

A000110


Bell or exponential numbers: number of ways to partition a set of n labeled elements.
(Formerly M1484 N0585)


888



1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147, 115975, 678570, 4213597, 27644437, 190899322, 1382958545, 10480142147, 82864869804, 682076806159, 5832742205057, 51724158235372, 474869816156751, 4506715738447323, 44152005855084346, 445958869294805289, 4638590332229999353, 49631246523618756274
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OFFSET

0,3


COMMENTS

The leading diagonal of its difference table is the sequence shifted, see Bernstein and Sloane (1995).  N. J. A. Sloane, Jul 04 2015
Also the number of equivalence relations that can be defined on a set of n elements.  Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005
a(n) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. Adjacent regions cannot have the same color.  David W. Wilson, Feb 22 2005
If an integer is squarefree and has n distinct prime factors then a(n) is the number of ways of writing it as a product of its divisors.  Amarnath Murthy, Apr 23 2001
Consider rooted trees of height at most 2. Letting each tree 'grow' into the next generation of n means we produce a new tree for every node which is either the root or at height 1, which gives the Bell numbers.  Jon Perry, Jul 23 2003
Begin with [1,1] and follow the rule that [1,k] > [1,k+1] and [1,k] k times, e.g., [1,3] is transformed to [1,4], [1,3], [1,3], [1,3]. Then a(n) is the sum of all components: [1,1] = 2; [1,2], [1,1] = 5; [1,3], [1,2], [1,2], [1,2], [1,1] = 15; etc.  Jon Perry, Mar 05 2004
Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (e.g., 'abba') such that the leftmost letter is always 'a' and no letter may be greater than one more than the greatest letter to its left. Thus 'aac' is not valid since 'c' is more than one greater than 'a'. For example, a(3)=5 because there are 5 rhyme schemes: aaa, aab, aba, abb, abc; also see example by Neven Juric.  Bill Blewett, Mar 23 2004
In other words, number of lengthn restricted growth strings (RGS) [s(0),s(1),...,s(n1)] where s(0)=0 and s(k)<=1+max(prefix) for k>=1, see example (cf. A080337 and A189845).  Joerg Arndt, Apr 30 2011
Number of partitions of {1, ...,n+1} into subsets of nonconsecutive integers, including the partition 12...n+1. E.g., a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers, namely, 1324, 1324, 1423, 1243, 1234.  Augustine O. Munagi, Mar 20 2005
Triangle (addition) scheme to produce terms, derived from the recurrence, from Oscar Arevalo (loarevalo(AT)sbcglobal.net), May 11 2005:
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
... [This is Aitken's array A011971]
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)!)) * (1/(Product_{j=1..d(i)} m(i,j)!))  Thomas Wieder, May 18 2005
a(n+1) is the number of binary relations on an nelement set that are both symmetric and transitive.  Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005
If the rule from Jon Perry, Mar 05 2004, is used, then a(n1) = [number of components used to form a(n)] / 2.  Daniel Kuan (dkcm(AT)yahoo.com), Feb 19 2006
a(n) is the number of functions f from {1,...,n} to {1,...,n,n+1} that satisfy the following two conditions for all x in the domain: (1) f(x)>x; (2)f(x)=n+1 or f(f(x))=n+1. E.g., a(3)=5 because there are exactly five functions that satisfy the two conditions: f1={(1,4),(2,4),(3,4)}, f2={(1,4),(2,3),(3,4)}, f3={(1,3),(2,4),(3,4)}, f4={(1,2),(2,4),(3,4)} and f5={(1,3),(2,3),(3,4)}.  Dennis P. Walsh, Feb 20 2006
Number of asynchronic siteswap patterns of length n which have no zerothrows (i.e., contain no 0's) and whose number of orbits (in the sense given by Allen Knutson) is equal to the number of balls. E.g., for n=4, the condition is satisfied by the following 15 siteswaps: 4444, 4413, 4242, 4134, 4112, 3441, 2424, 1344, 2411, 1313, 1241, 2222, 3131, 1124, 1111. Also number of ways to choose n permutations from identity and cyclic permutations (1 2), (1 2 3), ..., (1 2 3 ... n) so that their composition is identity. For n=3 we get the following five: id o id o id, id o (1 2) o (1 2), (1 2) o id o (1 2), (1 2) o (1 2) o id, (1 2 3) o (1 2 3) o (1 2 3). (To see the bijection, look at Ehrenborg and Readdy paper.)  Antti Karttunen, May 01 2006
