
Emeric Deutsch, Oct 28 2006
STATUS
approved