

A000670


Fubini numbers: number of preferential arrangements of n labeled elements; or number of weak orders on n labeled elements; or number of ordered partitions of [n].
(Formerly M2952 N1191)


280



1, 1, 3, 13, 75, 541, 4683, 47293, 545835, 7087261, 102247563, 1622632573, 28091567595, 526858348381, 10641342970443, 230283190977853, 5315654681981355, 130370767029135901, 3385534663256845323, 92801587319328411133, 2677687796244384203115
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OFFSET

0,3


COMMENTS

Number of ways n competitors can rank in a competition, allowing for the possibility of ties.
Also number of asymmetric generalized weak orders on n points.
Also called the ordered Bell numbers.
A weak order is a relation that is transitive and complete.
Called Fubini numbers by Comtet: counts formulas in Fubini theorem when switching the order of summation in multiple sums.  Olivier Gérard, Sep 30 2002
If the points are unlabeled then the answer is a(0) = 1, a(n) = 2^(n1) (cf. A011782).
For n>0, a(n) is the number of elements in the Coxeter complex of type A_{n1}. The corresponding sequence for type B is A080253 and there one can find a worked example as well as a geometric interpretation.  Tim Honeywill & Paul Boddington, Feb 10 2003
Also number of labeled (1+2)free posets.  Detlef Pauly, May 25 2003
Also the number of chains of subsets starting with the empty set and ending with a set of n distinct objects.  Andrew Niedermaier, Feb 20 2004
From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A007680(n) = [3,10,42,216,...] gives [3,13,75,541,...].
Stirling transform of a(n) = [1,3,13,75,...] is A083355(n) = [1,4,23,175,...].
Stirling transform of A000142(n) = [1,2,6,24,120,...] is a(n) = [1,3,13,75,...].
Stirling transform of A005359(n1) = [1,0,2,0,24,0,...] is a(n1) = [1,1,3,13,75,...].
Stirling transform of A005212(n1) = [0,1,0,6,0,120,0,...] is a(n1) = [0,1,3,13,75,...].
(End)
Unreduced denominators in convergent to log(2) = lim_{n>infinity} n*a(n1)/a(n).
a(n) is congruent to a(n+(p1)p^(h1)) (mod p^h) for n>=h (see Barsky).
StirlingBernoulli transform of 1/(1x^2).  Paul Barry, Apr 20 2005
This is the sequence of moments of the probability distribution of the number of tails before the first head in a sequence of fair coin tosses. The sequence of cumulants of the same probability distribution is A000629. That sequence is twice the result of deletion of the first term of this sequence.  Michael Hardy (hardy(AT)math.umn.edu), May 01 2005
With p(n) = the number of integer partitions of n, p(i) = the number of parts of the ith partition of n, d(i) = the number of different parts of the ith partition of n, p(j,i) = the jth part of the ith partition of n, m(i,j) = multiplicity of the jth part of the ith partition of n, one has: a(n) = Sum_{i=1..p(n)} (n!/(Product_{j=1..p(i)}p(i,j)!)) * (p(i)!/(Product_{j=1..d(i)} m(i,j)!)).  Thomas Wieder, May 18 2005
The number of chains among subsets of [n]. The summed term in the new formula is the number of such chains of length k.  Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Occurs also as first column of a matrixinversion occurring in a sumoflikepowers problem. Consider the problem for any fixed natural number m>2 of finding solutions to the equation Sum_{k=1..n}k^m = (k+1)^m. Erdős conjectured that there are no solutions for n,m>2. Let D be the matrix of differences of D[m,n] := Sum_{k=1..n} k^m  (k+1)^m. Then the generating functions for the rows of this matrix D constitute a set of polynomials in n (for varying n along columns) and the mth polynomial defining the mth row. Let GF_D be the matrix of the coefficients of this set of polynomials. Then the present sequence is the (unsigned) first column of GF_D^1.  Gottfried Helms, Apr 01 2007
Assuming A=log(2), D is d/dx and f(x)=x/(exp(x)1), we have a(n) = (n!/2A^(n+1)) Sum_{k=0..n} (A^k/k!) D^n f(A) which gives Wilf's asymptotic value when n tends to infinity. Equivalently, D^n f(a) = 2( A*a(n)  2*a(n1) ).  Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
List partition transform (see A133314) of (1,1,1,1,...).  Tom Copeland, Oct 24 2007
First column of A154921.  Mats Granvik, Jan 17 2009
A slightly more transparent interpretation of a(n) is as the number of 'factor sequences' of N for the case in which N is a product of n distinct primes. A factor sequence of N of length k is of the form 1=x(1),x(2),...,x(k)=N, where {x(i)} is an increasing sequence such that x(i) divides x(i+1), i=1,2,...,k1. For example, N=70 has the 13 factor sequences {1,70}, {1,2,70}, {1,5,70}, {1,7,70}, {1,10,70}, {1,14,70}, {1,35,70}, {1,2,10,70}, {1,2,14,70}, {1,5,10,70}, {1,5,35,70}, {1,7,14,70}, {1,7,35,70}.  Martin Griffiths, Mar 25 2009