a(n) is the number of permutations on [n] in which a 321 (scattered) pattern occurs only as part of a 3241 pattern. Example: a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" reference a(n) = number of permutation tableaux of size n (A000142) whose first row contains no 0's. Example: a(3)=5 counts {{}, {}, {}}, {{1}, {}}, {{1}, {0}}, {{1}, {1}}, {{1, 1}}.  David Callan, Oct 07 2006
Take the series 1^n/1! + 2^n/2! + 3^n/3! + 4^n/4! ... If n=1 then the result will be e, about 2.71828. If n=2, the result will be 2e. If n=3, the result will be 5e. This continues, following the pattern of the Bell numbers: e, 2e, 5e, 15e, 52e, 203e, etc.  Jonathan R. Love (japanada11(AT)yahoo.ca), Feb 22 2007
From Gottfried Helms, Mar 30 2007: (Start)
This sequence is also the first column in the matrixexponential of the (lower triangular) Pascalmatrix, scaled by exp(1): PE = exp(P) / exp(1) =
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
First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamardproduct Toeplitz(A000110) (X) P:
1
1 1
2 1 1
5 2 1 1
15 5 2 1 1 (X) P
52 15 5 2 1 1
203 52 15 5 2 1 1
877 203 52 15 5 2 1 1
(End)
The terms can also be computed with finite steps and precise integer arithmetic. Instead of exp(P)/exp(1) one can compute A = exp(P  I) where I is the identitymatrix of appropriate dimension since (PI) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the rowindex starting at 1.  Gottfried Helms, Apr 10 2007
Define a Bell pseudoprime to be a composite number n such that a(n) == 2 (mod n). W. F. Lunnon recently found the Bell pseudoprimes 21361 = 41*521 and C46 = 3*23*16218646893090134590535390526854205539989357 and conjectured that Bell pseudoprimes are extremely scarce. So the second Bell pseudoprime is unlikely to be known with certainty in the near future. I confirmed that 21361 is the first.  David W. Wilson, Aug 04 2007 and Sep 24 2007
This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314.  Tom Copeland, Oct 21 2007
Starting (1, 2, 5, 15, 52, ...), equals row sums and right border of triangle A136789. Also row sums of triangle A136790.  Gary W. Adamson, Jan 21 2008
This is the exponential transform of A000012.  Thomas Wieder, Sep 09 2008
From Abdullahi Umar, Oct 12 2008: (Start)
a(n) is also the number of idempotent orderdecreasing full transformations (of an nchain).
a(n) is also the number of nilpotent partial oneone orderdecreasing transformations (of an nchain).
a(n+1) is also the number of partial oneone orderdecreasing transformations (of an nchain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of npattern sequences [Cooper & Kennedy]. An npattern sequence is a sequence of integers (a_1,...,a_n) such that a_i = i or a_i = a_j for some j < i. For example, Bell(3) = 5 since the 3pattern sequences are (1,1,1), (1,1,3), (1,2,1), (1,2,2) and (1,2,3).
Bell(n) is the number of sequences of positive integers (N_1,...,N_n) of length n such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (see the comment by B. Blewett above). It is interesting to note that if we strengthen the latter condition to N_(i+1) <= 1 + N_i we get the Catalan numbers A000108 instead of the Bell numbers.
(End)
Equals the eigensequence of Pascal's triangle, A007318; and starting with offset 1, = row sums of triangles A074664 and A152431.  Gary W. Adamson, Dec 04 2008
The entries f(i, j) in the exponential of the infinite lowertriangular matrix of binomial coefficients b(i, j) are f(i, j) = b(i, j) e a(i  j).  David Pasino, Dec 04 2008
Equals Lim_{k>inf.} A071919^k.  Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1xexp(x)), the eigensequence of A007318^(1). Starting with offset 1 = A154108 convolved with (1,2,3,...) = row sums of triangle A154109.  Gary W. Adamson, Jan 04 2009
Repeated iterates of (binomial transform of [1,0,0,0,...]) will converge upon (1, 2, 5, 15, 52,...) when each result is prefaced with a "1"; such that the final result is the fixed limit: (binomial transform of [1,1,2,5,15,...] = (1,2,5,15,52,...).  Gary W. Adamson, Jan 14 2009
