cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A194689 a(n) = Sum_{k=0..n} binomial(n,k)*w(k)*w(n-k) where w() = A000296().

Original entry on oeis.org

1, 0, 2, 2, 14, 42, 222, 1066, 6078, 36490, 238046, 1653610, 12214270, 95361866, 784071966, 6764984362, 61066919230, 575200190986, 5640081557598, 57450510336234, 606773139773054, 6633515763375306, 74950634205257630, 873995513192234410, 10504736507220958142, 129983468625156713354
Offset: 0

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Author

N. J. A. Sloane, Sep 01 2011

Keywords

References

  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.5 (p. 771, Problem 37).

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n-k) * Binomial[n,k] * BellB[k,2] * 2^(n-k), {k, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jun 25 2022 *)
  • PARI
    N=66;  x='x+O('x^N);
    Q(k) = if (k>N, 1, 1 + x + x*k - x/(1 - 2*x*(k+1)/Q(k+1) ) );
    gf=1/Q(0);  Vec(Ser(gf))
    /* Joerg Arndt, Mar 07 2013 */
    
  • PARI
    my(N=66, x='x+O('x^N)); Vec(serlaplace(exp(2*(exp(x)-1-x)))) \\ Seiichi Manyama, Nov 20 2020

Formula

G.f.: 1/Q(0) where Q(k) = 1 + x + x*k - x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 07 2013
G.f.: 1/Q(0), where Q(k)= 1 - x*k - 2*x^2*(k+1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 06 2013
E.g.f.: exp(2*(exp(x) - 1 - x)). - Ilya Gutkovskiy, Apr 07 2018
a(0) = 1; a(n) = 2 * Sum_{k=1..n-1} binomial(n-1,k) * a(n-1-k). - Seiichi Manyama, Nov 20 2020
a(n) ~ 4 * n^(n-2) * exp(n/LambertW(n/2) - n - 2) / (sqrt(1 + LambertW(n/2)) * LambertW(n/2)^(n-2)). - Vaclav Kotesovec, Jun 26 2022

A113059 a(n) = n! * Sum_{k=0..n} A000296(k)/k!.

Original entry on oeis.org

1, 1, 3, 10, 44, 231, 1427, 10151, 81923, 740732, 7425042, 81773715, 981864897, 12767876941, 178774288331, 2681781213130, 42909715480460, 729474427239587, 13130613291110603, 249482261007109579, 4989650444408388515, 104782705832468197252, 2305219956684224457858
Offset: 0

Views

Author

Karol A. Penson, Oct 12 2005

Keywords

Comments

Number of set partitions of [n] where the k-th singletons are k-colored and all other blocks are unicolored. - Alois P. Heinz, Apr 29 2025

Crossrefs

Programs

  • Magma
    m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(Exp(x)-1-x)/(1-x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, May 23 2018
  • Maple
    b:= proc(n, k) option remember; `if`(n=0, 1, k*b(n-1, k+1)+
          add(b(n-j, k)*binomial(n-1, j-1), j=2..n))
        end:
    a:= n-> b(n, 1):
    seq(a(n), n=0..22);  # Alois P. Heinz, Apr 29 2025
    # second Maple program:
    b:= proc(n, k, m) option remember; `if`(n=0, k!, `if`(k>0,
          b(n-1, k-1, m+1)*k, 0)+m*b(n-1, k, m)+b(n-1, k+1, m))
        end:
    a:= n-> b(n, 0$2):
    seq(a(n), n=0..22);  # Alois P. Heinz, Apr 29 2025
  • Mathematica
    With[{nmax = 50}, CoefficientList[Series[Exp[Exp[x] - 1 - x]/(1 - x), {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, May 23 2018 *)
  • PARI
    x='x+O('x^30); Vec(serlaplace( exp(exp(x)-1-x)/(1-x))) \\ G. C. Greubel, May 23 2018
    

Formula

a(n) = (-1)^n*n!*Sum_{k >=0} LaguerreL(n, -n-1, k-1)/k!/exp(1), n>=0.
E.g.f.: exp(exp(x)-1-x)/(1-x).
a(n) ~ exp(exp(1)-2) * n!. - Vaclav Kotesovec, Jun 26 2022

A355137 Indices of primes in the sequence A000296.

Original entry on oeis.org

5, 6, 41, 119, 5118
Offset: 1

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Author

Vaclav Kotesovec, Jun 20 2022

Keywords

Comments

The next term, if it exists, is > 20000.

Examples

			5 is in the sequence because A000296(5) = 11 is prime.
		

Crossrefs

Programs

  • Mathematica
    s = 1; A000296 = Table[s = BellB[n] - s, {n, 0, 1000}]; Select[Range[1000], PrimeQ[A000296[[#]]]&]

A372525 Composite numbers k that divide A000296(k) + (-1)^k.

Original entry on oeis.org

6, 1991, 5459, 17391
Offset: 1

Views

Author

Amiram Eldar, Jun 07 2024

Keywords

Comments

Equivalently, composite numbers k such that Sum_{i=1..k-1} (-1)^i * Bell(i) == 1 (mod k), where Bell(i) = A000110(i).
The congruence holds for all the primes. This is a special case of Sun and Zagier's (2011) Theorem 1.1, when m = 1.
a(5) > 56000, if it exists.

Crossrefs

Programs

  • Mathematica
    f[k_] := f[k] = Sum[Binomial[k-1, i] * f[k-i-1], {i, 1, k-1}]; f[0] = 1; Select[Range[2000], CompositeQ[#] && Divisible[f[#] + (-1)^# , #] &]

A000110 Bell or exponential numbers: number of ways to partition a set of n labeled elements.

Original entry on oeis.org

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
Offset: 0

Views

Author

Keywords

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 length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] 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 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. - 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 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, 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 n-element 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(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, 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, May 01 2006
a(n) is the 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" 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
From Gottfried Helms, 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 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 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 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 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, Apr 10 2007
When n is prime, a(n) == 2 (mod n), but the converse is not always true. 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 order-decreasing full transformations (of an n-chain).
a(n) is also the number of nilpotent partial one-one order-decreasing transformations (of an n-chain).
a(n+1) is also the number of partial one-one order-decreasing transformations (of an n-chain). (End)
From Peter Bala, Oct 19 2008: (Start)
Bell(n) is the number of n-pattern sequences [Cooper & Kennedy]. An n-pattern 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 3-pattern 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 lower-triangular 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->oo} A071919^k. - Gary W. Adamson, Jan 02 2009
Equals A154107 convolved with A014182, where A014182 = expansion of exp(1-x-exp(-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 n-th derivative of 1/Gamma(1+x) evaluated at x=1:
a) produce a number of such derivatives through seq(subs(x=1, simplify((d^n/dx^n)GAMMA(1+x)^(-1))), n=1..5);
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)^2 - 4*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. Adams-Watters, Jul 09 2009
Asymptotic expansions (0!+1!+2!+...+(n-1)!)/(n-1)! = 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 n-th 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 0-1 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 monotonic-labeled forests on n vertices with rooted trees of height less than 2. We note that a labeled rooted tree is monotonic-labeled 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
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) 1-23, 3-21, 12-3, 32-1, and (ii) 1-32, 3-12, 21-3, 23-1. (See Claesson 2001 reference.) - David Callan, Oct 03 2013
Conjecture: No a(n) has the form x^m with m > 1 and x > 1. - Zhi-Wei Sun, Dec 02 2013
Sum_{n>=0} a(n)/n! = e^(e-1) = 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..oo} 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(N-k)); trivially X(k) = N-1 choose k-1. - 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 (n-1)-triangular honeycomb bishop graph. - Eric W. Weisstein, Aug 10 2017
Even-numbered entries represent the numbers of configurations of identity and non-identity for alleles of a gene in n diploid individuals with distinguishable maternal and paternal alleles. - Noah A Rosenberg, Jan 28 2019
Number of partial equivalence relations (PERs) on a set with n elements (offset=1), i.e., number of symmetric, transitive (not necessarily reflexive) relations. The idea is to add a dummy element D to the set, and then take equivalence relations on the result; anything equivalent to D is then removed for the partial equivalence relation. - David Spivak, Feb 06 2019
Number of words of length n+1 with no repeated letters, when letters are unlabeled. - Thomas Anton, Mar 14 2019
Named by Becker and Riordan (1948) after the Scottish-American mathematician and writer Eric Temple Bell (1883 - 1960). - Amiram Eldar, Dec 04 2020
Also the number of partitions of {1,2,...,n+1} with at most one n+1 singleton. E.g., a(3)=5: {13|24, 12|34, 123|4, 14|23, 1234}. - Yuchun Ji, Dec 21 2020
a(n) is the number of sigma algebras on the set of n elements. Note that each sigma algebra is generated by a partition of the set. For example, the sigma algebra generated by the partition {{1}, {2}, {3,4}} is {{}, {1}, {2}, {1,2}, {3,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}. - Jianing Song, Apr 01 2021
a(n) is the number of P_3-free graphs on n labeled nodes. - M. Eren Kesim, Jun 04 2021
a(n) is the number of functions X:([n] choose 2) -> {+,-} such that for any ordered 3-tuple abc we have X(ab)X(ac)X(bc) not in {+-+,++-,-++}. - Robert Lauff, Dec 09 2022
From Manfred Boergens, Mar 11 2025: (Start)
The partitions in the definition can be described as disjoint covers of the set. "Covers" in general give rise to the following amendments:
For disjoint covers which may include one empty set see A186021.
For arbitrary (including non-disjoint) covers see A003465.
For arbitrary (including non-disjoint) covers which may include one empty set see A000371. (End)

Examples

			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 one-to-one 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
		