Starting (1, 3, 13, 75, ...) = row sums of triangle A163204.  Gary W. Adamson, Jul 23 2009
Equals double inverse binomial transform of A007047: (1, 3, 11, 51, ...).  Gary W. Adamson, Aug 04 2009
If f(x)=Sum_{n>=0}c(n)*x^n converges for every x, then Sum_{n>=0}f(n*x)/2^(n+1) = Sum_{n>=0}c(n)*a(n)*x^n. Example: Sum_{n>=0}exp(n*x)/2^(n+1) = Sum_{n>=0}a(n)*x^n/n! = 1/(2exp(x)) = E.g.f.  Miklos Kristof, Nov 02 2009
Hankel transform is A091804.  Paul Barry, Mar 30 2010
It appears that the prime numbers greater than 3 in this sequence (13, 541, 47293, ...) are of the form 4n+1.  Paul Muljadi, Jan 28 2011
The Fi1 and Fi2 triangle sums of A028246 are given by the terms of this sequence. For the definitions of these triangle sums, see A180662.  Johannes W. Meijer, Apr 20 2011
The modified generating function A(x) = 1/(2exp(x))1 = x + 3*x^2/2! + 13*x^3/3! + ... satisfies the autonomous differential equation A' = 1 + 3*A + 2*A^2 with initial condition A(0) = 0. Applying [Bergeron et al., Theorem 1] leads to two combinatorial interpretations for this sequence: (A) a(n) gives the number of planeincreasing 012 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors. (B) a(n) gives the number of nonplaneincreasing 012 trees on n vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors. Examples are given below.  Peter Bala, Aug 31 2011
Starting with offset 1 = the eigensequence of A074909 (the beheaded Pascal's triangle), and row sums of triangle A208744.  Gary W. Adamson, Mar 05 2012
a(n) = number of words of length n on the alphabet of positive integers for which the letters appearing in the word form an initial segment of the positive integers. Example: a(2) = 3 counts 11, 12, 21. The map "record position of block containing i, 1<=i<=n" is a bijection from lists of sets on [n] to these words. (The lists of sets on [2] are 12, 1/2, 2/1.)  David Callan, Jun 24 2013
This sequence was the subject of one of the earliest uses of the database. Don Knuth, who had a computer printout of the database prior to the publication of the 1973 Handbook, wrote to N. J. A. Sloane on May 18, 1970, saying: "I have just had my first real 'success' using your index of sequences, finding a sequence treated by Cayley that turns out to be identical to another (a priori quite different) sequence that came up in connection with computer sorting." A000670 is discussed in Exercise 3 of Section 5.3.1 of The Art of Computer Programming, Vol. 3, 1973.  N. J. A. Sloane, Aug 21 2014
Ramanujan gives a method of finding a continued fraction of the solution x of an equation 1 = x + a2*x^2 + ... and uses log(2) as the solution of 1 = x + x^2/2 + x^3/6 + ... as an example giving the sequence of simplified convergents as 0/1, 1/1, 2/3, 9/13, 52/75, 375/541, ... of which the sequence of denominators is this sequence, while A052882 is the numerators.  Michael Somos, Jun 19 2015
For n>=1, a(n) is the number of Dyck paths (A000108) with (i) n+1 peaks (UDs), (ii) no UUDDs, and (iii) at least one valley vertex at every nonnegative height less than the height of the path. For example, a(2)=3 counts UDUDUD (of height 1 with 2 valley vertices at height 0), UDUUDUDD, UUDUDDUD. These paths correspond, under the "glove" or "accordion" bijection, to the ordered trees counted by Cayley in the 1859 reference, after a harmless pruning of the "long branches to a leaf" in Cayley's trees. (Cayley left the reader to infer the trees he was talking about from examples for small n and perhaps from his proof.)  David Callan, Jun 23 2015
From David L. Harden, Apr 09 2017: (Start)
Fix a set X and define two distance functions d,D on X to be metrically equivalent when d(x_1,y_1) <= d(x_2,y_2) iff D(x_1,y_1) <= D(x_2,y_2) for all x_1, y_1, x_2, y_2 in X.
Now suppose that we fix a function f from unordered pairs of distinct elements of X to {1,...,n}. Then choose positive real numbers d_1 <= ... <= d_n such that d(x,y) = d_{f(x,y)}; the set of all possible choices of the d_i's makes this an nparameter family of distance functions on X. (The simplest example of such a family occurs when n is a triangular number: When that happens, write n = (k 2). Then the set of all distance functions on X, when X = k, is such a family.) The number of such distance functions, up to metric equivalence, is a(n).
It is easy to see that an equivalence class of distance functions gives rise to a welldefined weak order on {d_1, ..., d_n}. To see that any weak order is realizable, choose distances from the set of integers {n1,...,2n2} so that the triangle inequality is automatically satisfied. (End)
a(n) is the number of rooted labeled forests on n nodes that avoid the patterns 213, 312, and 321.  Kassie Archer, Aug 30 2018


