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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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

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Author

N. J. A. Sloane

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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  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 210.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 92-93.
  • 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.
  • De Angelis, Valerio, and Dominic Marcello. "Wilf's Conjecture." The American Mathematical Monthly 123.6 (2016): 557-573.
  • N. G. de Bruijn, Asymptotic Methods in Analysis, Dover, 1981, Sections 3.3. Case b and 6.1-6.3.
  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 52, p. 19, Ellipses, Paris 2008.
  • G. Dobinski, Summierung der Reihe Sum(n^m/n!) für m = 1, 2, 3, 4, 5, ..., Grunert Archiv (Arch. f. Math. und Physik), 61 (1877) 333-336.
  • L. F. Epstein, A function related to the series for exp(exp(z)), J. Math. and Phys., 18 (1939), 153-173.
  • G. Everest, A. van der Poorten, I. Shparlinski and T. Ward, Recurrence Sequences, Amer. Math. Soc., 2003; see esp. p. 255.
  • Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 5.8, p. 321.
  • Flajolet, Philippe and Schott, Rene, Nonoverlapping partitions, continued fractions, Bessel functions and a divergent series, European J. Combin. 11 (1990), no. 5, 421-432.
  • Martin Gardner, Fractal Music, Hypercards and More (Freeman, 1992), Chapter 2.
  • H. W. Gould, Research bibliography of two special number sequences, Mathematica Monongaliae, Vol. 12, 1971.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, 2nd ed., p. 493.
  • Silvia Heubach and Toufik Mansour, Combinatorics of Compositions and Words, CRC Press, 2010.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 26.
  • D. E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.2.1.5 (p. 418).
  • Christian Kramp, Der polynomische Lehrsatz (Leipzig: 1796), 113.
  • Lehmer, D. H. Some recursive sequences. Proceedings of the Manitoba Conference on Numerical Mathematics (Univ. Manitoba, Winnipeg, Man., 1971), pp. 15--30. Dept. Comput. Sci., Univ. Manitoba, Winnipeg, Man., 1971. MR0335426 (49 #208)
  • J. Levine and R. E. Dalton, Minimum periods, modulo p, of first-order Bell exponential integers, Math. Comp., 16 (1962), 416-423.
  • 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.
  • Moser, Leo, and Max Wyman. An asymptotic formula for the Bell numbers. Trans. Royal Soc. Canada, 49 (1955), 49-53.
  • A. Murthy, Generalization of partition function, introducing Smarandache factor partition, Smarandache Notions Journal, Vol. 11, No. 1-2-3, Spring 2000.
  • 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.
  • P. Peart, Hankel determinants via Stieltjes matrices. Proceedings of the Thirty-first Southeastern International Conference on Combinatorics, Graph Theory and Computing (Boca Raton, FL, 2000). Congr. Numer. 144 (2000), 153-159.
  • A. M. Robert, A Course in p-adic Analysis, Springer-Verlag, 2000; p. 212.
  • G.-C. Rota, Finite Operator Calculus.
  • Frank Ruskey, Jennifer Woodcock and Yuji Yamauchi, Counting and computing the Rand and block distances of pairs of set partitions, Journal of Discrete Algorithms, Volume 16, October 2012, Pages 236-248.
  • 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. P. Stanley, Enumerative Combinatorics, Cambridge; see Section 1.4 and Example 5.2.4.
  • Abdullahi Umar, On the semigroups of order-decreasing finite full transformations, Proc. Roy. Soc. Edinburgh 120A (1992), 129-142.
  • Abdullahi Umar, On the semigroups of partial one-to-one order-decreasing finite transformations, Proc. Roy. Soc. Edinburgh 123A (1993), 355-363.

Links

Crossrefs

Equals row sums of triangle A008277 (Stirling subset numbers).
Partial sums give A005001. a(n) = A123158(n, 0).
See A061462 for powers of 2 dividing a(n).
Rightmost diagonal of triangle A121207. A144293 gives largest prime factor.
Cf. A000045, A000108, A000166, A000204, A000255, A000311, A000296, A003422, A024716, A029761, A049020, A058692, A060719, A084423, A087650, A094262, A103293, A165194, A165196, A173110, A227840, A182386.
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.).
Cf. A000371, A003465, A186021.