References

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  • J. Balogh, B. Bollobas and D. Weinreich, A jump to the Bell numbers for hereditary graph properties, J. Combin. Theory Ser. B 95 (2005), no. 1, 29-48.
  • R. E. Beard, On the coefficients in the expansion of exp(exp(t)) and exp(-exp(t)), J. Institute Actuaries, 76 (1951), 152-163.
  • H. W. Becker, Abstracts of two papers related to Bell numbers, Bull. Amer. Math. Soc., 52 (1946), p. 415.
  • E. T. Bell, The iterated exponential numbers, Ann. Math., 39 (1938), 539-557.
  • C. M. Bender, D. C. Brody and B. K. Meister, Quantum Field Theory of Partitions, J. Math. Phys., 40,7 (1999), 3239-45.
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  • John H. Conway et al., The Symmetries of Things, Peters, 2008, p. 207.
  • Colin Defant, Highly sorted permutations and Bell numbers, ECA 1:1 (2021) Article S2R6.
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  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., p. 493.
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  • Levinson, H.; Silverman, R. Topologies on finite sets. II. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 699--712, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561090 (81c:54006)
  • S. Linusson, The number of M-sequences and f-vectors, Combinatorica, 19 (1999), 255-266.
  • L. Lovasz, Combinatorial Problems and Exercises, North-Holland, 1993, pp. 14-15.
  • M. Meier, On the number of partitions of a given set, Amer. Math. Monthly, 114 (2007), p. 450.
  • Merris, Russell, and Stephen Pierce. "The Bell numbers and r-fold transitivity." Journal of Combinatorial Theory, Series A 12.1 (1972): 155-157.
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  • Amarnath Murthy and Charles Ashbacher, Generalized Partitions and Some New Ideas on Number Theory and Smarandache Sequences, Hexis, Phoenix; USA 2005. See Section 1.4,1.8.
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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.
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
Diagonal of A102661. - Manfred Boergens, Mar 11 2025
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), A085686 (inv. Euler Transf.).

Programs

  • 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
    
  • Julia
    function a(n)
        t = [zeros(BigInt, n+1) for _ in 1:n+1]
        t[1][1] = 1
        for i in 2:n+1
            t[i][1] = t[i-1][i-1]
            for j in 2:i
                t[i][j] = t[i-1][j-1] + t[i][j-1]
            end
        end
        return [t[i][1] for i in 1:n+1]
    end
    print(a(28)) # Paul Muljadi, May 07 2024
    
  • Magma
    [Bell(n): n in [0..40]]; // Vincenzo Librandi, Feb 07 2011
    
  • Maple
    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
    A000110:= n-> add(Stirling2(n, k), k=0..n): seq(A000110(n), n=0..22); # version 3, from Zerinvary Lajos, Jun 28 2007
    A000110 := n -> combinat[bell](n): # version 4, from Peter Luschny, Mar 30 2011
    spec:= [S, {S=Set(U, card >= 1), U=Set(Z, card >= 1)}, labeled]: G:={P=Set(Set(Atom, card>0))}: combstruct[gfsolve](G, unlabeled, x): seq(combstruct[count]([P, G, labeled], size=i), i=0..22);  # version 5, Zerinvary Lajos, Dec 16 2007
    BellList := proc(m) local A, P, n; A := [1, 1]; P := [1]; for n from 1 to m - 2 do
    P := ListTools:-PartialSums([A[-1], op(P)]); A := [op(A), P[-1]] od; A end: BellList(29); # Peter Luschny, Mar 24 2022
  • Mathematica
    f[n_] := Sum[ StirlingS2[n, k], {k, 0, n}]; Table[ f[n], {n, 0, 40}] (* Robert G. Wilson v *)
    Table[BellB[n], {n, 0, 40}] (* Harvey P. Dale, Mar 01 2011 *)
    B[0] = 1; B[n_] := 1/E Sum[k^(n - 1)/(k-1)!, {k, 1, Infinity}] (* Dimitri Papadopoulos, Mar 10 2015, edited by M. F. Hasler, Nov 30 2018 *)
    BellB[Range[0,40]] (* Eric W. Weisstein, Aug 10 2017 *)
    b[1] = 1; k = 1; Flatten[{1, Table[Do[j = k; k += b[m]; b[m] = j;, {m, 1, n-1}]; b[n] = k, {n, 1, 40}]}] (* Vaclav Kotesovec, Sep 07 2019 *)
    Table[j! Coefficient[Series[Exp[Exp[x] - 1], {x, 0, 20}], x, j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 01 2023 *)
    Table[(D[Exp[Exp[x]], {x, n}] /. x -> 0)/E, {n, 0, 20}] (* Joan Ludevid, Nov 05 2024 *)
  • Maxima
    makelist(belln(n),n,0,40); /* Emanuele Munarini, Jul 04 2011 */
    
  • 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
    
  • Perl
    use bignum;sub a{my($n)=@;my@t=map{[(0)x($n+1)]}0..$n;$t[0][0]=1;for my$i(1..$n){$t[$i][0]=$t[$i-1][$i-1];for my$j(1..$i){$t[$i][$j]=$t[$i-1][$j-1]+$t[$i][$j-1]}}return map{$t[$][0]}0..$n-1}print join(", ",a(28)),"\n" # Paul Muljadi, May 08 2024
  • 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(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[k-1] += A[k]
            R.append(A[0])
        return R
    A000110_list(40) # Peter Luschny, Jan 18 2011
    
  • 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
    
  • Python
    from sympy import bell
    print([bell(n) for n in range(27)]) # Michael S. Branicky, Dec 15 2021
    
  • Python
    from functools import cache
    @cache
    def a(n, k=0): return int(n < 1) or k*a(n-1, k) + a(n-1, k+1)
    print([a(n) for n in range(27)])  # Peter Luschny, Jun 14 2022
    
  • Sage
    from sage.combinat.expnums import expnums2; expnums2(30, 1) # Zerinvary Lajos, Jun 26 2008
    
  • Sage
    [bell_number(n) for n in (0..40)] # G. C. Greubel, Jun 13 2019
    

Formula

E.g.f.: exp(exp(x) - 1).
Recurrence: a(n+1) = Sum_{k=0..n} a(k)*binomial(n, k).
a(n) = Sum_{k=0..n} Stirling2(n, k).
a(n) = Sum_{j=0..n-1} (1/(n-1)!)*A000166(j)*binomial(n-1, j)*(n-j)^(n-1). - André F. Labossière, Dec 01 2004
G.f.: (Sum_{k>=0} 1/((1-k*x)*k!))/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, 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(b-n-1/2)*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, North-Holland, 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} (1-j*x)) (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(n-1, n-1) [i.e., a(n-1, n-1) of Aitken's array (A011971)]. - Gerald McGarvey, Jun 26 2004
a(n) = Sum_{k=1..n} (1/k!)*(Sum_{i=1..k} (-1)^(k-i)*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_{x=0..Pi} e^(e^(e^(ix))) sin(nx) dx ) where Im denotes imaginary part [Cesaro]. - David Callan, Sep 03 2005
O.g.f.: 1/(1-x-x^2/(1-2*x-2*x^2/(1-3*x-3*x^2/(.../(1-n*x-n*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 (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)=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..n-1} Sum_{i=0..k} binomial(k,i)*(2^(k-i))*a(i). - Yalcin Aktar, 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, 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,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 permutation 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)
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) = (n-1)! Sum_{k=1..n} a(n-k)/((n-k)! (k-1)!) 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(k-r). - 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)
Sum_{k=1..n-1} a(n)*binomial(n,k) = Sum_{j=1..n}(j+1)*Stirling2(n,j+1). - [Zhao] - R. J. Mathar, Jun 24 2024
From Thomas Wieder, Feb 25 2009: (Start)
a(n) = Sum_{k_1=0..n+1} Sum_{k_2=0..n}...Sum_{k_i=0..n-i}...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,i-1]=-1, A[i,j]:=binomial(j-1,i-1), (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/(1-x))*A(x/(1-x)) + 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 + (!(-1-1/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) = (1-1/(G(0)+1))/exp(1); G(k) = x*k^2 + (3*x-1)*k - 2 + x - (k+1)*(x*k+x-1)^2/G(k+1); (continued fraction Euler's kind, 1-step).
G.f.: W(x) = (1 + G(0)/(x^2-3*x+2))/exp(1); G(k) = 1 - (x*k+x-1)/( ((k+1)!) - (((k+1)!)^2)*(1-x-k*x+(k+1)!)/( ((k+1)!)*(1-x-k*x+(k+1)!) - (x*k+2*x-1)*(1-2*x-k*x+(k+2)!)/G(k+1))); (continued fraction).
G.f.: A(x) = 1/(1 - x/(1-x/(1 + x/G(0)))); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1); (continued fraction, 1-step).
G.f.: -1/U(0) where U(k) = x*k - 1 + x - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + x/U(0) where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1); (continued fraction, 1-step).
G.f.: 1 + 1/(U(0) - x) where U(k) = 1 + x - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
G.f.: 1 + x/(U(0)-x) where U(k) = 1 - x*(k+1)/(1 - x/U(k+1)); (continued fraction, 2-step).
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*k-1) - x*(2*k+1)*(2*k+3)*(2*x*k-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+x-1)/G(k+1) )); (continued fraction).
G.f.: -(1+2*x) * Sum_{k >= 0} x^(2*k)*(4*x*k^2-2*k-2*x-1) / ((2*k+1) * (2*x*k-1)) * A(k) / B(k) where A(k) = Product_{p=0..k} (2*p+1), B(k) = Product_{p=0..k} (2*p-1) * (2*x*p-x-1) * (2*x*p-2*x-1).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-k*x)/(1-x/(x-1/G(k+1) )); (continued fraction).
G.f.: 1 + x*(S-1) where S = Sum_{k>=0} ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k/Product_{i=0..k-1} (1-x-x*i)/(1-x).
G.f.: (G(0) - 2)/(2*x-1) where G(k) = 2 - 1/(1-k*x)/(1-x/(x-1/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/(1-k*x)/(1-x/(x+1/G(k+1) )); (continued fraction).
G.f.: 1/(x*(1-x)*G(0)) - 1/x where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/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/(1-x*Q(0)), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))); (continued fraction).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1-x*(k+1))*(1-x*(k+2))/Q(k+1) ); (continued fraction).
(End)
a(n) ~ exp(exp(W(n))-n-1)*n^n/W(n)^(n+1/2), where W(x) is the Lambert W-function. - Vladimir Reshetnikov, Nov 01 2015
a(n) ~ n^n * exp(n/W(n)-1-n) / (sqrt(1+W(n)) * W(n)^n). - Vaclav Kotesovec, Nov 13 2015
From Natalia L. Skirrow, Apr 13 2025: (Start)
By taking logarithmic derivatives of the equivalent to Kotesovec's asymptotic for Bell polynomials at x=1, we obtain properties of the nth row of A008277 as a statistical distribution (where W=W(n),X=W(n)+1)
a(n+1)/a(n) ~ n/W + W/(2*(W+1)^2) is 1 more than the expectation.
(2*a(n+1)+a(n+2))/a(n) - (a(n+1)/a(n))^2 - a(n+2)/a(n+1) ~ n/(W*X)+1/(2*X^2)-3/(2*X^3)+1/X^4 is 1 more than the variance.
(This is a complete asymptotic characterisation, since they converge to normal distributions; see Harper, 1967)
(End)
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
a(n) = Sum_{k=0..n} hypergeom([1, -k], [], 1)*Stirling2(n+1, k+1) = Sum_{k=0..n} A182386(k)*Stirling2(n+1, k+1). - Mélika Tebni, Jul 02 2022
For n >= 1, a(n) = Sum_{i=0..n-1} a(i)*A074664(n-i). - Davide Rotondo, Apr 21 2024
a(n) is the n-th derivative of e^e^x divided by e at point x=0. - Joan Ludevid, Nov 05 2024