REFERENCES

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Index entries for "core" sequences
Index entries for related partitioncounting sequences


FORMULA

a(n) = Sum_{k=1..n} k! StirlingS2(n, k) (whereas the Bell numbers A000110(n) = Sum_{k=1..n} StirlingS2(n, k)).
E.g.f.: 1/(2exp(x)).
a(n) = Sum_{k=1..n} binomial(n, k)*a(nk), a(0) = 1.
The e.g.f. y(x) satisfies y' = 2*y^2  y.
a(n) = A052856(n)  1, if n>0.
a(n) = A052882(n)/n, if n>0.
a(n) = A076726(n)/2.
a(n) is asymptotic to (1/2)*n!*log_2(e)^(n+1), where log_2(e) = 1.442695... [Barthelemy80, Wilf90].
For n >= 1, a(n) = (n!/2) * Sum_{k=infinity..infinity} of (log(2) + 2 Pi i k)^(n1).  Dean Hickerson
a(n) = ((x*d/dx)^n)(1/(2x)) evaluated at x=1.  Karol A. Penson, Sep 24 2001
For n>=1, a(n) = Sum_{k>=1} (k1)^n/2^k = A000629(n)/2.  Benoit Cloitre, Sep 08 2002
Value of the nth Eulerian polynomial (cf. A008292) at x=2.  Vladeta Jovovic, Sep 26 2003
First Eulerian transform of the powers of 2 [A000079]. See A000142 for definition of FET.  Ross La Haye, Feb 14 2005
a(n) = Sum_{k=0..n} (1)^k*k!*Stirling2(n+1, k+1)(1+(1)^k)/2.  Paul Barry, Apr 20 2005
a(n) + a(n+1) = 2*A005649(n).  Philippe Deléham, May 16 2005  Thomas Wieder, May 18 2005
Equals inverse binomial transform of A000629.  Gary W. Adamson, May 30 2005
a(n) = Sum_{k=0..n} k!*( Stirling2(n+2, k+2)  Stirling2(n+1, k+2) ).  Micha Hofri (hofri(AT)wpi.edu), Jul 01 2006
Recurrence: 2a(n)=(a+1)^n where superscripts are converted to subscripts after binomial expansion  reminiscent of Bernoulli numbers' B_n=(B+1)^n.  Martin Kochanski (mjk(AT)cardbox.com), May 10 2007
a(n) = (1)^n * n!*Laguerre(n,P((.),2)), umbrally, where P(j,t) are the polynomials in A131758.  Tom Copeland, Sep 27 2007
Formula in terms of the hypergeometric function, in Maple notation: a(n)=hypergeom([2,2...2],[1,1...1],1/2)/4, n=1,2..., where in the hypergeometric function there are n upper parameters all equal to 2 and n1 lower parameters all equal to 1 and the argument is equal to 1/2. Example: a(4)=evalf(hypergeom([2,2,2,2],[1,1,1],1/2)/4)=75.  Karol A. Penson, Oct 04 2007
a(n) = Sum_{k=0..n} A131689(n,k).  Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 01 2009: (Start)
Analogy with the Bernoulli numbers.
We enlarge upon the above comment of M. Kochanski.
The Bernoulli polynomials B_n(x), n = 0,1,..., are given by the formula