From Karol A. Penson, May 03 2009: (Start)
Relation between the Bell numbers B(n) and the nth derivative of 1/Gamma(1+x) of such derivatives through seq(subs(x=0, simplify((d^n/dx^n)GAMMA(1+x)^(1))), n=1..6);
b) leave them expressed in terms of digamma (Psi(k)) and polygamma (Psi(k,n)) functions and unevaluated;
Examples of such expressions, for n=1..5, are:
n=1: Psi(1),
n=2: (Psi(1)^2+Psi(1,1)),
n=3: Psi(1)^3+3*Psi(1)*Psi(1,1)Psi(2,1),
n=4: (Psi(1)^4+6*Psi(1)^2*Psi(1,1)3*Psi(1,1)^24*Psi(1)*Psi(2,1)+Psi(3, 1)),
n=5: Psi(1)^5 +10*Psi(1)^3*Psi(1,1) 15*Psi(1)*Psi(1,1)^2 10*Psi(1)^2*Psi(2,1) +10*Psi(1,1)*Psi(2,1) +5*Psi(1)*Psi(3,1) Psi(4,1);
c) for a given n, read off the sum of absolute values of coefficients of every term involving digamma or polygamma functions.
This sum is equal to B(n). Examples: B(1)=1, B(2)=1+1=2, B(3)=1+3+1=5, B(4)=1+6+3+4+1=15, B(5)=1+10+15+10+10+5+1=52;
d) Observe that this decomposition of the Bell number B(n) apparently does not involve the Stirling numbers of the second kind explicitly.
(End)
The numbers given above by Penson lead to the multinomial coefficients A036040.  Johannes W. Meijer, Aug 14 2009
Column 1 of A162663.  Franklin T. AdamsWatters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n1)!)/(n1)! = a(0) + a(1)/n + a(2)/n^2 + ... and (0!+1!+2!+...+n!)/n! = 1 + a(0)/n + a(1)/n^2 + a(2)/n^3 + ....  Michael Somos, Jun 28 2009
Starting with offset 1 = row sums of triangle A165194.  Gary W. Adamson, Sep 06 2009
a(n+1) = A165196(2^n); where A165196 begins: (1, 2, 4, 5, 7, 12, 14, 15, ...). such that A165196(2^3) = 15 = A000110(4).  Gary W. Adamson, Sep 06 2009
The divergent series g(x=1,m) = 1^m*1!  2^m*2! + 3^m*3!  4^m*4! + ..., m >= 1, which for m=1 dates back to Euler, is related to the Bell numbers. We discovered that g(x=1,m) = (1)^m * (A040027(m)  A000110(m+1) * A073003). We observe that A073003 is Gompertz's constant and that A040027 was published by Gould, see for more information A163940.  Johannes W. Meijer, Oct 16 2009
a(n)= E(X^n), i.e., the nth moment about the origin of a random variable X that has a Poisson distribution with (rate) parameter, lambda = 1.  Geoffrey Critzer, Nov 30 2009
Let A000110 = S(x), then S(x) = A(x)/A(x^2) when A(x) = A173110; or (1, 1, 2, 5, 15, 52, ...) = (1, 1, 3, 6, 20, 60, ...) / (1, 0, 1, 0, 3, 0, 6, 0, 20, ...).  Gary W. Adamson, Feb 09 2010
The Bell numbers serve as the upper limit for the number of distinct homomorphic images from any given finite universal algebra. Every algebra homomorphism is determined by its kernel, which must be a congruence relation. As the number of possible congruence relations with respect to a finite universal algebra must be a subset of its possible equivalence classes (given by the Bell numbers), it follows naturally.  Max Sills, Jun 01 2010
For a proof of the o.g.f. given in the R. Stephan comment see, e.g., the W. Lang link under A071919.  Wolfdieter Lang, Jun 23 2010
Let B(x) = (1 + x + 2x^2 + 5x^3 + ...). Then B(x) is satisfied by A(x)/A(x^2) where A(x) = polcoeff A173110: (1 + x + 3x^2 + 6x^3 + 20x^4 + 60x^5 + ...) = B(x) * B(x^2) * B(x^4) * B(x^8) * ....  Gary W. Adamson, Jul 08 2010
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1) equals the number of ways to choose 0 or more balls of each color without choosing any two colors the same positive number of times. (See related comments for A000108, A008277, A016098.)  Matthew Vandermast, Nov 22 2010
A binary counter with faulty bits starts at value 0 and attempts to increment by 1 at each step. Each bit that should toggle may or may not do so. a(n) is the number of ways that the counter can have the value 0 after n steps. E.g., for n=3, the 5 trajectories are 0,0,0,0; 0,1,0,0; 0,1,1,0; 0,0,1,0; 0,1,3,0.  David Scambler, Jan 24 2011
No Bell number is divisible by 8, and no Bell number is congruent to 6 modulo 8; see Theorem 6.4 and Table 1.7 in Lunnon, Pleasants and Stephens.  Jon Perry, Feb 07 2011, clarified by Eric Rowland, Mar 26 2014