Extensions

Edited by M. F. Hasler, Nov 30 2018

A000178 Superfactorials: product of first n factorials.

Original entry on oeis.org

1, 1, 2, 12, 288, 34560, 24883200, 125411328000, 5056584744960000, 1834933472251084800000, 6658606584104736522240000000, 265790267296391946810949632000000000, 127313963299399416749559771247411200000000000, 792786697595796795607377086400871488552960000000000000
Offset: 0

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a(n) is also the Vandermonde determinant of the numbers 1,2,...,(n+1), i.e., the determinant of the (n+1) X (n+1) matrix A with A[i,j] = i^j, 1 <= i <= n+1, 0 <= j <= n. - Ahmed Fares (ahmedfares(AT)my-deja.com), May 06 2001
a(n) = (1/n!) * D(n) where D(n) is the determinant of order n in which the (i,j)-th element is i^j. - Amarnath Murthy, Jan 02 2002
Determinant of S_n where S_n is the n X n matrix S_n(i,j) = Sum_{d|i} d^j. - Benoit Cloitre, May 19 2002
Appears to be det(M_n) where M_n is the n X n matrix with m(i,j) = J_j(i) where J_k(n) denote the Jordan function of row k, column n (cf. A059380(m)). - Benoit Cloitre, May 19 2002
a(2n+1) = 1, 12, 34560, 125411328000, ... is the Hankel transform of A000182 (tangent numbers) = 1, 2, 16, 272, 7936, ...; example: det([1, 2, 16, 272; 2, 16, 272, 7936; 16, 272, 7936, 353792; 272, 7936, 353792, 22368256]) = 125411328000. - Philippe Deléham, Mar 07 2004
Determinant of the (n+1) X (n+1) matrix whose i-th row consists of terms 1 to n+1 of the Lucas sequence U(i,Q), for any Q. When Q=0, the Vandermonde matrix is obtained. - T. D. Noe, Aug 21 2004
Determinant of the (n+1) X (n+1) matrix A whose elements are A(i,j) = B(i+j) for 0 <= i,j <= n, where B(k) is the k-th Bell number, A000110(k) [I. Mezo, JIS 14 (2011) # 11.1.1]. - T. D. Noe, Dec 04 2004
The Hankel transform of the sequence A090365 is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
Theorem 1.3, page 2, of Polynomial points, Journal of Integer Sequences, Vol. 10 (2007), Article 07.3.6, provides an example of an Abelian quotient group of order (n-1) superfactorial, for each positive integer n. The quotient is obtained from sequences of polynomial values. - E. F. Cornelius, Jr. (efcornelius(AT)comcast.net), Apr 09 2007
Starting with offset 1 this is a 'Matryoshka doll' sequence with alpha=1, the multiplicative counterpart to the additive A000292. seq(mul(mul(i,i=alpha..k), k=alpha..n),n=alpha..12). - Peter Luschny, Jul 14 2009
For n>0, a(n) is also the determinant of S_n where S_n is the n X n matrix, indexed from 1, S_n(i,j)=sigma_i(j), where sigma_k(n) is the generalized divisor sigma function: A000203 is sigma_1(n). - Enrique Pérez Herrero, Jun 21 2010
a(n) is the multiplicative Wiener index of the (n+1)-vertex path. Example: a(4)=288 because in the path on 5 vertices there are 3 distances equal to 2, 2 distances equal to 3, and 1 distance equal to 4 (2*2*2*3*3*4=288). See p. 115 of the Gutman et al. reference. - Emeric Deutsch, Sep 21 2011
a(n-1) = Product_{j=1..n-1} j! = V(n) = Product_{1 <= i < j <= n} (j - i) (a Vandermondian V(n), see the Ahmed Fares May 06 2001 comment above), n >= 1, is in fact the determinant of any n X n matrix M(n) with entries M(n;i,j) = p(j-1,x = i), 1 <= i, j <= n, where p(m,x), m >= 0, are monic polynomials of exact degree m with p(0,x) = 1. This is a special x[i] = i choice in a general theorem given in Vein-Dale, p. 59 (written for the transposed matrix M(n;j,x_i) = p(i-1,x_j) = P_i(x_j) in Vein-Dale, and there a_{k,k} = 1, for k=1..n). See the Aug 26 2013 comment under A049310, where p(n,x) = S(n,x) (Chebyshev S). - Wolfdieter Lang, Aug 27 2013
a(n) is the number of monotonic magmas on n elements labeled 1..n with a symmetric multiplication table. I.e., Product(i,j) >= max(i,j); Product(i,j) = Product(j,i). - Chad Brewbaker, Nov 03 2013
The product of the pairwise differences of n+1 integers is a multiple of a(n) [and this does not hold for any k > a(n)]. - Charles R Greathouse IV, Aug 15 2014
a(n) is the determinant of the (n+1) X (n+1) matrix M with M(i,j) = (n+j-1)!/(n+j-i)!, 1 <= i <= n+1, 1 <= j <= n+1. - Stoyan Apostolov, Aug 26 2014
All terms are in A064807 and all terms after a(2) are in A005101. - Ivan N. Ianakiev, Sep 02 2016
Empirical: a(n-1) is the determinant of order n in which the (i,j)-th entry is the (j-1)-th derivative of x^(x+i-1) evaluated at x=1. - John M. Campbell, Dec 13 2016
Empirical: If f(x) is a smooth, real-valued function on an open neighborhood of 0 such that f(0)=1, then a(n) is the determinant of order n+1 in which the (i,j)-th entry is the (j-1)-th derivative of f(x)/((1-x)^(i-1)) evaluated at x=0. - John M. Campbell, Dec 27 2016
Also the automorphism group order of the n-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017
Is the zigzag Hankel transform of A000182. That is, a(2*n+1) is the Hankel transform of A000182 and a(2*n+2) is the Hankel transform of A000182(n+1). - Michael Somos, Mar 11 2020
Except for n = 0, 1, superfactorial a(n) is never a square (proof in link Mabry and Cormick, FFF 4 p. 349); however, when k belongs to A349079 (see for further information), there exists m, 1 <= m <= k such that a(k) / m! is a square. - Bernard Schott, Nov 29 2021

Examples

			a(3) = (1/6)* | 1 1 1 | 2 4 8 | 3 9 27 |
a(7) = n! * a(n-1) = 7! * 24883200 = 125411328000.
a(12) = 1! * 2! * 3! * 4! * 5! * 6! * 7! * 8! * 9! * 10! * 11! * 12!
= 1^12 * 2^11 * 3^10 * 4^9 * 5^8 * 6^7 * 7^6 * 8^5 * 9^4 * 10^3 * 11^2 * 12^1
= 2^56 * 3^26 * 5^11 * 7^6 * 11^2.
G.f. = 1 + x + 2*x^2 + 12*x^3 + 288*x^4 + 34560*x^5 + 24883200*x^6 + ...
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 545.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 135-145.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 50.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 231.
  • H. J. Ryser, Combinatorial Mathematics. Mathematical Association of America, Carus Mathematical Monograph 14, 1963, p. 53.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer, 1999.

Crossrefs

Programs

  • Magma
    [&*[Factorial(k): k in [0..n]]: n in [0..20]]; // Bruno Berselli, Mar 11 2015
    
  • Maple
    A000178 := proc(n)
        mul(i!,i=1..n) ;
    end proc:
    seq(A000178(n),n=0..10) ; # R. J. Mathar, Oct 30 2015
  • Mathematica
    a[0] := 1; a[1] := 1; a[n_] := n!*a[n - 1]; Table[a[n], {n, 1, 12}] (* Stefan Steinerberger, Mar 10 2006 *)
    Table[BarnesG[n], {n, 2, 14}] (* Zerinvary Lajos, Jul 16 2009 *)
    FoldList[Times,1,Range[20]!] (* Harvey P. Dale, Mar 25 2011 *)
    RecurrenceTable[{a[n] == n! a[n - 1], a[0] == 1}, a, {n, 0, 12}] (* Ray Chandler, Jul 30 2015 *)
    BarnesG[Range[2, 20]] (* Eric W. Weisstein, Jul 14 2017 *)
  • Maxima
    A000178(n):=prod(k!,k,0,n)$ makelist(A000178(n),n,0,30); /* Martin Ettl, Oct 23 2012 */
    
  • PARI
    A000178(n)=prod(k=2,n,k!) \\ M. F. Hasler, Sep 02 2007
    
  • PARI
    a(n)=polcoeff(1-sum(k=0, n-1, a(k)*x^k/prod(j=1, k+1, (1+j!*x+x*O(x^n)) )), n) \\ Paul D. Hanna, Oct 02 2013
    