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

A008292 Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 1

Views

Author

N. J. A. Sloane, Mar 15 1996

Keywords

Comments

The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010
Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004
T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
Subtriangle for k>=1 and n>=1 of triangle A123125. - Philippe Deléham, Oct 22 2006
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.).
G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246.
(End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008
Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008
For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008
The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009
From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011
The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011
Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011
Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014
Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015
The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016
Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 with a Mar 31 2017 comment for a rewritten form. For the e.g.f see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017
For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018
Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4       5       6      7     8    9 10 ...
1:  1
2:  1    1
3:  1    4     1
4:  1   11    11      1
5:  1   26    66     26       1
6:  1   57   302    302      57       1
7:  1  120  1191   2416    1191     120      1
8:  1  247  4293  15619   15619    4293    247     1
9:  1  502 14608  88234  156190   88234  14608   502    1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - _Michael Somos_, Mar 17 2011
Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - _Vladimir Shevelev_, Jul 01 2011
Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
.   1o (1+t)         1o t         1o t
.   |                / \          / \
.   |               /   \        /   \
.   2o (1+t)      2o     3o    3o    2o
.   |
.   |
.   3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
  • Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
  • Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
  • K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.

Links

Crossrefs

Columns k=2..8 are A000295, A000460, A000498, A000505, A000514, A001243, A001244.
Cf. A019538, A028246, A048993, A048994, A049019, A086885, A090582, A129185, A131758, A139605, A173018.

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Sum([0..k],j->(-1)^j*(k-j)^n*Binomial(n+1,j))))); # Muniru A Asiru, Jun 29 2018
  • Haskell
    import Data.List (genericLength)
a008292 n k = a008292_tabl !! (n-1) !! (k-1)
a008292_row n = a008292_tabl !! (n-1)
a008292_tabl = iterate f [1] where
   f xs = zipWith (+)
     (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
     where ks = [1 .. 1 + genericLength xs]
-- Reinhard Zumkeller, May 07 2013
  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n,k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
  • Maple
    A008292 := proc(n,k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1,k)+(n-k+1)*procname(n-1,k-1) ; end if; end proc:
  • Mathematica
    t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)
Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)
Table[Tally[
   Count[#, x_ /; x > 0] & /@ (Differences /@
      Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */
  • PARI
    {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
  • PARI
    {A(n,c)=c^(n+c-1)+sum(i=1,c-1,(-1)^i/i!*(c-i)^(n+c-1)*prod(j=1,i,n+c+1-j))}
  • Python
    from sympy import binomial
def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017
  • R
    T <- function(n, k) {
  S <- numeric()
  for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
  return(sum(S))
}
for (n in 1:10){
  for (k in 1:n) print(T(n,k))
} # Indranil Ghosh, Nov 08 2017
  • Sage
    [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019

Formula

T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
Row sums = n! = A000142(n) unless n=0. - Michael Somos, Mar 17 2011
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011
For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002
From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002
Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Deléham's operator defined in A084938.
Sum_{k=1..n} T(n, k)*2^k = A000629(n). - Philippe Deléham, Jun 05 2004
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_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.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008
G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010
If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1.
The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0.
(End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011
T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011
E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - Geoffrey Critzer, Nov 10 2012
From Peter Bala, Mar 12 2013: (Start)
Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i).
(End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013
From Tom Copeland, Sep 18 2014: (Start)
A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271).
E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015
Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015
For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015
From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016
Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024

Extensions

Thanks to Michael Somos for additional comments.
Further comments from Christian G. Bower, May 12 2000

A001286 Lah numbers: a(n) = (n-1)*n!/2.

