A052889 -id:A052889 - cp's OEIS frontend

A056857 Triangle read by rows: T(n,c) = number of successive equalities in set partitions of n.

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

Offset: 1

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

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 123|4, 124|3 and 134|2. - Emeric Deutsch, Nov 10 2006
Let P be the lower-triangular Pascal-matrix (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 lower-triangular 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 n-th Bell polynomial, the n-th 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

			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)

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 [Apparently unpublished]

Alois P. Heinz, Rows n = 1..141, flattened
H. W. Becker, Rooks and rhymes, Math. Mag., 22 (1948/49), 23-26. See Table III.
H. W. Becker, Rooks and rhymes, Math. Mag., 22 (1948/49), 23-26. [Annotated scanned copy]
Fufa Beyene, Jörgen Backelin, Roberto Mantaci, and Samuel A. Fufa, Set Partitions and Other Bell Number Enumerated Objects, J. Int. Seq., Vol. 26 (2023), Article 23.1.8.
A. Hennessy and Paul Barry, Generalized Stirling Numbers, Exponential Riordan Arrays, and Orthogonal Polynomials, J. Int. Seq. 14 (2011) # 11.8.2, Corollary 17.
G. Hurst and A. Schultz, An elementary (number theory) proof of Touchard's congruence, arXiv:0906.0696v2 [math.CO], 2009.
A. O. Munagi, Set partitions with successions and separations, Intl. J. Math. Math. Sci. 2005 (2005) 451-463.
M. Spivey, A generalized recurrence for Bell numbers, J. Int. Seq., 11 (2008), no. 2, Article 08.2.5
W. Yang, Bell numbers and k-trees, Disc. Math. 156 (1996) 247-252.

Cf. Bell numbers A000110 (column c=0), A052889 (c=1), A105479 (c=2), A105480 (c=3).
Cf. A056858-A056863. 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.
T(2n,n) gives A297926.

with(combinat): A056857:=(n,c)->binomial(n-1,c)*bell(n-1-c): for n from 1 to 11 do seq(A056857(n,c),c=0..n-1) 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..n-1),j=0..n-1)])))):
A056857_matrix(8); # Peter Luschny, Apr 18 2011

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

B(n) = sum(k=0, n, stirling(n, k, 2));
tabl(nn)={for(n=1, nn, for(k=0, n - 1, print1(B(n - 1 - k) * binomial(n - 1, k),", ");); print(););};
tabl(12); \\ Indranil Ghosh, Mar 19 2017

from sympy import bell, binomial
for n in range(1,12):
    print([bell(n - 1 - k) * binomial(n - 1, k) for k in range(n)]) # Indranil Ghosh, Mar 19 2017

def a(n): return (-1)^n / factorial(n)
@cached_function
def p(n, m):
    R = PolynomialRing(QQ, "x")
    if n == 0: return R(a(m))
    return R((m + x)*p(n - 1, m) - (m + 1)*p(n - 1, m + 1))
for n in range(11): print(p(n, 0).list())  # Peter Luschny, Jun 18 2023

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*y-1). - Vladeta Jovovic, Feb 13 2003
G.f.: 1/(1-xy-x-x^2/(1-xy-2x-2x^2/(1-xy-3x-3x^2/(1-xy-4x-4x^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(n-k) and Sum_{k=0..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 n-th 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

More terms from David Wasserman, Apr 22 2002

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

Original entry on oeis.org

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

Emeric Deutsch, Oct 28 2006

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

			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)

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.

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.

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:

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 *)

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

A070071 a(n) = n*B(n), where B(n) are the Bell numbers, A000110.

Original entry on oeis.org

0, 1, 4, 15, 60, 260, 1218, 6139, 33120, 190323, 1159750, 7464270, 50563164, 359377681, 2672590508, 20744378175, 167682274352, 1408702786668, 12277382510862, 110822101896083, 1034483164707440, 9972266139291771, 99147746245841106, 1015496134666939958