  • PARI
    for(j=1,13, print1(prod(k=1,j,k^(j-k)),", ")) \\ Hugo Pfoertner, Apr 09 2020
    
  • Python
    A000178_list, n, m = [1], 1,1
    for i in range(1,100):
        m *= i
        n *= m
        A000178_list.append(n) # Chai Wah Wu, Aug 21 2015
    
  • Python
    from math import prod
    def A000178(n): return prod(i**(n-i+1) for i in range(2,n+1)) # Chai Wah Wu, Nov 26 2023
  • Ruby
    def mono_choices(a,b,n)
        n - [a,b].max
    end
    def comm_mono_choices(n)
        accum =1
        0.upto(n-1) do |i|
            i.upto(n-1) do |j|
                accum = accum * mono_choices(i,j,n)
            end
        end
        accum
    end
    1.upto(12) do |k|
        puts comm_mono_choices(k)
    end # Chad Brewbaker, Nov 03 2013
    

Formula

a(0) = 1, a(n) = n!*a(n-1). - Lee Hae-hwang, May 13 2003, corrected by Ilya Gutkovskiy, Jul 30 2016
a(0) = 1, a(n) = 1^n * 2^(n-1) * 3^(n-2) * ... * n = Product_{r=1..n} r^(n-r+1). - Amarnath Murthy, Dec 12 2003 [Formula corrected by Derek Orr, Jul 27 2014]
a(n) = sqrt(A055209(n)). - Philippe Deléham, Mar 07 2004
a(n) = Product_{i=1..n} Product_{j=0..i-1} (i-j). - Paul Barry, Aug 02 2008
log a(n) = 0.5*n^2*log n - 0.75*n^2 + O(n*log n). - Charles R Greathouse IV, Jan 13 2012
Asymptotic: a(n) ~ exp(zeta'(-1) - 3/4 - (3/4)*n^2 - (3/2)*n)*(2*Pi)^(1/2 + (1/2)*n)*(n+1)^((1/2)*n^2 + n + 5/12). For example, a(100) is approx. 0.270317...*10^6941. (See A213080.) - Peter Luschny, Jun 23 2012
G.f.: 1 + x/(U(0) - x) where U(k) = 1 + x*(k+1)! - x*(k+2)!/U(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 02 2012
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - 1/(1 + 1/((k+1)!*x*G(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Jun 14 2013
G.f.: 1 = Sum_{n>=0} a(n)*x^n / Product_{k=1..n+1} (1 + k!*x). - Paul D. Hanna, Oct 02 2013
A203227(n+1)/a(n) -> e, as n -> oo. - Daniel Suteu, Jul 30 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = G(n+2), where G(n) is the Barnes G-function.
a(n) ~ exp(1/12 - n*(3*n+4)/4)*n^(n*(n+2)/2 + 5/12)*(2*Pi)^((n+1)/2)/A, where A is the Glaisher-Kinkelin constant (A074962).
Sum_{n>=0} (-1)^n/a(n) = A137986. (End)
0 = a(n)*a(n+2)^3 + a(n+1)^2*a(n+2)^2 - a(n+1)^3*a(n+3) for all n in Z (if a(-1)=1). - Michael Somos, Mar 11 2020
Sum_{n>=0} 1/a(n) = A287013 = 1/A137987. - Amiram Eldar, Nov 19 2020
a(n) = Wronskian(1, x, x^2, ..., x^n). - Mohammed Yaseen, Aug 01 2023
From Andrea Pinos, Apr 04 2024: (Start)
a(n) = e^(Sum_{k=1..n} (Integral_{x=1..k+1} Psi(x) dx)).
a(n) = e^(Integral_{x=1..n+1} (log(sqrt(2*Pi)) - (x-1/2) + x*Psi(x)) dx).
a(n) = e^(Integral_{x=1..n+1} (log(sqrt(2*Pi)) - (x-1/2) + (n+1)*Psi(x) - log(Gamma(x))) dx).
Psi(x) is the digamma function. (End)

A005043 Riordan numbers: a(n) = (n-1)*(2*a(n-1) + 3*a(n-2))/(n+1).

Original entry on oeis.org

1, 0, 1, 1, 3, 6, 15, 36, 91, 232, 603, 1585, 4213, 11298, 30537, 83097, 227475, 625992, 1730787, 4805595, 13393689, 37458330, 105089229, 295673994, 834086421, 2358641376, 6684761125, 18985057351, 54022715451, 154000562758, 439742222071, 1257643249140
Offset: 0