Original entry on oeis.org

1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600, 199584000, 2634508800, 37362124800, 566658892800, 9153720576000, 156920924160000, 2845499424768000, 54420176498688000, 1094805903679488000, 23112569077678080000, 510909421717094400000
Offset: 2

Views

Author

N. J. A. Sloane

Keywords

Comments

Number of surjections from {1,...,n} to {1,...,n-1}. - Benoit Cloitre, Dec 05 2003
First Eulerian transform of 0,1,2,3,4,... . - Ross La Haye, Mar 05 2005
With offset 0 : determinant of the n X n matrix m(i,j)=(i+j+1)!/i!/j!. - Benoit Cloitre, Apr 11 2005
These numbers arise when expressing n(n+1)(n+2)...(n+k)[n+(n+1)+(n+2)+...+(n+k)] as sums of squares: n(n+1)[n+(n+1)] = 6(1+4+9+16+ ... + n^2), n(n+1)(n+2)(n+(n+1)+(n+2)) = 36(1+(1+4)+(1+4+9)+...+(1+4+9+16+ ... + n^2)), n(n+1)(n+2)(n+3)(n+(n+1)+(n+2)+(n+3)) = 240(...), ... . - Alexander R. Povolotsky, Oct 16 2006
a(n) is the number of edges in the Hasse diagram for the weak Bruhat order on the symmetric group S_n. For permutations p,q in S_n, q covers p in the weak Bruhat order if p,q differ by an adjacent transposition and q has one more inversion than p. Thus 23514 covers 23154 due to the transposition that interchanges the third and fourth entries. Cf. A002538 for the strong Bruhat order. - David Callan, Nov 29 2007
a(n) is also the number of excedances in all permutations of {1,2,...,n} (an excedance of a permutation p is a value j such p(j)>j). Proof: j is exceeded (n-1)! times by each of the numbers j+1, j+2, ..., n; now, Sum_{j=1..n} (n-j)(n-1)! = n!(n-1)/2. Example: a(3)=6 because the number of excedances of the permutations 123, 132, 312, 213, 231, 321 are 0, 1, 1, 1, 2, 1, respectively. - Emeric Deutsch, Dec 15 2008
(-1)^(n+1)*a(n) is the determinant of the n X n matrix whose (i,j)-th element is 0 for i = j, is j-1 for j>i, and j for j < i. - Michel Lagneau, May 04 2010
Row sums of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is the total number of ascents (descents) over all n-permutations. a(n) = Sum_{k=1..n} A008292(n,k)*k. - Geoffrey Critzer, Jan 06 2013
For m>=4, a(m-2) is the number of Hamiltonian cycles in a simple graph with m vertices which is complete, except for one edge. Proof: think of distinct round-table seatings of m persons such that persons "1" and "2" may not be neighbors; the count is (m-3)(m-2)!/2. See also A001710. - Stanislav Sykora, Jun 17 2014
Popularity of left (right) children in treeshelves. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n. See A278677, A278678 or A278679 for more definitions and examples. See A008292 for the distribution of the left (right) children in treeshelves. - Sergey Kirgizov, Dec 24 2016

Examples

			G.f. = x^2 + 6*x^3 + 36*x^4 + 240*x^5 + 1800*x^6 + 15120*x^7 + 141120*x^8 + ...
a(10) = (1+2+3+4+5+6+7+8+9)*(1*2*3*4*5*6*7*8*9) = 16329600. - _Reinhard Zumkeller_, May 15 2010

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 90, ex. 4.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
  • 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).

Links

Crossrefs

Cf. A001710, A052609, A062119, A075181, A060638, A060608, A060570, A060612, A135218, A019538, A053495, A051683, A213168, A278677, A278678, A278679, A008292.
A002868 is an essentially identical sequence.
Column 2 of |A008297|.
Third column (m=2) of triangle |A111596(n, m)|: matrix product of |S1|.S2 Stirling number matrices.
Cf. also A000110, A000111.
Cf. A001113, A001620, A091725, A099285.