a(n+1) is the number of (symmetric) positive semidefinite n X n 01 matrices. These correspond to equivalence relations on {1,...,n+1}, where matrix element M[i,j] = 1 if and only if i and j are equivalent to each other but not to n+1.  Robert Israel, Mar 16 2011
a(n) is the number of monotoniclabeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotoniclabeled if the label of any parent vertex is greater than the label of any offspring vertex. See link "Counting forests with Stirling and Bell numbers".  Dennis P. Walsh, Nov 11 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000772 and A094198.  Peter Bala, Nov 25 2011
B(n) counts the length n+1 rhyme schemes without repetitions. E.g., for n=2 there are 5 rhyme schemes of length 3 (aaa, aab, aba, abb, abc), and the 2 without repetitions are aba, abc. This is basically O. Munagi's result that the Bell numbers count partitions into subsets of nonconsecutive integers (see comment above dated Mar 20 2005).  Eric Bach, Jan 13 2012
Number n is prime if mod(a(n)2,n) = 0. Dmitry Kruchinin, Feb 14 2012
Right and left borders and row sums of A212431 = A000110 or a shifted variant.  Gary W. Adamson, Jun 21 2012
Number of maps f: [n] > [n] where f(x)<=x and f(f(x))=f(x) (projections).  Joerg Arndt, Jan 04 2013
Permutations of [n] avoiding any given one of the 8 dashed patterns in the equivalence classes (i) 123, 321, 123, 321, and (ii) 132, 312, 213, 231. (See Claesson 2001 reference.)  David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1.  ZhiWei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e1) = 5.57494152476... , see A234473.  Richard R. Forberg, Dec 26 2013 (This is the e.g.f. for x=1.  Wolfdieter Lang, Feb 02 2015)
Sum_{j=0..n} binomial(n,j)*a(j) = (1/e)*Sum_{k>=0} (k+1)^n/k! = (1/e) Sum_{k=1..infinity} k^(n+1)/k! = a(n+1), n >= 0, using the Dobinski formula. See the comment by Gary W. Adamson, Dec 04 2008 on the Pascal eigensequence.  Wolfdieter Lang, Feb 02 2015
In fact it is not really an eigensequence of the Pascal matrix; rather the Pascal matrix acts on the sequence as a shift. It is an eigensequence (the unique eigensequence with eigenvalue 1) of the matrix derived from the Pascal matrix by adding at the top the row [1, 0, 0, 0 ...]. The binomial sum formula may be derived from the definition in terms of partitions: label any element X of a set S of N elements, and let X(k) be the number of subsets of S containing X with k elements. Since each subset has a unique coset, the number of partitions p(N) of S is given by p(N) = Sum_{k=1..N} (X(k) p(Nk)); trivially X(k) = N1 choose k1.  Mason Bogue, Mar 20 2015
a(n) is the number of ways to nest n matryoshkas (Russian nesting dolls): we may identify {1, 2, ..., n} with dolls of ascending sizes and the sets of a set partition with stacks of dolls.  Carlo Sanna, Oct 17 2015
Number of permutations of [n] where the initial elements of consecutive runs of increasing elements are in decreasing order. a(4) = 15: `1234, `2`134, `23`14, `234`1, `24`13, `3`124, `3`2`14, `3`24`1, `34`12, `34`2`1, `4`123, `4`2`13, `4`23`1, `4`3`12, `4`3`2`1.  Alois P. Heinz, Apr 27 2016
Taking with alternating signs, the Bell numbers are the coefficients in the asymptotic expansion (Ramanujan): (1)^n*(A000166(n)  n!/exp(1)) ~ 1/n  2/n^2 + 5/n^3  15/n^4 + 52/n^5  203/n^6 + O(1/n^7).  Vladimir Reshetnikov, Nov 10 2016
Number of treeshelves avoiding pattern T231. See A278677 for definitions and examples.  Sergey Kirgizov, Dec 24 2016
Presumably this satisfies Benford's law, although the results in Hürlimann (2009) do not make this clear.  N. J. A. Sloane, Feb 09 2017
a(n) = Sum(# of standard immaculate tableaux of shape m, m is a composition of n), where this sum is over all integer compositions m of n > 0. This formula is easily seen to hold by identifying standard immaculate tableaux of size n with set partitions of { 1, 2, ..., n }. For example, if we sum over integer compositions of 4 lexicographically, we see that 1+1+2+1+3+3+3+1 = 15 = A000110(4).  John M. Campbell, Jul 17 2017
a(n) is also the number of independent vertex sets (and vertex covers) in the (n1)triangular honeycomb bishop graph.  Eric W. Weisstein, Aug 10 2017