Offset: 0

Karol A. Penson, Apr 19 2002

nonn

a(n) is the total number of successions among all partitions of {1,2,...,n+1}; a succession is a pair (i,i+1) of consecutive integers lying in a block. For example, a(3)=15 because {1,2,3,4} has 6 partitions with 1 succession - 1/2/34, 1/23/4, 12/3/4, 14/23, 134/2, 124/3, 3 partitions with 2 successions - 1/234, 123/4, 12/34 and 1 partition with 3 successions - 1234. Thus a(3) = 6*1 + 3*2 + 1*3 = 15. - Augustine O. Munagi, Jul 01 2008
a(n) is the number of occurrences of integers in a list of all partitions of the set {1,...,n}. For example, the list 123, 1/23, 2/13, 3/12, 1/2/3 of all partitions of the set {1,2,3} requires 15 occurrences of integers each belonging to that set. [From Michael Hardy (hardy(AT)math.umn.edu), Nov 08 2008]
The bijection between the two foregoing characterizations is as follows: Fix x in {1,2,...,n} and associate x with the succession (x,x+1) which appears in some partitions of {1,2,...,n+1}. Replace x,x+1 by x and partition the n-set {1,2,...,x,x+2,...,n+1}, giving B(n) partitions. Thus the succession (x,x+1) occurs among partitions of {1,2,...,n+1} exactly B(n) times. - Augustine O. Munagi, Jun 02 2010

Alois P. Heinz, Table of n, a(n) for n = 0..574 (terms n=0..200 from Vincenzo Librandi)
Augustine O. Munagi, Extended set partitions with successions, European J. Combin. 29(5) (2008), 1298--1308.

Cf. A000110, A052889, A105479, A105480, A105481, A175757.

Row sums of A270236, A270701, A270702, A286416, A319298, A319375.

[n*Bell(n): n in [0..25]]; // Vincenzo Librandi, Mar 15 2014

with(combinat): a:=n->sum(numbcomb (n,0)*bell(n), j=1..n): seq(a(n), n=0..21); # Zerinvary Lajos, Apr 25 2007
with(combinat): a:=n->sum(bell(n), j=1..n): seq(a(n), n=0..21); # Zerinvary Lajos, Apr 25 2007
a:=n->sum(sum(Stirling2(n, k), j=1..n), k=1..n): seq(a(n), n=0..21); # Zerinvary Lajos, Jun 28 2007

a[n_] := n!*Coefficient[Series[x E^(E^x+x-1), {x, 0, n}], x, n]
Table[Sum[BellB[n, 1], {i, 1, n}], {n, 0, 21}] (* Zerinvary Lajos, Jul 16 2009 *)
Table[n*BellB[n], {n, 0, 20}] (* Vaclav Kotesovec, Mar 13 2014 *)

a(n)=local(t); if(n<0,0,t=exp(x+O(x^n)); n!*polcoeff(x*t*exp(t-1),n))

[bell_number(n)*n for n in range(22) ] # Zerinvary Lajos, Mar 14 2009

E.g.f: x*exp(x)*exp(exp(x)-1).
Sum_{k=1..n} n*binomial(n-1, k-1)*Bell(n-k), n >= 2. - Zerinvary Lajos, Nov 22 2006
a(n) ~ n^(n+1) * exp(n/LambertW(n)-1-n) / (sqrt(1+LambertW(n)) * LambertW(n)^n). - Vaclav Kotesovec, Mar 13 2014
a(n) = Sum_{k=1..n} k * A175757(n,k). - Alois P. Heinz, Mar 03 2020
a(n) = Sum_{j=0..n} n * Stirling2(n,j). - Detlef Meya, Apr 11 2024

A105488 Number of partitions of {1...n} containing 2 detached pairs of consecutive integers, i.e., partitions in which only 1- or 2-strings of consecutive integers can appear in a block and there are exactly two 2-strings.

Original entry on oeis.org

1, 6, 30, 150, 780, 4263, 24556, 149040, 951615, 6378625, 44785620, 328660566, 2515643767, 20044428810, 165955025400, 1425299331992, 12678325080012, 116635133853189, 1108221018960830, 10862073229428120, 109694927532209481, 1140199081827172719

Offset: 4

Augustine O. Munagi, Apr 10 2005

Number of partitions enumerated by A105479 in which the maximal length of consecutive integers in a block is 2.
With offset 2t, number of partitions of {1...N} containing 2 detached strings of t consecutive integers, where N=n+2j, t=2+j, j = 0,1,2,..., i.e., partitions of [n] in which only v-strings of consecutive integers can appear in a block, where v=1 or v=t and there are exactly two t-strings.
Equals the minimum of the sum of the Rand distances over all A000110(n) set partitions of n elements. E.g. a(3) = 6 because over the 5 set partitions of {1, 2, 3} the sum of Rand distances from {{1}, {2}, {3}} to the rest is 6. - Andrey Goder (andy.goder(AT)gmail.com), Dec 08 2006
a(n+3) = A000110(n) * A000217(n) = Sum_{k=1..n} A285362(n,k) is the sum of the entries in all set partitions of [n]. - Alois P. Heinz, Apr 16 2017

			a(5)=6 because the partitions of {1,2,3,4,5} with 2 detached pairs of consecutive integers are 145/23,125/34,1245/3,12/34/5,12/3/45,1/23/45.