Programs

Formula

a(n) = Sum_{i=0..n-1} (-1)^(n-i-1) * i^n * binomial(n-1,i). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000 [corrected by Amiram Eldar, May 02 2022]
E.g.f.: x^2/[2(1-x)^2]. - Ralf Stephan, Apr 02 2004
a(n+1) = (-1)^(n+1)*det(M_n) where M_n is the n X n matrix M_(i,j)=max(i*(i+1)/2,j*(j+1)/2). - Benoit Cloitre, Apr 03 2004
Row sums of table A051683. - Alford Arnold, Sep 29 2006
5th binomial transform of A135218: (1, 1, 1, 25, 25, 745, 3145, ...). - Gary W. Adamson, Nov 23 2007
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j) then a(n)=(-1)^n*f(n,2,-2), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = A000217(n-1)*A000142(n-1). - Reinhard Zumkeller, May 15 2010
a(n) = (n+1)!*Sum_{k=1..n-1} 1/(k^2+3*k+2). - Gary Detlefs, Sep 14 2011
Sum_{n>=2} 1/a(n) = 2*(2 - exp(1) - gamma + Ei(1)) = 1.19924064599..., where gamma = A001620 and Ei(1) = A091725. - Ilya Gutkovskiy, Nov 24 2016
a(n+1) = a(n)*n*(n+1)/(n-1). - Chai Wah Wu, Apr 11 2018
Sum_{n>=2} (-1)^n/a(n) = 2*(gamma - Ei(-1)) - 2/e, where e = A001113 and Ei(-1) = -A099285. - Amiram Eldar, May 02 2022

A367955 Number T(n,k) of partitions of [n] whose block maxima sum to k, triangle T(n,k), n>=0, n<=k<=n*(n+1)/2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 5, 2, 3, 1, 1, 1, 2, 5, 10, 7, 7, 11, 3, 4, 1, 1, 1, 2, 5, 10, 23, 15, 23, 25, 37, 18, 14, 19, 4, 5, 1, 1, 1, 2, 5, 10, 23, 47, 39, 49, 81, 84, 129, 74, 78, 70, 87, 33, 23, 29, 5, 6, 1, 1, 1, 2, 5, 10, 23, 47, 103, 81, 129, 172, 261, 304, 431, 299, 325, 376, 317, 424, 196, 183, 144, 165, 52, 34, 41, 6, 7, 1
Offset: 0

Views

Author

Alois P. Heinz, Dec 05 2023

Keywords

Comments

Rows and also columns reversed converge to A365441.
T(n,k) is defined for all n,k >= 0. The triangle contains only the positive terms. T(n,k) = 0 if k < n or k > n*(n+1)/2.

Examples

			T(4,7) = 5: 123|4, 124|3, 13|24, 14|23, 1|2|34.
T(5,9) = 10: 1234|5, 1235|4, 124|35, 125|34, 134|25, 135|24, 14|235, 15|234, 1|23|45, 1|245|3.
T(5,13) = 3: 1|23|4|5, 1|24|3|5, 1|25|3|4.
T(5,14) = 4: 12|3|4|5, 13|2|4|5, 14|2|3|5, 15|2|3|4.
T(5,15) = 1: 1|2|3|4|5.
Triangle T(n,k) begins:
  1;
  .  1;
  .  .  1, 1;
  .  .  .  1, 1, 2, 1;
  .  .  .  .  1, 1, 2, 5, 2,  3,  1;
  .  .  .  .  .  1, 1, 2, 5, 10,  7,  7, 11,  3,  4,  1;
  .  .  .  .  .  .  1, 1, 2,  5, 10, 23, 15, 23, 25, 37, 18, 14, 19, 4, 5, 1;
  ...

Links

Crossrefs

Row sums give A000110.
Column sums give A204856.
Antidiagonal sums give A368102.
T(2n,3n) gives A365441.
T(n,2n) gives A368675.
Row maxima give A367969.
Row n has A000124(n-1) terms (for n>=1).
Cf. A000217, A124327 (the same for block minima), A200660, A278677.

Programs

  • Maple
    b:= proc(n, m) option remember; `if`(n=0, 1,
      b(n-1, m)*m + expand(x^n*b(n-1, m+1)))
    end:
T:= (n, k)-> coeff(b(n, 0), x, k):
seq(seq(T(n, k), k=n..n*(n+1)/2), n=0..10);
# second Maple program:
b:= proc(n, i, t) option remember; `if`(i*(i+1)/2 b(k, n, 0):
seq(seq(T(n, k), k=n..n*(n+1)/2), n=0..10);
  • Mathematica
    b[n_, i_, t_] := b[n, i, t] = If[i*(i + 1)/2 < n, 0, If[n == 0, t^i, If[t == 0, 0, t*b[n, i - 1, t]] + (t + 1)^Max[0, 2*i - n - 1]*b[n - i, Min[n - i, i - 1], t + 1]]];
T[0, 0] = 1; T[n_, k_] := b[k, n, 0];
Table[Table[T[n, k], {k, n, n*(n + 1)/2}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Oct 03 2024, after Alois P. Heinz's second Maple program *)

