0,3

Number of partitions of a set of n labeled elements.

a(n-1) = number of nonisomorphic colorings of a map consisting of a row of n+1 adjacent regions. - David W. Wilson, Feb 22, 2005

If an integer is square-free 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 (amarnath_murthy(AT)yahoo.com), 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 (perry(AT)globalnet.co.uk), 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],[1,1],[1,2]=15, etc... - Jon Perry (perry(AT)globalnet.co.uk), Mar 05 2004

Number of distinct rhyme schemes for a poem of n lines: a rhyme scheme is a string of letters (eg, '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. - Bill Blewett (BillBle(AT)microsoft.com), Mar 23 2004

Asymptotic expansion of (0!+1!+2!+...+n!)/n! = a(0)+a(1)/n+a(2)/n^2+... - Michael Somos, Aug 22 2004

Also the number of equivalence relations in (alternatively, or the number of partitions of) a set of n elements. - Federico Arboleda (federico.arboleda(AT)gmail.com), Mar 09 2005

Number of partitions of {1, ...,n+1} into subsets of nonconsecutive integers, including the partition 1|2|...|n+1. E.g. a(3)=5: there are 5 partitions of {1,2,3,4} into subsets of nonconsecutive integers namely 13|24, 13|2|4, 14|2|3, 1|24|3, 1|2|3|4. - A. O. Munagi (amunagi(AT)yahoo.com), 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 ... [This is Aitken's array A011971]

With p(n) = the number of integer partitions of n, p(i) = the number of parts of the i-th partition of n, d(i) = the number of different parts of the i-th partition of n, p(j,i) = the j-th part of the i-th partition of n, m(i,j) = multiplicity of the j-th part of the i-th partition of n, sum_{i=1}^{p(n)} = sum over i, and prod_{j=1}^{d(i)} = product over j one has: a(n)=sum_{i=1}^{p(n)} (n!/(prod_{j=1}^{p(i)}p(i,j)!)) * (1/(prod_{j=1}^{d(i)} m(i,j)!)) - Thomas Wieder (wieder.thomas(AT)t-online.de), May 18 2005

a(n+1) = the number of binary relations on an n-element set that are both symmetric and transitive. - Justin Witt (justinmwitt(AT)gmail.com), Jul 12 2005

If Jon Perry's rule is used, i.e. "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],[1,1],[1,2]=15, etc..." then a(n-1) = [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 (dwalsh(AT)mtsu.edu), Feb 20 2006

Number of asynchronic siteswap patterns of length n which have no zero-throws (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 (his-firstname.his-surname(AT)gmail.com), May 01 2006.

a(n) = number of permutations on [n] in which a 3-2-1 (scattered) pattern occurs only as part of a 3-2-4-1 pattern. Example. a(3) = 5 counts all permutations on [3] except 321. See "Eigensequence for Composition" ref. 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 (callan(AT)stat.wisc.edu), 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

Comment from Gottfried Helms (helms(AT)uni-kassel.de), Mar 30 2007. (Start) This sequence is also the first column in the matrix-exponential of the (lower triangular) Pascal-matrix, 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

..877...1421...1092...525...175...42

First 4 columns are A000110, A033306, A105479, A105480. The general case is mentioned in the two latter entries. PE is also the Hadamard-product 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

..877...203...52...15...5...2 (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 identity-matrix of appropriate dimension since (P-I) is nilpotent to the order of its dimension. Then a(n)=A[n,1] where n is the row-index starting at 1. - Gottfried Helms helms(at)uni-kassel.de, Apr 10 2007.

Comment from David W. Wilson (davidwwilson(AT)comcast.net), Aug 04 2007 and Sep 24 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.

This sequence and A000587 form a reciprocal pair under the list partition transform described in A133314. - Tom Copeland (tcjpn(AT)msn.com), 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 (qntmpkt(AT)yahoo.com), Jan 21 2008

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Index entries for sequences related to rooted trees

Index entries for "core" sequences

Index entries for sequences related to juggling

Index entries for related partition-counting sequences

E.g.f.: exp (exp(x)- 1). Recurrence: a(n+1) = Sum a(k)C(n, k). Also a(n) = Sum Stirling2(n, k), k=1..n.

a(n) = SUM(j = 0 to n-1, (1/(n-1)!) * A000166(j) * C(n-1, j) * (n-j)^(n-1)). - Andre F. Labossiere (boronali(AT)laposte.net), Dec 01 2004

G.f.: sum(1/((1-k*x)*k!), k = 0 .. infinity)/exp(1) = hypergeom([ -1/x], [(x-1)/x], 1)/exp(1)=((1-2*x)+LaguerreL(1/x, (x-1)/x, 1)+x*LaguerreL(1/x, (2*x-1)/x, 1))*Pi/(x^2*sin(Pi*(2*x-1)/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 (penson(AT)lptl.jussieu.fr), Mar 25, 2002.

a(n) = exp(-1)*sum(k=>0, k^n/k!) [Dobinski] - Benoit Cloitre (benoit7848c(AT)orange.fr), 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(b-n-1/2)/sqrt(ln(n)) where b satisfies b*ln(b) = n - 1/2 (see Graham, Knuth and Patashnik, Concrete Mathematics, 2nd ed., p. 493) - Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 23 2002