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Index entries for "core" sequences
Index entries for sequences related to juggling
Index entries for sequences related to partitions
Index entries for sequences related to rooted trees
Index entries for sequences related to Benford's law


FORMULA

E.g.f.: exp(exp(x)  1).
Recurrence: a(n+1) = Sum a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n1} (1/(n1)!)*A000166(j)*binomial(n1, j)*(nj)^(n1).  André F. Labossière, Dec 01 2004
G.f.: (Sum_{k=0..infinity} 1/((1k*x)*k!))/exp(1) = hypergeom([ 1/x], [(x1)/x], 1)/exp(1) = ((12*x)+LaguerreL(1/x, (x1)/x, 1)+x*LaguerreL(1/x, (2*x1)/x, 1))*Pi/(x^2*sin(Pi*(2*x1)/x)), where LaguerreL(mu, nu, z) =( GAMMA(mu+nu+1)/GAMMA(mu+1)/GAMMA(nu+1))* hypergeom([ mu], [nu+1], z) is the Laguerre function, the analytic extension of the Laguerre polynomials, for mu not equal to a nonnegative integer. This generating function has an infinite number of poles accumulating in the neighborhood of x=0. Karol A. Penson, Mar 25 2002
a(n) = exp(1)*Sum_{k >= 0} k^n/k! [Dobinski].  Benoit Cloitre, May 19 2002
a(n) is asymptotic to n!*(2 Pi r^2 exp(r))^(1/2) exp(exp(r)1) / r^n, where r is the positive root of r exp(r) = n. See, e.g., the Odlyzko reference.
a(n) is asymptotic to b^n*exp(bn1/2)*sqrt(b/(b+n)) where b satisfies b*log(b) = n  1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493).  Benoit Cloitre, Oct 23 2002, corrected by Vaclav Kotesovec, Jan 06 2013
Lovasz (Combinatorial Problems and Exercises, NorthHolland, 1993, Section 1.14, Problem 9) gives another asymptotic formula, quoted by Mezo and Baricz.  N. J. A. Sloane, Mar 26 2015
G.f.: Sum_{k>=0} x^k/(Product_{j=1..k} 1jx) (see Klazar for a proof).  Ralf Stephan, Apr 18 2004
a(n+1) = exp(1)*Sum_{k>=0} (k+1)^(n)/k!.  Gerald McGarvey, Jun 03 2004
For n>0, a(n) = Aitken(n1, n1) [i.e., a(n1, n1) of Aitken's array (A011971)].  Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (1)^(ki)*binomial(k, i)*i^n+0^n).  Paul Barry, Apr 18 2005
a(n) = A032347(n) + A040027(n+1).  Jon Perry, Apr 26 2005
a(n) = 2*n!/(Pi*e)*Im( integral_{0}^{Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro].  David Callan, Sep 03 2005
O.g.f.: 1/(1xx^2/(12*x2*x^2/(13*x3*x^2/(.../(1n*xn*x^2/(...)))))) (continued fraction due to Ph. Flajolet).  Paul D. Hanna, Jan 17 2006
From Karol A. Penson, Jan 14 2007: (Start)
Representation of Bell numbers B(n), n=1,2..., as special values of hypergeometric function of type (n1)F(n1), in Maple notation: B(n)=exp(1)*hypergeom([2,2,...,2],[1,1,...,1],1), n=1,2..., i.e. having n1 parameters all equal to 2 in the numerator, having n1 parameters all equal to 1 in the denominator and the value of the argument equal to 1.
Examples:
B(1)=exp(1)*hypergeom([],[],1)=1
B(2)=exp(1)*hypergeom([2],[1],1)=2
B(3)=exp(1)*hypergeom([2,2],[1,1],1)=5
B(4)=exp(1)*hypergeom([2,2,2],[1,1,1],1)=15
B(5)=exp(1)*hypergeom([2,2,2,2],[1,1,1,1],1)=52
(Warning: this formula is correct but its application by a computer may not yield exact results, especially with a large number of parameters.)
(End)
a(n+1) = 1 + Sum_{k=0..n1} Sum_{i=0..k} binomial(k,i))*(2^(ki))*a(i).  Yalcin Aktar, Feb 27 2007
a(n) = [1,0,0,...,0] T^(n1) [1,1,1,...,1]', where T is the n X n matrix with main diagonal {1,2,3,...,n}, 1's on the diagonal immediately above and 0's elsewhere. [Meier]
a(n) = ((2*n!)/(Pi * e)) * ImaginaryPart(Integral[from 0 to Pi](e^e^e^(i*theta))*sin(n*theta) dtheta).  Jonathan Vos Post, Aug 27 2007
From Tom Copeland, Oct 10 2007: (Start)
a(n) = T(n,1) = Sum_{j=0...n} S2(n,j) = Sum_{j=0...n} E(n,j) * Lag(n,1,jn) = Sum_{j=0...n} [ E(n,j)/n! ] * [ n!*Lag(n,1, jn) ] where T(n,x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m. Note that E(n,j)/n! = E(n,j) / (Sum_{k=0...n} E(n,k)).
The Eulerian numbers count the permutation ascents and the expression [n!*Lag(n,1, jn)] is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to n!*a(n) = Sum_{j=0..n} E(n,j) * [n!*Lag(n,1, jn)].
(End)
Define f_1(x), f_2(x), ... such that f_1(x)=e^x and for n=2,3,... f_{n+1}(x) = (d/dx)(x*f_n(x)). Then for Bell numbers B_n we have B_n=1/e*f_n(1).  Milan Janjic, May 30 2008
a(n) = (n1)! Sum_{k=1..n} a(nk)/((nk)! (k1)!) where a(0)=1.  Thomas Wieder, Sep 09 2008