Formula

Sum_{k=n..n*(n+1)/2} k * T(n,k) = A278677(n-1) for n>=1.
Sum_{k=n..n*(n+1)/2} (k-n) * T(n,k) = A200660(n) for n>=1.
T(n,n) = T(n,n*(n+1)/2) = 1.

A101053 a(n) = n! * Sum_{k=0..n} Bell(k)/k! (cf. A000110).

Original entry on oeis.org

1, 2, 6, 23, 107, 587, 3725, 26952, 219756, 1998951, 20105485, 221838905, 2666280457, 34689290378, 485840964614, 7288997427755, 116634438986227, 1982868327635663, 35692311974248093, 678159760252918824, 13563246929216611852, 284828660383365005643
Offset: 0

Views

Author

Karol A. Penson, Nov 29 2004

Keywords

Comments

Sequence was originally defined as an infinite sum involving generalized Laguerre polynomials: a(n) = ((-1)^n*n!/exp(1))*Sum_{k>=0} LaguerreL(n,-n-1,k)/k!, n=0,1... . It appears in the problem of normal ordering of functions of boson operators.
a(n) is the number of ways to linearly order the elements in a (possibly empty) subset S of {1,2,...,n} and then partition the complement of S. - Geoffrey Critzer, Aug 07 2015

Links

Crossrefs

Cf. A000110, A186755, A278677, A352270.

Programs

  • Magma
    m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!( Exp(Exp(x)-1)/(1-x) )); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Mar 31 2019
  • Maple
    with(combinat): a:=n->add(bell(j)*n!/j!,j=0..n): seq(a(n),n=0..20); # Zerinvary Lajos, Mar 19 2007
  • Mathematica
    nn = 21; Range[0, nn]! CoefficientList[Series[Exp[(Exp[x]-1)]/(1-x), {x, 0, nn}], x] (* Geoffrey Critzer, Aug 07 2015 *)
  • PARI
    egf(s)=my(v=Vec(s),i); while(polcoeff(s,i)==0,i++); i--; vector(i+#v,j,polcoeff(s,j+i)*(j+i)!)
egf(exp(exp(x)-1)/(1-x)) \\ Charles R Greathouse IV, Aug 07 2015
  • PARI
    my(x='x+O('x^30)); Vec(serlaplace( exp(exp(x)-1)/(1-x) )) \\ G. C. Greubel, Mar 31 2019
  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=1, i, ((j-1)!+1)*binomial(i-1, j-1)*v[i-j+1])); v; \\ Seiichi Manyama, Jul 14 2022
  • Sage
    m = 30; T = taylor(exp(exp(x)-1)/(1-x), x, 0, m); [factorial(n)*T.coefficient(x, n) for n in (0..m)] # G. C. Greubel, Mar 31 2019

Formula

E.g.f: exp(exp(x)-1)/(1-x).
a(n) ~ exp(exp(1)-1) * n!. - Vaclav Kotesovec, Jun 26 2022
a(0) = 1; a(n) = Sum_{k=1..n} ((k-1)! + 1) * binomial(n-1,k-1) * a(n-k). - Seiichi Manyama, Jul 14 2022

Extensions

New definition from Vladeta Jovovic, Dec 01 2004

A285595 Sum T(n,k) of the k-th entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 4, 2, 17, 10, 3, 76, 52, 18, 4, 362, 274, 111, 28, 5, 1842, 1500, 675, 200, 40, 6, 9991, 8614, 4185, 1380, 325, 54, 7, 57568, 51992, 26832, 9568, 2510, 492, 70, 8, 351125, 329650, 178755, 67820, 19255, 4206, 707, 88, 9, 2259302, 2192434, 1239351, 494828, 149605, 35382, 6629, 976, 108, 10
Offset: 1

Views

Author

Alois P. Heinz, Apr 22 2017

Keywords

Comments

T(n,k) is also k times the number of blocks of size >k in all set partitions of [n+1]. T(3,2) = 10 = 2 * 5 because there are 5 blocks of size >2 in all set partitions of [4], namely in 1234, 123|4, 124|3, 134|2, 1|234.