Alois P. Heinz, Table of n, a(n) for n = 4..577
A. O. Munagi, Set Partitions with Successions and Separations, IJMMS 2005:3 (2005), 451-463.
W. Rand, Objective criteria for the evaluation of clustering methods, J. Amer. Stat. Assoc., 66 (336): 846-850, 1971.

Cf. A000110, A000217, A052889, A105479, A105489, A105484, A271023, A285362.

seq(binomial(n-2,2)*combinat[bell](n-3),n=4..28);

a[n_] := Binomial[n-2, 2]*BellB[n-3];
Table[a[n], {n, 4, 25}] (* Jean-François Alcover, May 11 2019 *)

a(n) = binomial(n-2, 2)*Bell(n-3), which is the case r = 2 in the general case of r pairs, d(n, r)=binomial(n-r, r)*Bell(n-r-1), which is the case t=2 of the general formula d(n, r, t)=binomial(n-r*(t-1), r)*B(n-r*(t-1)-1).

A033306 Triangle of coefficients of ordered cycle-index polynomials: T(n,k) = binomial(n,k)Bell(k)Bell(n-k).

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 5, 6, 6, 5, 15, 20, 24, 20, 15, 52, 75, 100, 100, 75, 52, 203, 312, 450, 500, 450, 312, 203, 877, 1421, 2184, 2625, 2625, 2184, 1421, 877, 4140, 7016, 11368, 14560, 15750, 14560, 11368, 7016, 4140, 21147, 37260, 63144, 85260, 98280, 98280, 85260, 63144, 37260, 21147

Offset: 0

N. J. A. Sloane

			   1;
   1,  1;
   2,  2,   2;
   5,  6,   6,   5;
  15, 20,  24,  20, 15;
  52, 75, 100, 100, 75, 52;
  ...

J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 80.

Cf. A000110, row sums give A001861.
Columns include A000110 and A052889.
Cf. A000807.

A033306 := proc(n,k)
    if k < 0 or k > n then
        0;
    else
        binomial(n,k)*combinat[bell](k)*combinat[bell](n-k) ;
    end if;
end proc: # R. J. Mathar, Mar 21 2013
# second Maple program:
b:= proc(n) option remember; expand(`if`(n>0, add(
     (x^j+1)*b(n-j)*binomial(n-1, j-1), j=1..n), 1))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n)):
seq(T(n), n=0..10);  # Alois P. Heinz, Aug 30 2019

t[n_, k_] := Binomial[n, k] * BellB[k] * BellB[n-k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 14 2014 *)

E.g.f.: exp(exp(x*y)+exp(x)-2).

Sum_{k=0..2n} (-1)^k * T(2n,k) = A000807(n). - Alois P. Heinz, Feb 13 2024

Edited by Vladeta Jovovic, Sep 17 2003

A175757 Triangular array read by rows: T(n,k) is the number of blocks of size k in all set partitions of {1,2,...,n}.

Original entry on oeis.org

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

Offset: 1

Geoffrey Critzer, Dec 04 2010

The row sums of this triangle equal A005493. Equals A056857 without its leftmost column.

T(n,k) = binomial(n,k)*B(n-k) where B is the Bell number.

			The set {1,2,3} has 5 partitions, {{1, 2, 3}}, {{2, 3}, {1}}, {{1, 3}, {2}}, {{1, 2}, {3}}, and {{2}, {3}, {1}}, and there are a total of 3 blocks of size 2, so T(3,2)=3.
Triangle begins:
    1;
    2,   1;
    6,   3,   1;
   20,  12,   4,  1;
   75,  50,  20,  5, 1;
  312, 225, 100, 30, 6, 1;
  ...