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Also called Motzkin summands or ring numbers.
The old name was "Motzkin sums", used in certain publications. The sequence has the property that Motzkin(n) = A001006(n) = a(n) + a(n+1), e.g., A001006(4) = 9 = 3 + 6 = a(4) + a(5).
Number of 'Catalan partitions', that is partitions of a set 1,2,3,...,n into parts that are not singletons and whose convex hulls are disjoint when the points are arranged on a circle (so when the parts are all pairs we get Catalan numbers). - Aart Blokhuis (aartb(AT)win.tue.nl), Jul 04 2000
Number of ordered trees with n edges and no vertices of outdegree 1. For n > 1, number of dissections of a convex polygon by nonintersecting diagonals with a total number of n+1 edges. - Emeric Deutsch, Mar 06 2002
Number of Motzkin paths of length n with no horizontal steps at level 0. - Emeric Deutsch, Nov 09 2003
Number of Dyck paths of semilength n with no peaks at odd level. Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUDUDD, where U=(1,1), D=(1,-1). Number of Dyck paths of semilength n with no ascents of length 1 (an ascent in a Dyck path is a maximal string of up steps). Example: a(4)=3 because we have UUUUDDDD, UUDDUUDD and UUDUUDDD. - Emeric Deutsch, Dec 05 2003
Arises in Schubert calculus as follows. Let P = complex projective space of dimension n+1. Take n projective subspaces of codimension 3 in P in general position. Then a(n) is the number of lines of P intersecting all these subspaces. - F. Hirzebruch, Feb 09 2004
Difference between central trinomial coefficient and its predecessor. Example: a(6) = 15 = 141 - 126 and (1 + x + x^2)^6 = ... + 126*x^5 + 141*x^6 + ... (Catalan number A000108(n) is the difference between central binomial coefficient and its predecessor.) - David Callan, Feb 07 2004
a(n) = number of 321-avoiding permutations on [n] in which each left-to-right maximum is a descent (i.e., is followed by a smaller number). For example, a(4) counts 4123, 3142, 2143. - David Callan, Jul 20 2005
The Hankel transform of this sequence give A000012 = [1, 1, 1, 1, 1, 1, 1, ...]; example: Det([1, 0, 1, 1; 0, 1, 1, 3; 1, 1, 3, 6; 1, 3, 6, 15]) = 1. - Philippe Deléham, May 28 2005
The number of projective invariants of degree 2 for n labeled points on the projective line. - Benjamin J. Howard (bhoward(AT)ima.umn.edu), Nov 24 2006
Define a random variable X=trA^2, where A is a 2 X 2 unitary symplectic matrix chosen from USp(2) with Haar measure. The n-th central moment of X is E[(X+1)^n] = a(n). - Andrew V. Sutherland, Dec 02 2007
Let V be the adjoint representation of the complex Lie algebra sl(2). The dimension of the invariant subspace of the n-th tensor power of V is a(n). - Samson Black (sblack1(AT)uoregon.edu), Aug 27 2008
Starting with offset 3 = iterates of M * [1,1,1,...], where M = a tridiagonal matrix with [0,1,1,1,...] in the main diagonal and [1,1,1,...] in the super and subdiagonals. - Gary W. Adamson, Jan 08 2009
a(n) has the following standard-Young-tableaux (SYT) interpretation: binomial(n+1,k)*binomial(n-k-1,k-1)/(n+1)=f^(k,k,1^{n-2k}) where f^lambda equals the number of SYT of shape lambda. - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 02 2010
a(n) is also the sum of the numbers of standard Young tableaux of shapes (k,k,1^{n-2k}) for all 1 <= k <= floor(n/2). - Amitai Regev (amotai.regev(AT)weizmann.ac.il), Mar 10 2010
a(n) is the number of derangements of {1,2,...,n} having genus 0. The genus g(p) of a permutation p of {1,2,...,n} is defined by g(p)=(1/2)[n+1-z(p)-z(cp')], where p' is the inverse permutation of p, c = 234...n1 = (1,2,...,n), and z(q) is the number of cycles of the permutation q. Example: a(3)=1 because p=231=(123) is the only derangement of {1,2,3} with genus 0. Indeed, cp'=231*312=123=(1)(2)(3) and so g(p) = (1/2)(3+1-1-3)=0. - Emeric Deutsch, May 29 2010
Apparently: Number of Dyck 2n-paths with all ascents length 2 and no descent length 2. - David Scambler, Apr 17 2012
This is true. Proof: The mapping "insert a peak (UD) after each upstep (U)" is a bijection from all Dyck n-paths to those Dyck (2n)-paths in which each ascent is of length 2. It sends descents of length 1 in the n-path to descents of length 2 in the (2n)-path. But Dyck n-paths with no descents of length 1 are equinumerous with Riordan n-paths (Motzkin n-paths with no flatsteps at ground level) as follows. Given a Dyck n-path with no descents of length 1, split it into consecutive step pairs, then replace UU with U, DD with D, UD with a blue flatstep (F), DU with a red flatstep, and concatenate the new steps to get a colored Motzkin path. Each red F will be (immediately) preceded by a blue F or a D. In the latter case, transfer the red F so that it precedes the matching U of the D. Finally, erase colors to get the required Riordan path. For example, with lowercase f denoting a red flatstep, U^5 D^2 U D^4 U^4 D^3 U D^2 -> (U^2, U^2, UD, DU, D^2, D^2, U^2, U^2 D^2, DU, D^2) -> UUFfDDUUDfD -> UUFFDDUFUDD. - David Callan, Apr 25 2012
From Nolan Wallach, Aug 20 2014: (Start)
Let ch[part1, part2] be the value of the character of the symmetric group on n letters corresponding to the partition part1 of n on the conjucgacy class given by part2. Let A[n] be the set of (n+1) partitions of 2n with parts 1 or 2. Then deleting the first term of the sequence one has a(n) = Sum_{k=1..n+1} binomial(n,k-1)*ch[[n,n], A[n][[k]]])/2^n. This via the Frobenius Character Formula can be interpreted as the dimension of the SL(n,C) invariants in tensor^n (wedge^2 C^n).
Explanation: Let p_j denote sum (x_i)^j the sum in k variables. Then the Frobenius formula says then (p_1)^j_1 (p_2)^j_2 ... (p_r)^j_r is equal to sum(lambda, ch[lambda, 1^j_12^j_2 ... r^j_r] S_lambda) with S_lambda the Schur function corresponding to lambda. This formula implies that the coefficient of S([n,n]) in (((p_1)^1+p_2)/2)^n in its expansion in terms of Schur functions is the right hand side of our formula. If we specialize the number of variables to 2 then S[n,n](x,y)=(xy)^n. Which when restricted to y=x^(-1) is 1. That is it is 1 on SL(2).
On the other hand ((p_1)^2+p_2)/2 is the complete homogeneous symmetric function of degree 2 that is tr(S^2(X)). Thus our formula for a(n) is the same as that of Samson Black above since his V is the same as S^2(C^2) as a representation of SL(2). On the other hand, if we multiply ch(lambda) by sgn you get ch(Transpose(lambda)). So ch([n,n]) becomes ch([2,...,2]) (here there are n 2's). The formula for a(n) is now (1/2^n)*Sum_{j=0..n} ch([2,..,2], 1^(2n-2j) 2^j])*(-1)^j)*binomial(n,j), which calculates the coefficient of S_(2,...,2) in (((p_1)^2-p_2)/2)^n. But ((p_1)^2-p_2)/2 in n variables is the second elementary symmetric function which is the character of wedge^2 C^n and S_(2,...,2) is 1 on SL(n).
(End)
a(n) = number of noncrossing partitions (A000108) of [n] that contain no singletons, also number of nonnesting partitions (A000108) of [n] that contain no singletons. - David Callan, Aug 27 2014
From Tom Copeland, Nov 02 2014: (Start)
Let P(x) = x/(1+x) with comp. inverse Pinv(x) = x/(1-x) = -P[-x], and C(x)= [1-sqrt(1-4x)]/2, an o.g.f. for the shifted Catalan numbers A000108, with inverse Cinv(x) = x * (1-x).
Fin(x) = P[C(x)] = C(x)/[1 + C(x)] is an o.g.f. for the Fine numbers, A000957 with inverse Fin^(-1)(x) = Cinv[Pinv(x)] = Cinv[-P(-x)].
Mot(x) = C[P(x)] = C[-Pinv(-x)] gives an o.g.f. for shifted A005043, the Motzkin or Riordan numbers with comp. inverse Mot^(-1)(x) = Pinv[Cinv(x)] = (x - x^2) / (1 - x + x^2) (cf. A057078).
BTC(x) = C[Pinv(x)] gives A007317, a binomial transform of the Catalan numbers, with BTC^(-1)(x) = P[Cinv(x)].
Fib(x) = -Fin[Cinv(Cinv(-x))] = -P[Cinv(-x)] = x + 2 x^2 + 3 x^3 + 5 x^4 + ... = (x+x^2)/[1-x-x^2] is an o.g.f. for the shifted Fibonacci sequence A000045, so the comp. inverse is Fib^(-1)(x) = -C[Pinv(-x)] = -BTC(-x) and Fib(x) = -BTC^(-1)(-x).
Various relations among the o.g.f.s may be easily constructed, such as Fib[-Mot(-x)] = -P[P(-x)] = x/(1-2*x) a generating fct for 2^n.
Generalizing to P(x,t) = x /(1 + t*x) and Pinv(x,t) = x /(1 - t*x) = -P(-x,t) gives other relations to lattice paths, such as the o.g.f. for A091867, C[P[x,1-t]], and that for A104597, Pinv[Cinv(x),t+1]. (End)
Consistent with David Callan's comment above, A249548, provides a refinement of the Motzkin sums into the individual numbers for the non-crossing partitions he describes. - Tom Copeland, Nov 09 2014
The number of lattice paths from (0,0) to (n,0) that do not cross below the x-axis and use up-step=(1,1) and down-steps=(1,-k) where k is a positive integer. For example, a(4) = 3: [(1,1)(1,1)(1,-1)(1,-1)], [(1,1)(1,-1)(1,1)(1,-1)] and [(1,1)(1,1)(1,1)(1,-3)]. - Nicholas Ham, Aug 19 2015
A series created using 2*(a(n) + a(n+1)) + (a(n+1) + a(n+2)) has Hankel transform of F(2n), offset 3, F being a Fibonacci number, A001906 (Empirical observation). - Tony Foster III, Jul 30 2016
The series a(n) + A001006(n) has Hankel transform F(2n+1), offset n=1, F being the Fibonacci bisection A001519 (empirical observation). - Tony Foster III, Sep 05 2016
The Rubey and Stump reference proves a refinement of a conjecture of René Marczinzik, which they state as: "The number of 2-Gorenstein algebras which are Nakayama algebras with n simple modules and have an oriented line as associated quiver equals the number of Motzkin paths of length n. Moreover, the number of such algebras having the double centraliser property with respect to a minimal faithful projective-injective module equals the number of Riordan paths, that is, Motzkin paths without level-steps at height zero, of length n." - Eric M. Schmidt, Dec 16 2017
A connection to the Thue-Morse sequence: (-1)^a(n) = (-1)^A010060(n) * (-1)^A010060(n+1) = A106400(n) * A106400(n+1). - Vladimir Reshetnikov, Jul 21 2019
Named by Bernhart (1999) after the American mathematician John Riordan (1903-1988). - Amiram Eldar, Apr 15 2021

Examples

			a(5)=6 because the only dissections of a polygon with a total number of 6 edges are: five pentagons with one of the five diagonals and the hexagon with no diagonals.
G.f. = 1 + x^2 + x^3 + 3*x^4 + 6*x^5 + 15*x^6 + 36*x^7 + 91*x^8 + 232*x^9 + ...
From _Gus Wiseman_, Nov 15 2022: (Start)
The a(0) = 1 through a(6) = 15 lone-child-avoiding (no vertices of outdegree 1) ordered rooted trees with n + 1 vertices (ranked by A358376):
  o  .  (oo)  (ooo)  (oooo)   (ooooo)   (oooooo)
                     ((oo)o)  ((oo)oo)  ((oo)ooo)
                     (o(oo))  ((ooo)o)  ((ooo)oo)
                              (o(oo)o)  ((oooo)o)
                              (o(ooo))  (o(oo)oo)
                              (oo(oo))  (o(ooo)o)
                                        (o(oooo))
                                        (oo(oo)o)
                                        (oo(ooo))
                                        (ooo(oo))
                                        (((oo)o)o)
                                        ((o(oo))o)
                                        ((oo)(oo))
                                        (o((oo)o))
                                        (o(o(oo)))
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of triangle A020474, first differences of A082395.
First diagonal of triangular array in A059346.
Binomial transform of A126930. - Philippe Deléham, Nov 26 2009
The Hankel transform of a(n+1) is A128834. The Hankel transform of a(n+2) is floor((2*n+4)/3) = A004523(n+2). - Paul Barry, Mar 08 2011
The Kn11 triangle sums of triangle A175136 lead to A005043(n+2), while the Kn12(n) = A005043(n+4)-2^(n+1), Kn13(n) = A005043(n+6)-(n^2+9*n+56)*2^(n-2) and the Kn4(n) = A005043(2*n+2) = A099251(n+1) triangle sums are related to the sequence given above. For the definitions of these triangle sums see A180662. - Johannes W. Meijer, May 06 2011
Cf. A187306 (self-convolution), A348210 (column 1).
Bisections: A099251, A099252.

Programs

  • Haskell
    a005043 n = a005043_list !! n
    a005043_list = 1 : 0 : zipWith div
       (zipWith (*) [1..] (zipWith (+)
           (map (* 2) $ tail a005043_list) (map (* 3) a005043_list))) [3..]
    -- Reinhard Zumkeller, Jan 31 2012
    
  • Maple
    A005043 := proc(n) option remember; if n <= 1 then 1-n else (n-1)*(2*A005043(n-1)+3*A005043(n-2))/(n+1); fi; end;
    Order := 20: solve(series((x-x^2)/(1-x+x^2),x)=y,x); # outputs g.f.
  • Mathematica
    a[0]=1; a[1]=0; a[n_]:= a[n] = (n-1)*(2*a[n-1] + 3*a[n-2])/(n+1); Table[ a[n], {n, 0, 30}] (* Robert G. Wilson v, Jun 14 2005 *)
    Table[(-3)^(1/2)/6 * (-1)^n*(3*Hypergeometric2F1[1/2,n+1,1,4/3]+ Hypergeometric2F1[1/2,n+2,1,4/3]), {n,0,32}] (* cf. Mark van Hoeij in A001006 *) (* Wouter Meeussen, Jan 23 2010 *)
    RecurrenceTable[{a[0]==1,a[1]==0,a[n]==(n-1) (2a[n-1]+3a[n-2])/(n+1)},a,{n,30}] (* Harvey P. Dale, Sep 27 2013 *)
    a[ n_]:= SeriesCoefficient[2/(1+x +Sqrt[1-2x-3x^2]), {x, 0, n}]; (* Michael Somos, Aug 21 2014 *)
    a[ n_]:= If[n<0, 0, 3^(n+3/2) Hypergeometric2F1[3/2, n+2, 2, 4]/I]; (* Michael Somos, Aug 21 2014 *)
    Table[3^(n+3/2) CatalanNumber[n] (4(5+2n)Hypergeometric2F1[3/2, 3/2, 1/2-n, 1/4] -9 Hypergeometric2F1[3/2, 5/2, 1/2 -n, 1/4])/(4^(n+3) (n+1)), {n, 0, 31}] (* Vladimir Reshetnikov, Jul 21 2019 *)
    Table[Sqrt[27]/8 (3/4)^n CatalanNumber[n] Hypergeometric2F1[1/2, 3/2, 1/2 - n, 1/4], {n, 0, 31}] (* Jan Mangaldan, Sep 12 2021 *)
  • Maxima
    a[0]:1$
    a[1]:0$
    a[n]:=(n-1)*(2*a[n-1]+3*a[n-2])/(n+1)$
    makelist(a[n],n,0,12); /* Emanuele Munarini, Mar 02 2011 */
    