Examples

			T(3,2) = 10 because the sum of the second entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+2+3+3+0  = 10.
Triangle T(n,k) begins:
      1;
      4,     2;
     17,    10,     3;
     76,    52,    18,    4;
    362,   274,   111,   28,    5;
   1842,  1500,   675,  200,   40,   6;
   9991,  8614,  4185, 1380,  325,  54,  7;
  57568, 51992, 26832, 9568, 2510, 492, 70, 8;
  ...

Links

Crossrefs

Column k=1 gives A124325(n+1).
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Main diagonal and first lower diagonal give: A000027, A028552.
Cf. A007318, A175757, A278677, A283424, A285362, A285793, A286897.

Programs

  • Maple
    T:= proc(h) option remember; local b; b:=
      proc(n, l) option remember; `if`(n=0, [1, 0],
        (p-> p+[0, (h-n+1)*p[1]*x^1])(b(n-1, [l[], 1]))+
         add((p-> p+[0, (h-n+1)*p[1]*x^(l[j]+1)])(b(n-1,
         sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
      end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2])
    end:
seq(T(n), n=1..12);
# second Maple program:
b:= proc(n) option remember; `if`(n=0, [1, 0],
      add((p-> p+[0, p[1]*add(x^k, k=1..j-1)])(
         b(n-j)*binomial(n-1, j-1)), j=1..n))
    end:
T:= n-> (p-> seq(coeff(p, x, i)*i, i=1..n))(b(n+1)[2]):
seq(T(n), n=1..12);
  • Mathematica
    b[n_] := b[n] = If[n == 0, {1, 0}, Sum[# + {0, #[[1]]*Sum[x^k, {k, 1, j-1} ]}&[b[n - j]*Binomial[n - 1, j - 1]], {j, 1, n}]];
T[n_] := Table[Coefficient[#, x, i]*i, {i, 1, n}] &[b[n + 1][[2]]];
Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, May 23 2018, translated from 2nd Maple program *)

Formula

T(n,k) = k * Sum_{j=k+1..n+1} binomial(n+1,j)*A000110(n+1-j).
T(n,k) = k * Sum_{j=k+1..n+1} A175757(n+1,j).
Sum_{k=1..n} T(n,k)/k = A278677(n-1).

A278678 Popularity of left children in treeshelves avoiding pattern T321.

Original entry on oeis.org

1, 4, 19, 94, 519, 3144, 20903, 151418, 1188947, 10064924, 91426347, 887296422, 9164847535, 100398851344, 1162831155151, 14198949045106, 182317628906283, 2455925711626404, 34632584722468115, 510251350142181470, 7840215226100517191, 125427339735162102104
Offset: 2

Views

Author

Sergey Kirgizov, Nov 26 2016

Keywords

Comments

Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Classical Françon's bijection maps bijectively treeshelves into permutations. Pattern T321 illustrated below corresponds to a treeshelf constructed from permutation 321. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n.

Examples

			Treeshelves of size 3:
      1  1          1    1       1        1
     /    \        /      \     / \      / \
    2      2      /        \   2   \    /   2
   /        \    2          2       3  3
  3          3    \        /
                   3      3
Pattern T321:
      1
     /
    2
   /
  3
Treeshelves of size 3 that avoid pattern T321:
  1          1    1       1        1
   \        /      \     / \      / \
    2      /        \   2   \    /   2
     \    2          2       3  3
      3    \        /
            3      3
Popularity of left children is 4.

Links

Crossrefs

Cf. A000110, A000111, A000142, A001286, A008292, A131178, A278677, A278679.