Alois P. Heinz, Rows n = 1..141, flattened

Cf. A000110, A056860, A052889, A070071, A105479, A105480, A105481, A105482, A285595.

b:= proc(n) option remember; `if`(n=0, [1, 0],
      add((p-> p+[0, p[1]*x^j])(b(n-j)*
      binomial(n-1, j-1)), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n)[2]):
seq(T(n), n=1..12);  # Alois P. Heinz, Apr 24 2017

Table[Table[Length[Select[Level[SetPartitions[m],{2}],Length[#]==n&]],{n,1,m}],{m,1,10}]//Grid

E.g.f. for column k is x^k/k!*exp(exp(x)-1).

Sum_{k=1..n} k * T(n,k) = A070071(n). - Alois P. Heinz, Mar 03 2020

A177255 a(n) = Sum_{j=1..n} j*B(j-1), where B(k) = A000110(k) are the Bell numbers.

Original entry on oeis.org

0, 1, 3, 9, 29, 104, 416, 1837, 8853, 46113, 257583, 1533308, 9676148, 64452909, 451475027, 3314964857, 25442301577, 203604718076, 1695172374548, 14654631691569, 131309475792709, 1217516798735521, 11664652754184043, 115319114738472472, 1174967255260496776

Offset: 0

Emeric Deutsch, May 07 2010

nonn

Number of adjacent blocks in all partitions of the set {1,2,...,n}. An adjacent block is a block of the form (i, i+1, i+2, ...). Example: a(3)=9 because in 1-2-3, 1-23, 12-3, 13-2, and 123 we have 3, 2, 2, 1, and 1 adjacent blocks, respectively.

Harvey P. Dale, Table of n, a(n) for n = 0..575

Cf. A000110, A177254, A177256, A177257.

Partial sums of A052889.

[n eq 0 select 0 else (&+[j*Bell(j-1): j in [1..n]]): n in [0..30]]; // G. C. Greubel, May 11 2024

with(combinat): a := proc (n) options operator, arrow: sum(j*bell(j-1), j = 1 .. n) end proc; seq(a(n), n = 0 .. 23);

With[{nn=30},Join[{0},Accumulate[BellB[Range[0,nn-1]]Range[nn]]]] (* Harvey P. Dale, Nov 10 2014 *)

[sum(j*bell_number(j-1) for j in range(1,1+n)) for n in range(31)] # G. C. Greubel, May 11 2024

a(n) = Sum_{k=0..n} k * A177254(n,k).

A184174 Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having k adjacent blocks of size 2, i.e., blocks of the form (i,i+1) (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 10, 4, 1, 35, 14, 3, 139, 54, 9, 1, 611, 224, 38, 4, 2925, 1027, 171, 16, 1, 15128, 5112, 822, 80, 5, 83903, 27352, 4279, 415, 25, 1, 495929, 156392, 23826, 2272, 145, 6, 3108129, 950285, 141039, 13252, 855, 36, 1, 20565721, 6107540, 883982, 81692, 5257, 238, 7

Offset: 0

Emeric Deutsch, Feb 09 2011

Row n contains 1 + floor(n/2) entries.
Sum of entries in row n = A000110(n) (the Bell numbers).
T(n,0) = A184175(n).
Sum_{k>=0} k*T(n,k) = A052889(n-1).

			T(4,1)=4 because we have 12-3-4, 1-23-4, 1-2-34, 14-23. T(4,2)=1 because we have 12-34.
Triangle starts:
1;
1;
1, 1;
3, 2;
10, 4, 1;
35, 14, 3;
139, 54, 9, 1;
611, 224, 38, 4;
2925, 1027, 171, 16, 1;
15128, 5112, 822, 80, 5;
83903, 27352, 4279, 415, 25, 1;
495929, 156392, 23826, 2272, 145, 6;
3108129, 950285, 141039, 13252, 855, 36, 1; ...

Alois P. Heinz, Rows n = 0..200, flattened

Cf. A000110, A184175, A052889, A124323, A184176, A184177.

with(combinat): q := 2: a := proc (n, k) options operator, arrow: sum((-1)^(k+j)*binomial(j, k)*binomial(n+j-j*q, j)*bell(n-j*q), j = k .. floor(n/q)) end proc: for n from 0 to 13 do seq(a(n, k), k = 0 .. floor(n/q)) end do; # yields sequence in triangular form

T[n_, k_] := Sum[(-1)^(k+j)*Binomial[j, k]*Binomial[n-j, j]*BellB[n-2j], {j, k, Floor[n/2]}]; Table[T[n, k], {n, 0, 13}, {k, 0, Floor[n/2]}] // Flatten (* Jean-François Alcover, Feb 21 2017 *)