a(n+k) = Sum_{m=0..n} Stirling2(n,m) Sum_{r=0..k} binomial(k,r) m^r a(kr).  David Pasino (davepasino(AT)yahoo.com), Jan 25 2009. (Umbrally, this may be written as a(n+k) = Sum_{m=0..n} Stirling2(n,m) (a+m)^k.  N. J. A. Sloane, Feb 07 2009.)
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..ni}...Sum_{k_n=0..1}
delta(k_1,k_2,...,k_i,...,k_n)
where delta(k_1,k_2,...,k_i,...,k_n) = 0 if any k_i > k_(i+1) and k_(i+1) <> 0
and delta(k_1,k_2,...,k_i,...,k_n) = 1 otherwise.
(End)
Let A be the upper Hessenberg matrix of order n defined by: A[i,i1]=1, A[i,j]:=binomial(j1,i1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A).  Milan Janjic, Jul 08 2010
G.f. satisfies A(x)=(x/(1x))*A(x/(1x)) + 1.  Vladimir Kruchinin, Nov 28 2011
G.f.: 1 / (1  x / (1  1*x / (1  x / (1  2*x / (1  x / (1  3*x / ... )))))).  Michael Somos, May 12 2012
a(n+1) = Sum_{m=0..n} Stirling2(n, m)*(m+1), n >= 0. Compare with the third formula for a(n) above. Here Stirling2 = A048993.  Wolfdieter Lang, Feb 03 2015
G.f.: (1)^(1/x)*((1/x)!/e + (!(11/x))/x) where z! and !z are factorial and subfactorial generalized to complex arguments.  Vladimir Reshetnikov, Apr 24 2013
The following formulas were proposed during the period Dec 2011  Oct 2013 by Sergei N. Gladkovskii. (Start)
E.g.f.: exp(exp(x)1) = 1+x/(G(0)x); G(k) = (k+1)*Bell(k)+x*Bell(k+1)x*(k+1)*Bell(k)*Bell(k+2)/G(k+1) (continued fraction).
G.f.: W(x)=(11/(G(0)+1))/exp(1) ; G(k)= x*k^2 + (3*x1)*k  2 + x  (k+1)*(x*k+x1)^2/G(k+1); (continued fraction Euler's kind, 1step).
G.f.: W(x)= (1 + G(0)/(x^23*x+2))/exp(1); G(k)= 1 (x*k+x1)/( ((k+1)!) (((k+1)!)^2)*(1xk*x+(k+1)!)/( ((k+1)!)*(1xk*x+(k+1)!)  (x*k+2*x1)*(12*xk*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x)= 1/(1  x/(1x/(1 + x/G(0)))); G(k)= x  1 + x*k + x*(x1+x*k)/G(k+1); (continued fraction, 1step).
G.f.: 1/U(0) where U(k)= x*k  1 + x  x^2*(k+1)/U(k+1); (continued fraction, 1step).
G.f.: 1 + x/U(0) where U(k) = 1  x*(k+2)  x^2*(k+1)/U(k+1); (continued fraction, 1step).
G.f.: 1 + 1/(U(0)  x) where U(k) = 1 + x  x*(k+1)/(1  x/U(k+1)); (continued fraction, 2step).
G.f.: 1 + x/(U(0)x) where U(k) = 1  x*(k+1)/(1  x/U(k+1)); (continued fraction, 2step).
G.f.: 1/G(0) where G(k) = 1  x/(1  x*(2*k+1)/(1  x/(1  x*(2*k+2)/G(k+1) ))); (continued fraction).
G.f.: G(0)/(1+x) where G(k) = 1  2*x*(k+1)/((2*k+1)*(2*x*k1)  x*(2*k+1)*(2*k+3)*(2*x*k1)/(x*(2*k+3)  2*(k+1)*(2*x*k+x1)/G(k+1) )); (continued fraction).
G.f.: (1+2*x) * Sum(k >= 0, x^(2*k)*(4*x*k^22*k2*x1) / ((2*k+1) * (2*x*k1)) * A(k) / B(k) where A(k) = prod(p=0...k, (2*p+1)), B(k) = prod(p=0..k, (2*p1) * (2*x*px1) * (2*x*p2*x1)).
G.f.: (G(0)  1)/(x1) where G(k) = 1  1/(1k*x)/(1x/(x1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S1) where S=sum(k>=0, ( 1 + (1x)/(1xx*k) )*(x/(1x))^k/prod(i=0..k1, (1xx*i)/(1x) ) ).
G.f.: (G(0)  2)/(2*x1) where G(k) = 2  1/(1k*x)/(1x/(x1/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 1  (x*k  2)/(x*k  1  x*(x*k  1)/(x + (x*k  2)/G(k+1) )); (continued fraction).
G.f.: G(0) where G(k) = 2  (2*x*k  1)/(x*k  1  x*(x*k  1)/(x + (2*x*k  1)/G(k+1) )); (continued fraction).
G.f.: (G(0)  1)/(1+x) where G(k) = 1 + 1/(1k*x)/(1x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1x)*G(0))  1/x where G(k) = 1  x/(x  1/(1 + 1/(x*k1)/G(k+1) )); (continued fraction).
G.f.: 1 + x/( Q(0)  x ) where Q(k) = 1 + x/( x*k  1 )/Q(k+1); (continued fraction).
G.f.: 1+x/Q(0), where Q(k)= 1  x  x/(1  x*(k+1)/Q(k+1)); (continued fraction).
G.f.: 1/(1x*Q(0)), where Q(k)= 1 + x/(1  x + x*(k+1)/(x  1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1x), where Q(k) = 1  x^2*(k+1)/( x^2*(k+1)  (1x*(k+1))*(1x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))n1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert Wfunction.  Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/LambertW(n)1n) / (sqrt(1+LambertW(n)) * LambertW(n)^n).  Vaclav Kotesovec, Nov 13 2015
a(n) are the coefficients in the asymptotic expansion of exp(1)*(1)^x*x*Gamma(x,0,1), where Gamma(a,z0,z1) is the generalized incomplete Gamma function.  Vladimir Reshetnikov, Nov 12 2015
a(n) = 1 + floor(exp(1) * Sum_{k=1..2*n} k^n/k!).  Vladimir Reshetnikov, Nov 13 2015
a(p^m) ≡ m+1 (mod p) when p is prime and m >= 1 (see Lemma 3.1 in the Hurst/Schultz reference).  Seiichi Manyama, Jun 01 2016
Sum_{n>=0} (1)^n*a(n)/n! = exp(exp(1)1).  Ilya Gutkovskiy, Jun 01 2016