Programs

  • Maple
    b:= proc(u, o) option remember; `if`(u+o=0, 1,
      add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> (n+1)*b(n+1, 0)-b(n+2, 0):
seq(a(n), n=2..25);  # Alois P. Heinz, Oct 27 2017
  • Mathematica
    b[u_, o_] := b[u, o] = If[u+o == 0, 1, Sum[b[o-1+j, u-j], {j, 1, u}]];
a[n_] := (n+1)*b[n+1, 0] - b[n+2, 0];
Table[a[n], {n, 2, 25}] (* Jean-François Alcover, Nov 06 2017, after Alois P. Heinz *)
  • Python
    # by Taylor expansion
from sympy import *
from sympy.abc import z
h = (-sin(z) + 1 + (z-1)*cos(z))/ (1-sin(z))**2
NUMBER_OF_COEFFS = 20
coeffs = Poly(series(h,n = NUMBER_OF_COEFFS)).coeffs()
coeffs.reverse()
# and remove first coefficient 1 that corresponds to O(n**k)
coeffs.pop(0)
print([coeffs[n]*factorial(n+2) for n in range(len(coeffs))])

Formula

E.g.f.: (-sin(z) + 1 + (z-1)*cos(z))/ (1-sin(z))^2.
a(n) = (n+1)*e(n) - e(n+1), where e(n) is the n-th Euler number (see A000111).
Asymptotic: a(n) ~ 8*(Pi-2) / Pi^3 * n^2 * (2/Pi)^n.

A278679 Popularity of left children in treeshelves avoiding pattern T213.

Original entry on oeis.org

1, 5, 24, 128, 770, 5190, 38864, 320704, 2894544, 28382800, 300575968, 3419882304, 41612735632, 539295974000, 7417120846080, 107904105986048, 1655634186628352, 26721851169634560, 452587550053179392, 8026445538106839040, 148751109541600495104
Offset: 2

Views

Author

Sergey Kirgizov, Nov 26 2016

Keywords

Comments

Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Classical Françon's bijection maps bijectively treeshelves into permutations. Pattern T213 illustrated below corresponds to a treeshelf constructed from permutation 213. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n.

Examples

			Treeshelves of size 3:
      1  1          1    1       1        1
     /    \        /      \     / \      / \
    2      2      /        \   2   \    /   2
   /        \    2          2       3  3
  3          3    \        /
                   3      3
Pattern T213:
    1
   / \
  2   \
       3
Treeshelves of size 3 that avoid pattern T213:
      1  1          1    1        1
     /    \        /      \      / \
    2      2      /        \    /   2
   /        \    2          2  3
  3          3    \        /
                   3      3
Popularity of left children is 5.

Links

Crossrefs

Cf. A000110, A000111, A000142, A001286, A008292, A131178, A278677, A278678.

Programs

  • Mathematica
    terms = 21;
egf = (E^(Sqrt[2] z)(4z - 4) - (Sqrt[2] - 2) E^(2 Sqrt[2] z) + Sqrt[2] + 2)/((Sqrt[2] - 2) E^(Sqrt[2] z) + 2 + Sqrt[2])^2;
CoefficientList[egf + O[z]^(terms + 2), z]*Range[0, terms + 1]! // Round // Drop[#, 2]& (* Jean-François Alcover, Jan 26 2019 *)
  • Python
    ## by Taylor expansion
from sympy import *
from sympy.abc import z
h = (exp(sqrt(2)*z) * (4*z-4) - (sqrt(2)-2)*exp(2*sqrt(2)*z) + sqrt(2) + 2) / ((sqrt(2)-2)*exp(sqrt(2)*z) + 2 + sqrt(2))**2
NUMBER_OF_COEFFS = 20
coeffs = Poly(series(h,n = NUMBER_OF_COEFFS)).coeffs()
coeffs.reverse()
## and remove first coefficient 1 that corresponds to O(n**k)
coeffs.pop(0)
print([coeffs[n]*factorial(n+2) for n in range(len(coeffs))])

Formula

E.g.f.: (e^(sqrt(2)*z) * (4*z-4) - (sqrt(2)-2)*e^(2*sqrt(2)*z) + sqrt(2) + 2) / ((sqrt(2)-2)*e^(sqrt(2)*z) + 2 + sqrt(2))^2.
Asymptotic: n * (sqrt(2) / log(2*sqrt(2)+3) )^(n+1).

Extensions

More terms from Alois P. Heinz, Oct 27 2017

A286897 Sum T(n,k) of the k-th last entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.