(1)... B_n(x) := Sum_{k=0..n} binomial(n,k)*B(k)*x^(nk),
where B(n) denotes the sequence of Bernoulli numbers B(0) = 1,
B(1) = 1/2, B(2) = 1/6, B(3) = 0, ....
By analogy, we associate with the present sequence an Appell sequence of polynomials {P_n(x)}n>=0 defined by
(2)... P_n(x) := Sum_{k=0..n} binomial(n,k)*a(k)*x^(nk).
These polynomials have similar properties to the Bernoulli polynomials.
The first few values are P_0(x) = 1, P_1(x) = x + 1,
P_2(x) = x^2 + 2*x + 3, P_3(x) = x^3 + 3*x^2 + 9*x + 13 and
P_4(x) = x^4 + 4*x^3 + 18*x^2 + 52*x + 75. See A154921 for the triangle of coefficients of these polynomials.
The e.g.f. for this polynomial sequence is
(3)... exp(x*t)/(2  exp(t)) = 1 + (x + 1)*t + (x^2 + 2*x + 3)*t^2/2! + ....
The polynomials satisfy the difference equation
(4)... 2*P_n(x  1)  P_n(x) = (x  1)^n,
and so may be used to evaluate the weighted sums of powers of integers
(1/2)*1^m + (1/2)^2*2^m + (1/2)^3*3^m + ... + (1/2)^(n1)*(n1)^m
via the formula
(5)... Sum_{k=1..n1} (1/2)^k*k^m = 2*P_m(0)  (1/2)^(n1)*P_m(n),
analogous to the evaluation of the sums 1^m + 2^m + ... + (n1)^m in terms of Bernoulli polynomials.
This last result can be generalized to
(6)... Sum_{k=1..n1} (1/2)^k*(k+x)^m = 2*P_m(x)(1/2)^(n1)*P_m(x+n).
For more properties of the polynomials P_n(x), refer to A154921.
For further information on weighted sums of powers of integers and the associated polynomial sequences, see A162312.
The present sequence also occurs in the evaluation of another sum of powers of integers. Define
(7)... S_m(n) := Sum_{k=1..n1} (1/2)^k*((nk)*k)^m, m = 1,2,....
Then
(8)... S_m(n) = (1)^m *[2*Q_m(n)  (1/2)^(n1)*Q_m(n)],
where Q_m(x) are polynomials in x given by
(9)... Q_m(x) = Sum_{k=0..m} a(m+k)*binomial(m,k)*x^(mk).
The first few values are Q_1(x) = x + 3, Q_2(x) = 3*x^2 + 26*x + 75
and Q_3(x) = 13*x^3 + 225*x^2 + 1623*x + 4683.
For example, m = 2 gives
(10)... S_2(n) := Sum_{k=1..n1} (1/2)^k*((nk)*k)^2
= 2*(3*n^2  26*n + 75)  (1/2)^(n1)*(3*n^2 + 26*n + 75).
(End)
G.f.: 1/(1x/(12x/(12x/(14x/(13x/(16x/(14x/(18x/(15x/(110x/(16x/(1... (continued fraction); coefficients of continued fraction are given by floor((n+2)/2)*(3(1)^n)/2 (A029578(n+2)).  Paul Barry, Mar 30 2010
G.f.: 1/(1x2x^2/(14x8x^2/(17x18x^2/(110x32x^2/(1../(1(3n+1)x2(n+1)^2x^2/(1... (continued fraction).  Paul Barry, Jun 17 2010