  • PARI
    {a(n) = if( n<0, 0, n++; polcoeff( serreverse( (x - x^3) / (1 + x^3) + x * O(x^n)), n))}; /* Michael Somos, May 31 2005 */
    
  • PARI
    my(N=66); Vec(serreverse(x/(1+x*sum(k=1,N,x^k))+O(x^N))) \\ Joerg Arndt, Aug 19 2012
    
  • Python
    from functools import cache
    @cache
    def A005043(n: int) -> int:
        if n <= 1: return 1 - n
        return (n - 1) * (2 * A005043(n - 1) + 3 * A005043(n - 2)) // (n + 1)
    print([A005043(n) for n in range(32)]) # Peter Luschny, Nov 20 2022
  • Sage
    A005043 = lambda n: (-1)^n*jacobi_P(n,1,-n-3/2,-7)/(n+1)
    [simplify(A005043(n)) for n in (0..29)]
    # Peter Luschny, Sep 23 2014
    
  • Sage
    def ms():
        a, b, c, d, n = 0, 1, 1, -1, 1
        yield 1
        while True:
            yield -b + (-1)^n*d
            n += 1
            a, b = b, (3*(n-1)*n*a+(2*n-1)*n*b)/((n+1)*(n-1))
            c, d = d, (3*(n-1)*c-(2*n-1)*d)/n
    A005043 = ms()
    print([next(A005043) for  in range(32)]) # _Peter Luschny, May 16 2016
    

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*A000108(k). a(n) = (1/(n+1)) * Sum_{k=0..ceiling(n/2)} binomial(n+1, k)*binomial(n-k-1, k-1), for n > 1. - Len Smiley. [Comment from Amitai Regev (amitai.regev(AT)weizmann.ac.il), Mar 02 2010: the latter sum should be over the range k=1..floor(n/2).]
G.f.: (1 + x - sqrt(1-2*x-3*x^2))/(2*x*(1+x)).
G.f.: 2/(1+x+sqrt(1-2*x-3*x^2)). - Paul Peart (ppeart(AT)fac.howard.edu), May 27 2000
a(n+1) + (-1)^n = a(0)*a(n) + a(1)*a(n-1) + ... + a(n)*a(0). - Bernhart
a(n) = (1/(n+1)) * Sum_{i} (-1)^i*binomial(n+1, i)*binomial(2*n-2*i, n-i). - Bernhart
G.f. A(x) satisfies A = 1/(1+x) + x*A^2.
E.g.f.: exp(x)*(BesselI(0, 2*x) - BesselI(1, 2*x)). - Vladeta Jovovic, Apr 28 2003
a(n) = A001006(n-1) - a(n-1).
a(n+1) = Sum_{k=0..n} (-1)^k*A026300(n, k), where A026300 is the Motzkin triangle.
a(n) = Sum_{k=0..n} (-1)^k*binomial(n, k)*binomial(k, floor(k/2)). - Paul Barry, Jan 27 2005
a(n) = Sum_{k>=0} A086810(n-k, k). - Philippe Deléham, May 30 2005
a(n+2) = Sum_{k>=0} A064189(n-k, k). - Philippe Deléham, May 31 2005
Moment representation: a(n) = (1/(2*Pi))*Int(x^n*sqrt((1+x)(3-x))/(1+x),x,-1,3). - Paul Barry, Jul 09 2006
Inverse binomial transform of A000108 (Catalan numbers). - Philippe Deléham, Oct 20 2006
a(n) = (2/Pi)* Integral_{x=0..Pi} (4*cos(x)^2-1)^n*sin(x)^2 dx. - Andrew V. Sutherland, Dec 02 2007
G.f.: 1/(1-x^2/(1-x-x^2/(1-x-x^2/(1-x-x^2/(1-... (continued fraction). - Paul Barry, Jan 22 2009
G.f.: 1/(1+x-x/(1-x/(1+x-x/(1-x/(1+x-x/(1-... (continued fraction). - Paul Barry, May 16 2009
G.f.: 1/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-x/(1-x^2/(1-x/(1-... (continued fraction). - Paul Barry, Mar 02 2010
a(n) = -(-1)^n * hypergeom([1/2, n+2],[2],4/3) / sqrt(-3). - Mark van Hoeij, Jul 02 2010
a(n) = (-1)^n*hypergeometric([-n,1/2],[2],4). - Peter Luschny, Aug 15 2012
Let A(x) be the g.f., then x*A(x) is the reversion of x/(1 + x^2*Sum_{k>=0} x^k); see A215340 for the correspondence to Dyck paths without length-1 ascents. - Joerg Arndt, Aug 19 2012 and Apr 16 2013
a(n) ~ 3^(n+3/2)/(8*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 02 2012
G.f.: 2/(1+x+1/G(0)), where G(k) = 1 + x*(2+3*x)*(4*k+1)/( 4*k+2 - x*(2+3*x)*(4*k+2)*(4*k+3)/(x*(2+3*x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 05 2013
D-finite (an alternative): (n+1)*a(n) = 3*(n-2)*a(n-3) + (5*n-7)*a(n-2) + (n-2)*a(n-1), n >= 3. - Fung Lam, Mar 22 2014
Asymptotics: a(n) = (3^(n+2)/(sqrt(3*n*Pi)*(8*n)))*(1-21/(16*n) + O(1/n^2)) (with contribution by Vaclav Kotesovec). - Fung Lam, Mar 22 2014
a(n) = T(2*n-1,n)/n, where T(n,k) = triangle of A180177. - Vladimir Kruchinin, Sep 23 2014
a(n) = (-1)^n*JacobiP(n,1,-n-3/2,-7)/(n+1). - Peter Luschny, Sep 23 2014
a(n) = Sum_{k=0..n} C(n,k)*(C(k,n-k)-C(k,n-k-1)). - Peter Luschny, Oct 01 2014
Conjecture: a(n) = A002426(n) - A005717(n), n > 0. - Mikhail Kurkov, Feb 24 2019 [The conjecture is true. - Amiram Eldar, May 17 2024]
a(n) = A309303(n) + A309303(n+1). - Vladimir Reshetnikov, Jul 22 2019
From Peter Bala, Feb 11 2022: (Start)
a(n) = A005773(n+1) - 2*A005717(n) for n >= 1.
Conjectures: for n >= 1, n divides a(2*n+1) and 2*n-1 divides a(2*n). (End)

Extensions

Thanks to Laura L. M. Yang (yanglm(AT)hotmail.com) for a correction, Aug 29 2004
Name changed to Riordan numbers following a suggestion from Ira M. Gessel. - N. J. A. Sloane, Jul 24 2020

A005493 2-Bell numbers: a(n) = number of partitions of [n+1] with a distinguished block.

Original entry on oeis.org

1, 3, 10, 37, 151, 674, 3263, 17007, 94828, 562595, 3535027, 23430840, 163254885, 1192059223, 9097183602, 72384727657, 599211936355, 5150665398898, 45891416030315, 423145657921379, 4031845922290572, 39645290116637023, 401806863439720943, 4192631462935194064
Offset: 0

Views

Author

Keywords

Comments

Number of Boolean sublattices of the Boolean lattice of subsets of {1..n}.
a(n) = p(n+1) where p(x) is the unique degree n polynomial such that p(k) = A000110(k+1) for k = 0, 1, ..., n. - Michael Somos, Oct 07 2003
With offset 1, number of permutations beginning with 12 and avoiding 21-3.
Rows sums of Bell's triangle (A011971). - Jorge Coveiro, Dec 26 2004
Number of blocks in all set partitions of an (n+1)-set. Example: a(2)=10 because the set partitions of {1,2,3} are 1|2|3, 1|23, 12|3, 13|2 and 123, with a total of 10 blocks. - Emeric Deutsch, Nov 13 2006
Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 2. - Olivier Gérard, Oct 29 2007
See page 29, Theorem 5.6 of my paper on the arXiv: These numbers are the dimensions of the homogeneous components of the operad called ComTrip associated with commutative triplicial algebras. (Triplicial algebras are related to even trees and also to L-algebras, see A006013.) - Philippe Leroux, Nov 17 2007
Number of set partitions of (n+2) elements where two specific elements are clustered separately. Example: a(1)=3 because 1/2/3, 1/23, 13/2 are the 3 set partitions with 1, 2 clustered separately. - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
Equals A008277 * [1,2,3,...], i.e., the product of the Stirling number of the second kind triangle and the natural number vector. a(n+1) = row sums of triangle A137650. - Gary W. Adamson, Jan 31 2008
Prefaced with a "1" = row sums of triangle A152433. - Gary W. Adamson, Dec 04 2008
Equals row sums of triangle A159573. - Gary W. Adamson, Apr 16 2009
Number of embedded coalitions in an (n+1)-person game. - David Yeung (wkyeung(AT)hkbu.edu.hk), May 08 2008
If prefixed with 0, gives first differences of Bell numbers A000110 (cf. A106436). - N. J. A. Sloane, Aug 29 2013
Sum_{n>=0} a(n)/n! = e^(e+1) = 41.19355567... (see A235214). Contrast with e^(e-1) = Sum_{n>=0} A000110(n)/n!. - Richard R. Forberg, Jan 05 2014

Examples

			For example, a(1) counts (12), (1)-2, 1-(2) where dashes separate blocks and the distinguished block is parenthesized.
		

