

A056857


Triangle read by rows: T(n,c) = number of successive equalities in set partitions of n.


33



1, 1, 1, 2, 2, 1, 5, 6, 3, 1, 15, 20, 12, 4, 1, 52, 75, 50, 20, 5, 1, 203, 312, 225, 100, 30, 6, 1, 877, 1421, 1092, 525, 175, 42, 7, 1, 4140, 7016, 5684, 2912, 1050, 280, 56, 8, 1, 21147, 37260, 31572, 17052, 6552, 1890, 420, 72, 9, 1, 115975, 211470, 186300, 105240, 42630, 13104, 3150, 600, 90, 10, 1
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OFFSET

1,4


COMMENTS

Number of successive equalities s_i = s_{i+1} in a set partition {s_1, ..., s_n} of {1, ..., n}, where s_i is the subset containing i, s(1) = 1 and s(i) <= 1 + max of previous s(j)'s.
T(n,c) = number of set partitions of the set {1,2,...,n} in which the size of the block containing the element 1 is k+1. Example: T(4,2)=3 because we have 1234, 1243 and 1342.  Emeric Deutsch, Nov 10 2006
Let P be the lowertriangular Pascalmatrix (A007318), Then this is exp(P) / exp(1).  Gottfried Helms, Mar 30 2007. [This comment was erroneously attached to A011971, but really belongs here.  N. J. A. Sloane, May 02 2015]
From David Pasino (davepasino(AT)yahoo.com), Apr 15 2009: (Start)
As an infinite lowertriangular matrix (with offset 0 rather than 1, so the entries would be B(n  c)*binomial(n, c), B() a Bell number, rather than B(n  1  c)*binomial(n  1, c) as below), this array is S P S^1 where P is the Pascal matrix A007318, S is the Stirling2 matrix A048993, and S^1 is the Stirling1 matrix A048994.
Also, S P S^1 = (1/e)*exp(P). (End)
Exponential Riordan array [exp(exp(x)1),x]. Equal to A007318*A124323.  Paul Barry, Apr 23 2009
Equal to A049020*A048994 as infinite lower triangular matrices.  Philippe Deléham, Nov 19 2011
Build a superset Q[n] of set partitions of {1,2,...,n} by distinguishing "some" (possibly none or all) of the singletons. Indexed from n>=0, 0<=k<=n, T(n,k) is the number of elements in Q[n] that have exactly k distinguished singletons. A singleton is a subset containing one element. T(3,1) = 6 because we have {{1}'{2,3}}, {{1,2}{3}'}, {{1,3}{2}'}, {{1}'{2}{3}}, {{1}{2}'{3}}, {{1}{2}{3}'}.  Geoffrey Critzer, Nov 10 2012
Let Bell(n,x) denote the nth Bell polynomial, the nth row polynomial of A048993. Then this is the triangle of connection constants when expressing the basis polynomials Bell(n,x + 1) in terms of the basis polynomials Bell(n,x). For example, row 3 is (5, 6, 3, 1) and 5 + 6*Bell(1,x) + 3*Bell(2,x) + Bell(3,x) = 5 + 6*x + 3*(x + x^2) + (x + 3*x^2 + x^3) = 5 + 10*x + 6*x^2 + x^3 = (x + 1) + 3*(x + 1)^2 + (x + 1)^3 = Bell(3,x + 1).  Peter Bala, Sep 17 2013


REFERENCES

W. C. Yang, Conjectures on some sequences involving set partitions and Bell numbers, preprint, 2000. [Apparently unpublished]


LINKS

Alois P. Heinz, Rows n = 1..141, flattened
H. W. Becker, Rooks and rhymes, Math. Mag., 22 (1948/49), 2326. See Table III.
H. W. Becker, Rooks and rhymes, Math. Mag., 22 (1948/49), 2326. [Annotated scanned copy]
G. Hurst and A. Schultz, An elementary (number theory) proof of Touchard's congruence, arXiv:0906.0696v2 [math.CO]
A. O. Munagi, Set partitions with successions and separations, Intl. J. Math. Math. Sci. 2005 (2005) 451463.
M. Spivey, A generalized recurrence for Bell numbers, J. Int. Seq., 11 (2008), no. 2, Article 08.2.5
W. Yang, Bell numbers and ktrees, Disc. Math. 156 (1996) 247252.