G.f.: sum{k>=0, x^k/prod[l=1..k, 1-lx]}. - R. Stephan, Apr 18 2004

a(n+1) = exp(-1)*sum(k=>0, (k+1)^(n)/k!) - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Jun 03 2004

For n>0, a(n) = Aitken(n-1, n-1) [i.e. a(n-1, n-1) of Aiken's array (A011971)] - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Jun 26 2004

Bell(n) = n + 2*C(n-2, 1) + 6*C(n-3, 1) + C(n-2, 2) + 14*C(n-4, 1) + 12*C(n-3, 2) + 30*C(n-5, 1) + 61*C(n-4, 2) + 10*C(n-3, 3) + 62*C(n-6, 1) + 240*C(n-5, 2) + 124*C(n-4, 3) + 3*C(n-3, 4) + 126*C(n-7, 1) + 841*C(n-6, 2) + 890*C(n-5, 3) + 131*C(n-4, 4) + 254*C(n-8, 1) + 2772*C(n-7, 2) + 5060*C(n-6, 3) + 1830*C(n-5, 4) + 70*C(n-4, 5) + 510*C(n-9, 1) + 8821*C(n-8, 2) + 25410*C(n-7, 3) + 16990*C(n-6, 4) + 2226*C(n-5, 5) + 15*C(n-4, 6) + 1022*C(n-10, 1) + ..... . - Andre F. Labossiere (boronali(AT)laposte.net), Feb 22 2005

a(n)=sum{k=1..n, (1/k!)*sum{i=1..k, (-1)^(k-i)*binomial(k, i)*i^n}}+0^n; - Paul Barry (pbarry(AT)wit.ie), Apr 18 2005

a(n) = A032347(n) + A040027(n+1) - Jon Perry (perry(AT)globalnet.co.uk), 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 (callan(AT)stat.wisc.edu), Sep 03 2005

O.g.f.: A(x) = 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(1-... -n*x-n*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna (pauldhanna(AT)juno.com), Jan 17 2006

Representation of Bell numbers B(n), n=1,2..., as special values of hypergeometric function of type (n-1)F(n-1), in Maple notation: B(n)=exp(-1)*hypergeom([2,2...2],[1,1...1],1), n=1,2..., i.e. having n-1 parameters all equal to 2 in the numerator, having n-1 parameters all equal to 1 in the denominator and the value of the argument equal to 1. Examples: B(1)=evalf(exp(-1)*hypergeom([],[],1))= 1 B(2)=evalf(exp(-1)*hypergeom([2],[1],1))= 2 B(3)=evalf(exp(-1)*hypergeom([2,2],[1,1],1))= 5 B(4)=evalf(exp(-1)*hypergeom([2,2,2],[1,1,1],1))= 15 B(5)=evalf(exp(-1)*hypergeom([2,2,2,2],[1,1,1,1],1))=52 - Karol A. Penson (penson(AT)lptl.jussieu.fr), Jan 14 2007

a(n+1)=e+sum(sum((binomial(k,i))*(2^(k-i))*(a(i)),i=0..k),k=0..n-1) - Yalcin Aktar (aktaryalcin(AT)msn.com), Feb 27 2007

a(n) = [1,0,0,...,0] T^(n-1) [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 (jvospost2(AT)yahoo.com), Aug 27 2007

Formulae and comments 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,j-n) = sum(j=0,...,n) [ E(n,j)/n! ] * [ n!*Lag(n,-1, j-n) ] 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 permuatation ascents and the expression [n!*Lag(n,-1, j-n)] 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, j-n)]} . (End)

A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1, i)*A000110(n-1-i), i=0..n-1); fi; end; # version 1

A := series(exp(exp(x)-1), x, 60); A000110 := n->n!*coeff(A, x, n); # version 2

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(n-1, i)*a[n-1-i], i=0..n-1); 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)];

a:=n->sum(stirling2(n, k), k=0..n): seq(a(n), n=1..22); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 28 2007

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 (zerinvarylajos(AT)yahoo.com), Dec 16 2007

f[n_] := Sum[ StirlingS2[n, k], {k, 1, n}]; Table[ f[n], {n, 0, 21}] (from Robert G. Wilson v)

(PARI) a(n)=if(n<0, 0, n!*polcoeff(exp(exp(x+x*O(x^n))-1), n)) (from Michael Somos)

(PARI) a(n)=local(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)) (from 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)=exp(-1)*suminf(k=0, 1.0*k^(n-1)/k!) - Gottfried Helms, Mar 30 2007

(PARI) m=matpascal(5)-matid(6); pe=matid(6)+m/1! + m^2/2!+m^3/3!+m^4/4!+m^5/5! ; a(n) = pe[n, 1] - Gottfried Helms, Apr 10 2007

Partial sums give A005001. Cf. A049020. See A061462 for powers of 2 dividing a(n).

Right-most diagonal of triangle A121207.

Cf. A000311, A103293, A084423.

Cf. A005001, A087650, A029761, A024716, A000296, A058692, A060719.

Cf. A008277, A000166, A000255, A000108, A000045, A000204.

Cf. A094262, A008277, A005001, A003422, A000166, A000204, A000045, A000108.

a(n) = A123158(n,0).

Adjacent sequences: A000107 A000108 A000109 this_sequence A000111 A000112 A000113

Sequence in context: A099262 A108305 A099263 this_sequence A134381 A107589 A006790

core,nonn,easy,nice

njas