References

  • Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation. Unpublished as of 2017.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

A row or column of the array A108087.
Row sums of triangle A143494. - Wolfdieter Lang, Sep 29 2011. And also of triangle A362924. - N. J. A. Sloane, Aug 10 2023

Programs

  • Maple
    with(combinat): seq(bell(n+2)-bell(n+1),n=0..22); # Emeric Deutsch, Nov 13 2006
    seq(add(binomial(n, k)*(bell(n-k)), k=1..n), n=1..23); # Zerinvary Lajos, Dec 01 2006
    A005493  := proc(n) local a,b,i;
    a := [seq(3,i=1..n)]; b := [seq(2,i=1..n)];
    2^n*exp(-x)*hypergeom(a,b,x); round(evalf(subs(x=1,%),66)) end:
    seq(A005493(n),n=0..22); # Peter Luschny, Mar 30 2011
    BT := proc(n,k) option remember; if n = 0 and k = 0 then 1
    elif k = n then BT(n-1,0) else BT(n,k+1)+BT(n-1,k) fi end:
    A005493 := n -> add(BT(n,k),k=0..n):
    seq(A005493(i),i=0..22); # Peter Luschny, Aug 04 2011
    # For Maple code for r-Bell numbers, etc., see A232472. - N. J. A. Sloane, Nov 27 2013
  • Mathematica
    a=Exp[x]-1; Rest[CoefficientList[Series[a Exp[a],{x,0,20}],x] * Table[n!,{n,0,20}]]
    a[ n_] := If[ n<0, 0, With[ {m = n+1}, m! SeriesCoefficient[ # Exp@# &[ Exp@x - 1], {x, 0, m}]]]; (* Michael Somos, Nov 16 2011 *)
    Differences[BellB[Range[30]]] (* Harvey P. Dale, Oct 16 2014 *)
  • PARI
    {a(n) = if( n<0, 0, n! * polcoeff( exp( exp( x + x * O(x^n)) + 2*x - 1), n))}; /* Michael Somos, Oct 09 2002 */
    
  • PARI
    {a(n) = if( n<0, 0, n+=2; subst( polinterpolate( Vec( serlaplace( exp( exp( x + O(x^n)) - 1) - 1))), x, n))}; /* Michael Somos, Oct 07 2003 */
    
  • Python
    # requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
    from itertools import accumulate
    A005493_list, blist, b = [], [1], 1
    for _ in range(1001):
        blist = list(accumulate([b]+blist))
        b = blist[-1]
        A005493_list.append(blist[-2])
    # Chai Wah Wu, Sep 02 2014, updated Chai Wah Wu, Sep 20 2014

Formula

a(n-1) = Sum_{k=1..n} k*Stirling2(n, k) for n>=1.
E.g.f.: exp(exp(x) + 2*x - 1). First differences of Bell numbers (if offset 1). - Michael Somos, Oct 09 2002
G.f.: Sum_{k>=0} (x^k/Product_{l=1..k} (1-(l+1)x)). - Ralf Stephan, Apr 18 2004
a(n) = Sum_{i=0..n} 2^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n). - Fred Lunnon, Aug 04 2007 [Written umbrally, a(n) = (B+2)^n. - N. J. A. Sloane, Feb 07 2009]
Representation as an infinite series: a(n-1) = Sum_{k>=2} (k^n*(k-1)/k!)/exp(1), n=1, 2, ... This is a Dobinski-type summation formula. - Karol A. Penson, Mar 14 2002
Row sums of A011971 (Aitken's array, also called Bell triangle). - Philippe Deléham, Nov 15 2003
a(n) = exp(-1)*Sum_{k>=0} ((k+2)^n)/k!. - Gerald McGarvey, Jun 03 2004
Recurrence: a(n+1) = 1 + Sum_{j=1..n} (1+binomial(n, j))*a(j). - Jon Perry, Apr 25 2005
a(n) = A000296(n+3) - A000296(n+1). - Philippe Deléham, Jul 31 2005
a(n) = B(n+2) - B(n+1), where B(n) are Bell numbers (A000110). - Franklin T. Adams-Watters, Jul 13 2006
a(n) = A123158(n,2). - Philippe Deléham, Oct 06 2006
Binomial transform of Bell numbers 1, 2, 5, 15, 52, 203, 877, 4140, ... (see A000110).
Define f_1(x), f_2(x), ... such that f_1(x)=x*e^x, f_{n+1}(x) = (d/dx)(x*f_n(x)), for n=2,3,.... Then a(n-1) = e^(-1)*f_n(1). - Milan Janjic, May 30 2008
Representation of numbers a(n), n=0,1..., as special values of hypergeometric function of type (n)F(n), in Maple notation: a(n)=exp(-1)*2^n*hypergeom([3,3...3],[2.2...2],1), n=0,1..., i.e., having n parameters all equal to 3 in the numerator, having n parameters all equal to 2 in the denominator and the value of the argument equal to 1. Examples: a(0)= 2^0*evalf(hypergeom([],[],1)/exp(1))=1 a(1)= 2^1*evalf(hypergeom([3],[2],1)/exp(1))=3 a(2)= 2^2*evalf(hypergeom([3,3],[2,2],1)/exp(1))=10 a(3)= 2^3*evalf(hypergeom([3,3,3],[2,2,2],1)/exp(1))=37 a(4)= 2^4*evalf(hypergeom([3,3,3,3],[2,2,2,2],1)/exp(1))=151 a(5)= 2^5*evalf(hypergeom([3,3,3,3,3],[2,2,2,2,2],1)/exp(1)) = 674. - Karol A. Penson, Sep 28 2007
Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i <= j), and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = (-1)^(n)charpoly(A,-2). - Milan Janjic, Jul 08 2010
a(n) = D^(n+1)(x*exp(x)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A003128, A052852 and A009737. - Peter Bala, Nov 25 2011
From Sergei N. Gladkovskii, Oct 11 2012 to Jan 26 2014: (Start)
Continued fractions:
G.f.: 1/U(0) where U(k) = 1 - x*(k+3) - x^2*(k+1)/U(k+1).
G.f.: 1/(U(0)-x) where U(k) = 1 - x - x*(k+1)/(1 - x/U(k+1)).
G.f.: G(0)/(1-x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )).
G.f.: (G(0) - 1)/(x-1) where G(k) = 1 - 1/(1-2*x-k*x)/(1-x/(x-1/G(k+1) )).
G.f.: -G(0)/x where G(k) = 1 - 1/(1-k*x-x)/(1-x/(x-1/G(k+1) )).
G.f.: 1 - 2/x + (1/x-1)*S where S = sum(k>=0, ( 1 + (1-x)/(1-x-x*k) )*(x/(1-x))^k / prod(i=0..k-1, (1-x-x*i)/(1-x) ) ).
G.f.: (1-x)/x/(G(0)-x) - 1/x where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ).
G.f.: (1/G(0) - 1)/x^3 where G(k) = 1 - x/(x - 1/(1 + 1/(x*k-1)/G(k+1) )).
G.f.: 1/Q(0), where Q(k)= 1 - 2*x - x/(1 - x*(k+1)/Q(k+1)).
G.f.: G(0)/(1-3*x), where G(k) = 1 - x^2*(k+1)/( x^2*(k+1) - (1 - x*(k+3))*(1 - x*(k+4))/G(k+1) ). (End)
a(n) ~ exp(n/LambertW(n) + 3*LambertW(n)/2 - n - 1) * n^(n + 1/2) / LambertW(n)^(n+1). - Vaclav Kotesovec, Jun 09 2020
a(0) = 1; a(n) = 2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - Ilya Gutkovskiy, Jul 02 2020
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
a(n) = Sum_{k=0..n} 3^k*A124323(n, k). - Mélika Tebni, Jun 02 2022

Extensions

Definition revised by David Callan, Oct 11 2005

A001610 a(n) = a(n-1) + a(n-2) + 1, with a(0) = 0 and a(1) = 2.

Original entry on oeis.org

0, 2, 3, 6, 10, 17, 28, 46, 75, 122, 198, 321, 520, 842, 1363, 2206, 3570, 5777, 9348, 15126, 24475, 39602, 64078, 103681, 167760, 271442, 439203, 710646, 1149850, 1860497, 3010348, 4870846, 7881195, 12752042, 20633238, 33385281, 54018520
Offset: 0

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Author

Keywords

Comments

For prime p, p divides a(p-1). - T. D. Noe, Apr 11 2009 [This result follows immediately from the fact that A032190(n) = (1/n)*Sum_{d|n} a(d-1)*phi(n/d). - Petros Hadjicostas, Sep 11 2017]
Generalization. If a(0,x)=0, a(1,x)=2 and, for n>=2, a(n,x)=a(n-1,x)+x*a(n-2,x)+1, then we obtain a sequence of polynomials Q_n(x)=a(n,x) of degree floor((n-1)/2), such that p is prime iff all coefficients of Q_(p-1)(x) are multiple of p (sf. A174625). Thus a(n) is the sum of coefficients of Q_(n-1)(x). - Vladimir Shevelev, Apr 23 2010
Odd composite numbers n such that n divides a(n-1) are in A005845. - Zak Seidov, May 04 2010; comment edited by N. J. A. Sloane, Aug 10 2010
a(n) is the number of ways to modify a circular arrangement of n objects by swapping one or more adjacent pairs. E.g., for 1234, new arrangements are 2134, 2143, 1324, 4321, 1243, 4231 (taking 4 and 1 to be adjacent) and a(4) = 6. - Toby Gottfried, Aug 21 2011
For n>2, a(n) equals the number of Markov equivalence classes with skeleton the cycle on n+1 nodes. See Theorem 2.1 in the article by A. Radhakrishnan et al. below. - Liam Solus, Aug 23 2018
From Gus Wiseman, Feb 12 2019: (Start)
For n > 0, also the number of nonempty subsets of {1, ..., n + 1} containing no two cyclically successive elements (cyclically successive means 1 succeeds n + 1). For example, the a(5) = 17 stable subsets are:
{1}, {2}, {3}, {4}, {5}, {6},
{1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {2,6}, {3,5}, {3,6}, {4,6},
{1,3,5}, {2,4,6}.
(End)
Also the rank of the n-Lucas cube graph. - Eric W. Weisstein, Aug 01 2023