G.f.: A(x) = Sum_{n>=0} n!*x^n / Product_{k=1..n} (1k*x).  Paul D. Hanna, Jul 20 2011
a(n)=A074206(q_1*q_2*...*q_n), where {q_i} are distinct primes.  Vladimir Shevelev, Aug 05 2011
The adjusted e.g.f. A(x) := 1/(2exp(x))1, has inverse function A(x)^1 = Integral_{t=0..x} 1/((1+t)*(1+2*t)). Applying [Dominici, Theorem 4.1] to invert the integral yields a formula for a(n): Let f(x) = (1+x)*(1+2*x). Let D be the operator f(x)*d/dx. Then a(n) = D^(n1)(f(x)) evaluated at x = 0. Compare with A050351.  Peter Bala, Aug 31 2011
G.f.: 1+x/(1x+2x(x1)/(1+3x(2x1)/(1+4x(3x1)/(1+5x(4x1)/(1+... or 1+x/(U(0)x), U(k)=1+(k+2)(kx+x1)/U(k+1); (continued fraction).  Sergei N. Gladkovskii, Oct 30 2011
a(n) = D^n(1/(1x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A052801.  Peter Bala, Nov 25 2011
E.g.f.: 1 + x/(G(0)2*x) where G(k)= x + k + 1  x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1step).  Sergei N. Gladkovskii, Jul 11 2012
E.g.f. (2  2*x)*(1  2*x^3/(8*x^2  4*x + (x^2  4*x + 2)*G(0)))/(x^2  4*x + 2) where G(k)= k^2 + k*(x+4) + 2*x + 3  x*(k+1)*(k+3)^2 /G(k+1) ; (continued fraction, Euler's 1st kind, 1step).  Sergei N. Gladkovskii, Oct 01 2012
G.f.: 1 + x/G(0) where G(k) = 1  3*x*(k+1)  2*x^2*(k+1)*(k+2)/G(k+1); (continued fraction).  Sergei N. Gladkovskii, Jan 11 2013.
G.f.: 1/G(0) where G(k) = 1  x*(k+1)/( 1  2*x*(k+1)/G(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Mar 23 2013
a(n) is always odd. For odd prime p and n >= 1, a((p1)*n) = 0 (mod p).  Peter Bala, Sep 18 2013
G.f.: 1 + x/Q(0), where Q(k) = 1  3*x*(2*k+1)  2*x^2*(2*k+1)*(2*k+2)/( 1  3*x*(2*k+2)  2*x^2*(2*k+2)*(2*k+3)/Q(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Sep 23 2013
G.f.: T(0)/(1x), where T(k) = 1  2*x^2*(k+1)^2/( 2*x^2*(k+1)^2  (1x3*x*k)*(14*x3*x*k)/T(k+1) ); (continued fraction).  Sergei N. Gladkovskii, Oct 14 2013
a(n) = log(2)* Integral_{x>=0} floor(x)^n * 2^(x) dx.  Peter Bala, Feb 06 2015
For n > 0, a(n) = Re(polygamma(n, i log(2)/(2 Pi))/(2 Pi i)^(n+1))  n!/(2 log(2)^(n+1)).  Vladimir Reshetnikov, Oct 15 2015
a(n) = Sum_{k=1..n}(k*b2(k1)*(k)!*stirling2(n, k)), n>0, a(0)=1, where b2(n) is the nth Bernoulli number of the second kind.  Vladimir Kruchinin, Nov 21 2016
a(n) = Sum_{k=0..2^(n1)1} A284005(k), n>0, a(0)=1.  Mikhail Kurkov, Jul 08 2018