Original entry on oeis.org

1, 5, 1, 23, 6, 1, 109, 33, 7, 1, 544, 182, 45, 8, 1, 2876, 1034, 284, 59, 9, 1, 16113, 6122, 1815, 420, 75, 10, 1, 95495, 37927, 11931, 2987, 595, 93, 11, 1, 597155, 246030, 81205, 21620, 4665, 814, 113, 12, 1, 3929243, 1669941, 573724, 160607, 36900, 6979, 1082, 135, 13, 1
Offset: 1

Views

Author

Alois P. Heinz, May 15 2017

Keywords

Examples

			T(3,2) = 6 because the sum of the second last entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+1+1+2 = 6.
Triangle T(n,k) begins:
      1;
      5,     1;
     23,     6,     1;
    109,    33,     7,    1;
    544,   182,    45,    8,   1;
   2876,  1034,   284,   59,   9,  1;
  16113,  6122,  1815,  420,  75, 10,  1;
  95495, 37927, 11931, 2987, 595, 93, 11, 1;
  ...

Links

Crossrefs

Column k=1 gives A278677(n-1).
Row sums give A000110(n) * A000217(n) = A105488(n+3).
Cf. A285595.

Programs

  • Maple
    b:= proc(n, l) option remember; `if`(n=0, [1, 0],
      (p-> p+[0, n*p[1]*x^1])(b(n-1, [l[], 1]))+
       add((p-> p+[0, n*p[1]*x^(l[j]+1)])(b(n-1,
       sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n, [])[2]):
seq(T(n), n=1..14);
  • Mathematica
    b[0, ] = {1, 0}; b[n, l_] := b[n, l] = Function[p, p + {0, n*p[[1]]*x^1} ][b[n - 1, Append[l, 1]]] + Sum[Function[p, p + {0, n*p[[1]]*x^(l[[j]] + 1)}][b[n - 1, Reverse @ Sort[ReplacePart[l, j -> l[[j]] + 1]]]], {j, 1, Length[l]}];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, n}]][b[n, {}][[2]]];
Table[T[n], {n, 1, 14}] // Flatten (* Jean-François Alcover, May 26 2018, from Maple *)

A367850 Total sum of the block maxima minus the block minima over all partitions of [n].

Original entry on oeis.org

0, 0, 1, 6, 33, 182, 1034, 6122, 37927, 246030, 1669941, 11844324, 87644672, 675494180, 5413500801, 45040155758, 388441330457, 3467619369538, 31998729152474, 304846692965822, 2994781617653439, 30304301968015582, 315536869771786501, 3377398077726963112
Offset: 0

Views

Author

Alois P. Heinz, Dec 15 2023

Keywords

Examples

			a(3) = 6 = 2 + 1 + 2 + 1 + 0: 123, 12|3, 13|2, 1|23, 1|2|3.

Links

Crossrefs

Cf. A000110, A002538 (the same for permutations), A002620, A120325, A124325, A278677, A368338.

Programs

  • Maple
    b:= proc(n, m, t) option remember; `if`(n=0, [1, 0], (p->
      p+[0, p[1]*(n-t)])(b(n-1, m+1, t+1))+m*b(n-1, m, t+1))
    end:
a:= n-> b(n, 0, 1)[2]:
seq(a(n), n=0..23);
# second Maple program:
egf:= (z-2)*exp(2*z+exp(z)-1)+(2*z+1)*exp(z+exp(z)-1)+exp(exp(z)-1):
a:= n-> n!*coeff(series(egf, z, n+1), z, n):
seq(a(n), n=0..23);

Formula

E.g.f.: (z-2)*exp(2*z+exp(z)-1)+(2*z+1)*exp(z+exp(z)-1)+exp(exp(z)-1).
a(n) = A278677(n-1) - A124325(n+1) for n>=1.
a(n) = Bell(n+1)+(n+1)*Bell(n)-Bell(n+2)+Sum_{k=0..n} Stirling2(n+1,k)*(n+1-k).
a(n) = Sum_{k=0..A002620(n)} k * A368338(n,k).
a(n) mod 2 = A120325(n).
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