FORMULA

T(n, c) = B(n  1  c)*binomial(n  1, c), where T(n, c) is the number of set partitions of {1, ..., n} that have c successive equalities and B() is a Bell number.
E.g.f.: exp(exp(x)+x*y1).  Vladeta Jovovic, Feb 13 2003
G.f.: 1/(1xyxx^2/(1xy2x2x^2/(1xy3x3x^2/(1xy4x4x^2/(1... (continued fraction).  Paul Barry, Apr 23 2009
Considered as triangle T(n,k), 0<=k<=n : T(n,k)= A007318(n,k)*A000110(nk) and Sum_{k, 0<=k<=n} T(n,k)*x^k = A000296(n), A000110(n), A000110(n+1), A005493(n), A005494(n), A045379(n) for x = 1, 0, 1, 2, 3, 4 respectively.  Philippe Deléham, Dec 13 2009
Let R(n,x) denote the nth row polynomial of the triangle. Then A000110(n+j) = Bell(n+j,1) = sum {k = 1..n} R(j,k)*Stirling2(n,k) (Spivey).  Peter Bala, Sep 17 2013


EXAMPLE

For example {1, 2, 1, 2, 2, 3} is a set partition of {1, 2, 3, 4, 5, 6} and has 1 successive equality, at i = 4.
Triangle begins:
1;
1,1;
2,2,1;
5,6,3,1;
15,20,12,4,1;
52,75,50,20,5,1;
203,312,225,100,30,6,1;
...
From Paul Barry, Apr 23 2009: (Start)
Production matrix is
1, 1,
1, 1, 1,
1, 2, 1, 1,
1, 3, 3, 1, 1,
1, 4, 6, 4, 1, 1,
1, 5, 10, 10, 5, 1, 1,
1, 6, 15, 20, 15, 6, 1, 1,
1, 7, 21, 35, 35, 21, 7, 1, 1,
1, 8, 28, 56, 70, 56, 28, 8, 1, 1 (End)


MAPLE

with(combinat): A056857:=(n, c)>binomial(n1, c)*bell(n1c): for n from 1 to 11 do seq(A056857(n, c), c=0..n1) od; # yields sequence in triangular form; Emeric Deutsch, Nov 10 2006
with(linalg): # Yields sequence in matrix form:
A056857_matrix := n > subs(exp(1)=1, exponential(exponential(
matrix(n, n, [seq(seq(`if`(j=k+1, j, 0), k=0..n1), j=0..n1)])))):
A056857_matrix(8); # Peter Luschny, Apr 18 2011


MATHEMATICA

t[n_, k_] := BellB[n1k]*Binomial[n1, k]; Flatten[ Table[t[n, k], {n, 1, 11}, {k, 0, n1}]](* JeanFrançois Alcover, Apr 25 2012, after Emeric Deutsch *)


CROSSREFS

Cf. Bell numbers A000110 (column c=0), A052889 (c=1), A105479 (c=2), A105480 (c=3).
Cf. A056858A056863. Essentially same as A056860, where the rows are read from right to left.
Cf. also A007318, A005493, A270953.
See A259691 for another version.
T(2n+1,n+1) gives A124102.
Sequence in context: A171670 A124644 A259691 * A175579 A129100 A162382
Adjacent sequences: A056854 A056855 A056856 * A056858 A056859 A056860


KEYWORD

easy,nonn,tabl,nice


AUTHOR

Winston C. Yang (winston(AT)cs.wisc.edu), Aug 31 2000


EXTENSIONS

More terms from David Wasserman, Apr 22 2002


STATUS

approved