EXAMPLE

Let the points be labeled 1,2,3,...
a(2) = 3: 1<2, 2<1, 1=2.
a(3) = 13 from the 13 arrangements
1<2<3,
1<3<2,
2<1<3,
2<3<1,
3<1<2,
3<2<1,
1=2<3
1=3<2,
2=3<1,
1<2=3,
2<1=3,
3<1=2,
1=2=3.
Three competitors can finish in 13 ways: 1,2,3; 1,3,2; 2,1,3; 2,3,1; 3,1,2; 3,2,1; 1,1,3; 2,2,1; 1,3,1; 2,1,2; 3,1,1; 1,2,2; 1,1,1.
a(3) = 13. The 13 plane increasing 012 trees on 3 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 2 colors, are:
........................................................
........1 (x3 colors).....1(x2 colors)....1(x2 colors)..
......................../.\............./.\............
........2 (x3 colors)...2...3...........3...2...........
.......................................................
........3...............................................
......====..............====............====............
.Totals 9......+..........2....+..........2....=..13....
........................................................
a(4) = 75. The 75 nonplane increasing 012 trees on 4 vertices, where vertices of outdegree 1 come in 3 colors and vertices of outdegree 2 come in 4 colors, are:
...............................................................
.....1 (x3).....1(x4).......1(x4).....1(x4)........1(x3).......
............../.\........./.\......./.\......................
.....2 (x3)...2...3.(x3)..3...2(x3).4...2(x3)......2(x4).......
..................\...........\.........\......../.\..........
.....3.(x3).........4...........4.........3......3...4.........
..............................................................
.....4.........................................................
....====......=====........====......====.........====.........
Tots 27....+....12......+...12....+...12.......+...12...=...75.
From Joerg Arndt, Mar 18 2014: (Start)
The a(3) = 13 strings on the alphabet {1,2,3} containing all letters up to the maximal value appearing and the corresponding ordered set partitions are:
01: [ 1 1 1 ] { 1, 2, 3 }
02: [ 1 1 2 ] { 1, 2 } < { 3 }
03: [ 1 2 1 ] { 1, 3 } < { 2 }
04: [ 2 1 1 ] { 2, 3 } < { 1 }
05: [ 1 2 2 ] { 1 } < { 2, 3 }
06: [ 2 1 2 ] { 2 } < { 1, 3 }
07: [ 2 2 1 ] { 3 } < { 1, 2 }
08: [ 1 2 3 ] { 1 } < { 2 } < { 3 }
09: [ 1 3 2 ] { 1 } < { 3 } < { 2 }
00: [ 2 1 3 ] { 2 } < { 1 } < { 3 }
11: [ 2 3 1 ] { 3 } < { 1 } < { 2 }
12: [ 3 1 2 ] { 2 } < { 3 } < { 1 }
13: [ 3 2 1 ] { 3 } < { 2 } < { 1 }
(End)


MAPLE

A000670 := proc(n) option remember; local k; if n <=1 then 1 else add(binomial(n, k)*A000670(nk), k=1..n); fi; end;
with(combstruct); SeqSetL := [S, {S=Sequence(U), U=Set(Z, card >= 1)}, labeled]; seq(count(SeqSetL, size=j), j=1..12);
with(combinat): a:=n>add(add((1)^(ki)*binomial(k, i)*i^n, i=0..n), k=0..n): seq(a(n), n=0..18); # Zerinvary Lajos, Jun 03 2007
a := n > add(combinat:eulerian1(n, k)*2^k, k=0..n): # Peter Luschny, Jan 02 2015
a := n > (polylog(n, 1/2)+`if`(n=0, 1, 0))/2: seq(round(evalf(a(n), 32)), n=0..20); # Peter Luschny, Nov 03 2015