EXAMPLE

G.f. = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 203*x^6 + 877*x^7 + 4140*x^8 + ...
From Neven Juric, Oct 19 2009: (Start)
The a(4)=15 rhyme schemes for n=4 are
aaaa, aaab, aaba, aabb, aabc, abaa, abab, abac, abba, abbb, abbc, abca, abcb, abcc, abcd
The a(5)=52 rhyme schemes for n=5 are
aaaaa, aaaab, aaaba, aaabb, aaabc, aabaa, aabab, aabac, aabba, aabbb, aabbc, aabca, aabcb, aabcc, aabcd, abaaa, abaab, abaac, ababa, ababb, ababc, abaca, abacb, abacc, abacd, abbaa, abbab, abbac, abbba, abbbb, abbbc, abbca, abbcb, abbcc, abbcd, abcaa, abcab, abcac, abcad, abcba, abcbb, abcbc, abcbd, abcca, abccb, abccc, abccd, abcda, abcdb, abcdc, abcdd, abcde
(End)
From Joerg Arndt, Apr 30 2011: (Start)
Restricted growth strings (RGS):
For n=0 there is one empty string;
for n=1 there is one string [0];
for n=2 there are 2 strings [00], [01];
for n=3 there are a(3)=5 strings [000], [001], [010], [011], and [012];
for n=4 there are a(4)=15 strings
1: [0000], 2: [0001], 3: [0010], 4: [0011], 5: [0012], 6: [0100], 7: [0101], 8: [0102], 9: [0110], 10: [0111], 11: [0112], 12: [0120], 13: [0121], 14: [0122], 15: [0123].
These are onetoone with the rhyme schemes (identify a=0, b=1, c=2, etc.).
(End)
Consider the set S = {1, 2, 3, 4}. The a(4) = 1 + 3 + 6 + 4 + 1 = 15 partitions are: P1 = {{1}, {2}, {3}, {4}}; P21 .. P23 = {{a,4}, S\{a,4}} with a = 1, 2, 3; P24 .. P29 = {{a}, {b}, S\{a,b}} with 1 <= a < b <= 4; P31 .. P34 = {S\{a}, {a}} with a = 1 .. 4; P4 = {S}. See the Bottomley link for a graphical illustration.  M. F. Hasler, Oct 26 2017


MAPLE

A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n1, i)*A000110(n1i), i=0..n1); fi; end; # version 1
A := series(exp(exp(x)1), x, 60); A000110 := n>n!*coeff(A, x, n); # version 2
with(combinat); A000110:=n>sum(stirling2(n, k), k=0..n): seq(A000110(n), n=1..22); # version 3, from Zerinvary Lajos, Jun 28 2007
A000110 := n > combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
a:=array(0..200); a[0]:=1; a[1]:=1; lprint(0, 1); lprint(1, 1); M:=200; for n from 2 to M do a[n]:=add(binomial(n1, i)*a[n1i], i=0..n1); lprint(n, a[n]); od:
with(combstruct); spec := [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]; [seq(combstruct[count](spec, size=n), n=0..40)]; G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22); # Zerinvary Lajos, Dec 16 2007
A000110 := proc(n::integer) local k, Resultat; if n = 0 then Resultat:=1: return Resultat; end if; Resultat:=0: for k from 1 to n do Resultat:=Resultat+A000110(nk)/((nk)!*(k1)!): od; Resultat:=Resultat*(n1)!; return Resultat; end proc; # Thomas Wieder, Sep 09 2008