References

  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • GAP
    List([0..40], n-> Lucas(1,-1,n+1)[2] -1); # G. C. Greubel, Jul 12 2019
  • Haskell
    a001610 n = a001610_list !! n
    a001610_list =
       0 : 2 : map (+ 1) (zipWith (+) a001610_list (tail a001610_list))
    -- Reinhard Zumkeller, Aug 21 2011
    
  • Magma
    I:=[0,2]; [n le 2 select I[n] else Self(n-1)+Self(n-2)+1: n in [1..40]]; // Vincenzo Librandi, Mar 20 2015
    
  • Magma
    [Lucas(n+1) -1: n in [0..40]]; // G. C. Greubel, Jul 12 2019
    
  • Mathematica
    t = {0, 2}; Do[AppendTo[t, t[[-1]] + t[[-2]] + 1], {n, 2, 40}]; t
    RecurrenceTable[{a[n] == a[n - 1] +a[n - 2] +1, a[0] == 0, a[1] == 2}, a, {n, 0, 40}] (* Robert G. Wilson v, Apr 13 2013 *)
    CoefficientList[Series[x (2 - x)/((1 - x - x^2) (1 - x)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
    Table[Fibonacci[n] + Fibonacci[n + 2] - 1, {n, 0, 40}] (* Eric W. Weisstein, Feb 13 2018 *)
    LinearRecurrence[{2, 0, -1}, {2, 3, 6}, 20] (* Eric W. Weisstein, Feb 13 2018 *)
    Table[LucasL[n] - 1, {n, 20}] (* Eric W. Weisstein, Aug 01 2023 *)
    LucasL[Range[20]] - 1 (* Eric W. Weisstein, Aug 01 2023 *)
  • PARI
    a(n)=([0,1,0; 0,0,1; -1,0,2]^n*[0;2;3])[1,1] \\ Charles R Greathouse IV, Sep 08 2016
    
  • PARI
    vector(40, n, f=fibonacci; f(n+1)+f(n-1)-1) \\ G. C. Greubel, Jul 12 2019
    
  • Sage
    [lucas_number2(n+1,1,-1) -1 for n in (0..40)] # G. C. Greubel, Jul 12 2019
    

Formula

a(n) = A000204(n)-1 = A000032(n+1)-1 = A000071(n+1) + A000045(n).
G.f.: x*(2-x)/((1-x-x^2)*(1-x)) = (2*x-x^2)/(1-2*x+x^3). [Simon Plouffe in his 1992 dissertation]
a(n) = F(n) + F(n+2) - 1 where F(n) is the n-th Fibonacci number. - Zerinvary Lajos, Jan 31 2008
a(n) = A014217(n+1) - A000035(n+1). - Paul Curtz, Sep 21 2008
a(n) = Sum_{i=1..floor((n+1)/2)} ((n+1)/i)*C(n-i,i-1). In more general case of polynomials Q_n(x)=a(n,x) (see our comment) we have Q_n(x) = Sum_{i=1..floor((n+1)/2)}((n+1)/i)*C(n-i,i-1)*x^(i-1). - Vladimir Shevelev, Apr 23 2010
a(n) = Sum_{k=0..n-1} Lucas(k), where Lucas(n) = A000032(n). - Gary Detlefs, Dec 07 2010
a(0)=0, a(1)=2, a(2)=3; for n>=3, a(n) = 2*a(n-1) - a(n-3). - George F. Johnson, Jan 28 2013
For n > 1, a(n) = A048162(n+1) + 3. - Toby Gottfried, Apr 13 2013
For n > 0, a(n) = A169985(n + 1) - 1. - Gus Wiseman, Feb 12 2019

A052841 Expansion of e.g.f.: 1/(exp(x)*(2-exp(x))).

Original entry on oeis.org

1, 0, 2, 6, 38, 270, 2342, 23646, 272918, 3543630, 51123782, 811316286, 14045783798, 263429174190, 5320671485222, 115141595488926, 2657827340990678, 65185383514567950, 1692767331628422662, 46400793659664205566, 1338843898122192101558, 40562412499252036940910
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

From Michael Somos, Mar 04 2004: (Start)
Stirling transform of A005359(n)=[0,2,0,24,0,720,...] is a(n)=[0,2,6,38,270,...].
Stirling transform of -(-1)^n*A052657(n-1)=[0,0,2,-6,48,-240,...] is a(n-1)=[0,0,2,6,38,270,...].
Stirling transform of -(-1)^n*A052558(n-1)=[1,-1,4,-12,72,-360,...] is a(n-1)=[1,0,2,6,38,270,...].
Stirling transform of 2*A052591(n)=[2,4,24,96,...] is a(n+1)=[2,6,38,270,...].
(End)
Also the central moments of a Geometric(1/2) random variable (for example the number of coin tosses until the first head). - Svante Janson, Dec 10 2012
Also the number of ordered set partitions of {1..n} with no cyclical adjacencies (successive elements in the same block, where 1 is a successor of n). - Gus Wiseman, Feb 13 2019
Also the number of ordered set partitions of {1..n} with an even number of blocks. - Geoffrey Critzer, Jul 04 2020

Examples

			From _Gus Wiseman_, Feb 13 2019: (Start)
The a(4) = 38 ordered set partitions with no cyclical adjacencies:
  {{1}{2}{3}{4}}  {{1}{24}{3}}  {{13}{24}}
  {{1}{2}{4}{3}}  {{1}{3}{24}}  {{24}{13}}
  {{1}{3}{2}{4}}  {{13}{2}{4}}
  {{1}{3}{4}{2}}  {{13}{4}{2}}
  {{1}{4}{2}{3}}  {{2}{13}{4}}
  {{1}{4}{3}{2}}  {{2}{4}{13}}
  {{2}{1}{3}{4}}  {{24}{1}{3}}
  {{2}{1}{4}{3}}  {{24}{3}{1}}
  {{2}{3}{1}{4}}  {{3}{1}{24}}
  {{2}{3}{4}{1}}  {{3}{24}{1}}
  {{2}{4}{1}{3}}  {{4}{13}{2}}
  {{2}{4}{3}{1}}  {{4}{2}{13}}
  {{3}{1}{2}{4}}
  {{3}{1}{4}{2}}
  {{3}{2}{1}{4}}
  {{3}{2}{4}{1}}
  {{3}{4}{1}{2}}
  {{3}{4}{2}{1}}
  {{4}{1}{2}{3}}
  {{4}{1}{3}{2}}
  {{4}{2}{1}{3}}
  {{4}{2}{3}{1}}
  {{4}{3}{1}{2}}
  {{4}{3}{2}{1}}
(End)
		

Crossrefs

Main diagonal of A122101.
Inverse binomial transform of A000670.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 40);
    Coefficients(R!(Laplace( Exp(-x)/(2-Exp(x)) ))); // G. C. Greubel, Jun 11 2024
    
  • Maple
    spec := [S,{B=Prod(C,C),C=Set(Z,1 <= card),S=Sequence(B)},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
    P := proc(n,x) option remember; if n = 0 then 1 else
    (n*x+2*(1-x))*P(n-1,x)+x*(1-x)*diff(P(n-1,x),x); expand(%) fi end:
    A052841 := n -> subs(x=2, P(n,x)):
    seq(A052841(n), n=0..21); # Peter Luschny, Mar 07 2014
    h := n -> add(combinat:-eulerian1(n, k)*2^k, k=0..n):
    a := n -> (h(n)+(-1)^n)/2: seq(a(n), n=0..21); # Peter Luschny, Sep 19 2015
    b := proc(n, m) option remember; if n = 0 then 1 else
         (m - 1)*b(n - 1, m) + (m + 1)*b(n - 1, m + 1) fi end:
    a := n -> b(n, 0): seq(a(n), n = 0..21); # Peter Luschny, Jun 23 2023
  • Mathematica
    a[n_] := If[n == 0, 1, (PolyLog[-n, 1/2]/2 + (-1)^n)/2]; (* or *)
    a[n_] := HurwitzLerchPhi[1/2, -n, -1]/2; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Feb 19 2016, after Vladeta Jovovic *)
    With[{nn=30},CoefficientList[Series[1/(Exp[x](2-Exp[x])),{x,0,nn}],x] Range[ 0,nn]!] (* Harvey P. Dale, Apr 08 2019 *)
  • PARI
    a(n)=if(n<0,0,n!*polcoeff(subst(1/(1-y^2),y,exp(x+x*O(x^n))-1),n))
    
  • PARI
    {a(n)=polcoeff(sum(m=0,n,(2*m)!*x^(2*m)/prod(k=1,2*m,1-k*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 20 2011 */
    
  • SageMath
    def A052841_list(prec):
        P. = PowerSeriesRing(QQ, prec)
        return P( exp(-x)/(2-exp(x)) ).egf_to_ogf().list()
    A052841_list(40) # G. C. Greubel, Jun 11 2024

Formula

O.g.f.: Sum_{n>=0} (2*n)! * x^(2*n) / Product_{k=1..2*n} (1-k*x). - Paul D. Hanna, Jul 20 2011
a(n) = (A000670(n) + (-1)^n)/2 = Sum_{k>=0} (k-1)^n/2^(k+1). - Vladeta Jovovic, Feb 02 2003
Also, a(n) = Sum_{k=0..[n/2]} (2k)!*Stirling2(n, 2k). - Ralf Stephan, May 23 2004
a(n) = D^n*(1/(1-x^2)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A000670 and A005649. - Peter Bala, Nov 25 2011
E.g.f.: 1/(2*G(0)), where G(k) = 1 - 2^k/(2 - 4*x/(2*x - 2^k*(k+1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
a(n) ~ n!/(4*(log(2))^(n+1)). - Vaclav Kotesovec, Aug 10 2013
a(n) = (h(n)+(-1)^n)/2 where h(n) = Sum_{k=0..n} E(n,k)*2^k and E(n,k) the Eulerian numbers A173018 (see also A156365). - Peter Luschny, Sep 19 2015
a(n) = (-1)^n + Sum_{k=0..n-1} binomial(n,k) * a(k). - Ilya Gutkovskiy, Jun 11 2020

Extensions

Edited by N. J. A. Sloane, Sep 06 2013
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