{T(n,k) = my(A = sum(m=0,n, x^m/prod(k=0,m,1 - k*x + (1-y)*x^2 +x*O(x^n)))); polcoeff(polcoeff(A,n,x),k,y)}
for(n=0,12,for(k=0,n\2,print1(T(n,k),", "));print("")) \\ Paul D. Hanna, Sep 03 2017

T(n,k) = Sum_{j=k..floor(n/2)}(-1)^(k+j)*C(j,k)*C(n-j,j)*Bell(n-2j).

G.f.: A(x,y) = Sum_{n>=0} x^n / Product_{k=0..n} (1 - k*x + (1-y)*x^2). - Paul D. Hanna, Sep 03 2017

A184176 Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having k adjacent blocks of size 3, i.e., blocks of the form (i,i+1,i+2) (0 <= k <= floor(n/3)).

Original entry on oeis.org

1, 1, 2, 4, 1, 13, 2, 46, 6, 184, 18, 1, 805, 69, 3, 3840, 288, 12, 19775, 1324, 47, 1, 109180, 6578, 213, 4, 642382, 35136, 1032, 20, 4007712, 200398, 5390, 96, 1, 26399764, 1214136, 30027, 505, 5, 182939900, 7778856, 177744, 2792, 30, 1329327991, 52501052, 1112969, 16362, 170, 1

Offset: 0

Emeric Deutsch, Feb 09 2011

Number of entries in row n is 1 + floor(n/3).
Sum of entries in row n = A000110(n) (the Bell numbers).
T(n,0) = A184177(n).
Sum_{k>=0}k*T(n,k) = A052889(n-2).

			T(4,1) = 2 because we have 123-4 and 1-234.
Triangle starts:
    1;
    1;
    2;
    4,  1;
   13,  2;
   46,  6;
  184, 18,  1;

Alois P. Heinz, Rows n = 0..250, flattened

Cf. A000110, A052889, A184174, A184175, A184177.

with(combinat): q := 3: a := proc (n, k) options operator, arrow: sum((-1)^(k+j)*binomial(j, k)*binomial(n+j-j*q, j)*bell(n-j*q), j = k .. floor(n/q)) end proc: for n from 0 to 15 do seq(a(n, k), k = 0 .. floor(n/q)) end do; # yields sequence in triangular form

q = 3; a[n_, k_] := Sum[(-1)^(k+j)*Binomial[j, k]*Binomial[n+j-j*q, j]* BellB[n-j*q], {j, k, Floor[n/q]}]; Table[a[n, k], {n, 0, 15}, {k, 0, Floor[n/q]}] // Flatten (* Jean-François Alcover, Feb 22 2017, translated from Maple *)

T(n,k) = Sum_{j=k..floor(n/3)}(-1)^(k+j) * C(j,k) * C(n-2j,j) * Bell(n-3j).

A127741 a(n) = (n+1) * A005493(n).

Original entry on oeis.org

1, 6, 30, 148, 755, 4044, 22841, 136056, 853452, 5625950, 38885297, 281170080, 2122313505, 16688829122, 136457754030, 1158155642512, 10186602918035, 92711977180164, 871936904575985, 8462913158427580, 84668764368102012, 872196382566014506, 9241557859113581689

Offset: 0

Gary W. Adamson, Jan 27 2007

nonn

			a(n) = sum of terms in n-th row of A127740. a(2) = 30 = (6 + 9 + 15).

Chai Wah Wu, Table of n, a(n) for n = 0..250

Cf. A005493, A052889, A011971, A127740 (row sums).

lim=24;A005493=Differences[BellB[Range[lim]]];Array[(#+1)*A005493[[#+1]]&,lim-1,0] (* James C. McMahon, Jan 02 2025 *)

# requires Python 3.2 or higher. Otherwise use def'n of accumulate in Python docs.
from itertools import accumulate
A127741_list, blist, b = [], [1], 1
for n in range(1,1001):
    blist = list(accumulate([b]+blist))
    b = blist[-1]
    A127741_list.append(blist[-2]*n) # Chai Wah Wu, Sep 20 2014

a(n) = (n+1) * A005493(n).

Edited by Jon E. Schoenfield, May 27 2019

cp's OEIS Frontend

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