MATHEMATICA

f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 21}] (* Robert G. Wilson v *)
Table[BellB[n], {n, 0, 26}] (* Harvey P. Dale, Mar 01 2011 *)
B[0] = 1; B[n_] := 1/E Sum[k^(n  1)/(k1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
BellB[Range[0, 26]] (* Eric W. Weisstein, Aug 10 2017 *)


PROG

(PARI) {a(n) = my(m); if( n<0, 0, m = contfracpnqn( matrix(2, n\2, i, k, if( i==1, k*x^2, 1  (k+1)*x))); polcoeff(1 / (1  x + m[2, 1] / m[1, 1]) + x * O(x^n), n))}; /* Michael Somos */
(PARI) {a(n) = polcoeff( sum( k=0, n, prod( i=1, k, x / (1  i*x)), x^n * O(x)), n)}; /* Michael Somos, Aug 22 2004 */
(PARI) a(n)=round(exp(1)*suminf(k=0, 1.0*k^n/k!)) \\ Gottfried Helms, Mar 30 2007  WARNING! For illustration only: Gives silently a wrong result for n = 42 and an error for n > 42, with standard precision of 38 digits.  M. F. Hasler, Nov 30 2018
(PARI) {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n))  1), n))}; /* Michael Somos, Jun 28 2009 */
(PARI) Vec(serlaplace(exp(exp('x+O('x^66))1))) \\ Joerg Arndt, May 26 2012
(PARI) A000110(n)=sum(k=0, n, stirling(n, k, 2)) \\ M. F. Hasler, Nov 30 2018
(Sage) from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
(Python) # The objective of this implementation is efficiency.
# m > [a(0), a(1), ..., a(m)] for m > 0.
def A000110_list(m):
....A = [0 for i in range(0, m)]
....A[0] = 1
....R = [1, 1]
....for n in range(1, m):
........A[n] = A[0]
........for k in range(n, 0, 1):
............A[k1] += A[k]
........R.append(A[0])
....return R
A000110_list(100) # example call  Peter Luschny, Jan 18 2011
(MAGMA) [Bell(n): n in [0..80]]; // Vincenzo Librandi, Feb 07 2011
(Maxima) makelist(belln(n), n, 0, 80); /* Emanuele Munarini, Jul 04 2011 */
(Haskell)
type N = Integer
n_partitioned_k :: N > N > N
1 `n_partitioned_k` 1 = 1
1 `n_partitioned_k` _ = 0
n `n_partitioned_k` k = k * (pred n `n_partitioned_k` k) + (pred n `n_partitioned_k` pred k)
n_partitioned :: N > N
n_partitioned 0 = 1
n_partitioned n = sum $ map (\k > n `n_partitioned_k` k) $ [1 .. n]
 Felix Denis, Oct 16 2012
(Haskell)
a000110 = sum . a048993_row  Reinhard Zumkeller, Jun 30 2013
(Python)
# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A000110, blist, b = [1, 1], [1], 1
for _ in range(20):
....blist = list(accumulate([b]+blist))
....b = blist[1]
....A000110.append(b) # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 19 2014


CROSSREFS

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Cf. A000045, A000108, A000166, A000204, A000255, A000311, A000296, A003422, A024716, A029761, A049020, A058692, A060719, A084423, A087650, A094262, A103293, A165194, A165196, A173110, A227840.
Equals row sums of triangle A152432.
Row sums, right and left borders of A212431.
A diagonal of A011971.  N. J. A. Sloane, Jul 31 2012
Cf. A054767 (period of this sequence mod n).
Row sums are A048993.  Wolfdieter Lang, Oct 16 2014
Sequences in the Erné (1974) paper: A000110, A000798, A001035, A001927, A001929, A006056, A006057, A006058, A006059.
Bell polynomials B(n,x): A001861 (x=2), A027710 (x=3), A078944 (x=4), A144180 (x=5), A144223 (x=6), A144263 (x=7), A221159 (x=8).
Cf. A243991 (sum of reciprocals).
Sequence in context: A203645 A203646 A292935 * A303924 A186001 A134381
Adjacent sequences: A000107 A000108 A000109 * A000111 A000112 A000113


KEYWORD

core,nonn,easy,nice,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

Edited by M. F. Hasler, Nov 30 2018


STATUS

approved