MATHEMATICA

Table[(PolyLog[z, 1/2] + KroneckerDelta[z])/2, {z, 0, 20}] (* Wouter Meeussen *)
a[0] = 1; a[n_] := a[n] = Sum[Binomial[n, k]*a[n  k], {k, 1, n}]; Table[a[n], {n, 0, 30}] (* Roger L. Bagula and Gary W. Adamson, Sep 13 2008 *)
t = 30; Range[0, t]! CoefficientList[Series[1/(2  Exp[x]), {x, 0, t}], x] (* Vincenzo Librandi, Mar 16 2014 *)
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1 / (2  Exp@x), {x, 0, n}]]; (* Michael Somos, Jun 19 2015 *)
Table[Sum[k^n/2^(k+1), {k, 0, Infinity}], {n, 0, 20}] (* Vaclav Kotesovec, Jun 26 2015 *)
Table[HurwitzLerchPhi[1/2, n, 0]/2, {n, 0, 20}] (* JeanFrançois Alcover, Jan 31 2016 *)
Fubini[n_, r_] := Sum[k!*Sum[(1)^(i+k+r)*((i+r)^(nr)/(i!*(kir)!)), {i, 0, kr}], {k, r, n}]; Fubini[0, 1] = 1; Table[Fubini[n, 1], {n, 0, 20}] (* JeanFrançois Alcover, Mar 31 2016 *)


PROG

(PARI) {a(n) = if( n<0, 0, n! * polcoeff( subst( 1 / (1  y), y, exp(x + x*O(x^n))  1), n))}; /* Michael Somos, Mar 04 2004 */
(PARI) Vec(serlaplace(1/(2exp('x+O('x^66))))) /* Joerg Arndt, Jul 10 2011 */
(PARI) {a(n)=polcoeff(sum(m=0, n, m!*x^m/prod(k=1, m, 1k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */
(PARI) {a(n) = if( n<1, n==0, sum(k=1, n, binomial(n, k) * a(nk)))}; /* Michael Somos, Jul 16 2017 */
(Maxima) makelist(sum(stirling2(n, k)*k!, k, 0, n), n, 0, 12); /* Emanuele Munarini, Jul 07 2011 */
(Maxima) a[0]:1$ a[n]:=sum(binomial(n, k)*a[nk], k, 1, n)$ A000670(n):=a[n]$ makelist(A000670(n), n, 0, 30); /* Martin Ettl, Nov 05 2012 */
(Sage)
@CachedFunction
def A000670(n) : return 1 if n == 0 else add(A000670(k)*binomial(n, k) for k in range(n))
[A000670(n) for n in (0..20)] # Peter Luschny, Jul 14 2012
(Haskell)
a000670 n = a000670_list !! n
a000670_list = 1 : f [1] (map tail $ tail a007318_tabl) where
f xs (bs:bss) = y : f (y : xs) bss where y = sum $ zipWith (*) xs bs
 Reinhard Zumkeller, Jul 26 2014


CROSSREFS

See A240763 for a list of the actual preferential arrangements themselves.
A000629, this sequence, A002050, A032109, A052856, A076726 are all moreorless the same sequence.  N. J. A. Sloane, Jul 04 2012
Binomial transform of A052841. Inverse binomial transform of A000629.
Asymptotic to A034172.
Cf. A002144, A002869, A004121, A004122, A007047, A007318, A048144, A053525, A080253, A080254, A011782, A154921, A162312, A163204, A242280, A261959, A290376.
Row r=1 of A094416. Row 0 of array in A226513. Row n=1 of A262809.
Main diagonal of A135313 and A261781 and A276890.
Row sums of triangles A019538, A131689, A208744 and A276891.
A217389 and A239914 give partial sums.
Sequence in context: A276900 A276930 A034172 * A032036 A305535 A300793
Adjacent sequences: A000667 A000668 A000669 * A000671 A000672 A000673


KEYWORD

nonn,core,nice,easy


AUTHOR

N. J. A. Sloane


STATUS

approved



