A185422 -id:A185422 - cp's OEIS frontend

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1

Offset: 1

N. J. A. Sloane

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
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 f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
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, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by:  E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016

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Index entries for "core" sequences

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810-A039813, A048994, A087107-A087111, A087127, A094262, A127701.
See also A331155.
Cf. A038207, A086885, A132440, A138378, A173018, A238385.
Cf. A055154, A163353, A256894.
Cf. A000110 (row sums), A102661 (partial row sums).

a008277 n k = a008277_tabl !! (n-1) !! (k-1)
a008277_row n = a008277_tabl !! (n-1)
a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012

n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016

[[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019

seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);

Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
B = BellMatrix[1&, rows];
Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
a[n_, n_] := 1; a[n_, 1] := 1;
a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
With[{m = 11},
 Flatten@MapIndexed[Take[#, #2[[1]]] &,
   Transpose@
    Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)

create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */

for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013

Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012

stirling_number2 # Danny Rorabaugh, Oct 11 2015

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct 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, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
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)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A080635 Number of permutations on n letters without double falls and without initial falls.

Original entry on oeis.org

1, 1, 1, 3, 9, 39, 189, 1107, 7281, 54351, 448821, 4085883, 40533129, 435847959, 5045745069, 62594829027, 828229153761, 11644113200031, 173331882039141, 2723549731505163, 45047085512477049, 782326996336904679, 14233537708408467549, 270733989894887810547

Offset: 0

Emanuele Munarini, Feb 28 2003

A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
exp(x*(1-y+y^2)*D_y)*f(y)|_{y=0} = f(1-E(-x)) for any function f with a Taylor series. D_y means differentiation with respect to y and E(x) is the e.g.f. given below. For a proof of exp(x*g(y)*D_y)*f(y) = f(F^{-1}(x+F(y))) with the compositional inverse F^{-1} of F(y)=int(1/g(y),y) with F(0)=0 see, e.g., the Datolli et al. reference.
Number of increasing ordered trees on vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2. - David Callan, Mar 30 2007
Number of increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2. - Wenjin Woan, May 21 2011

			E.g.f. = 1 + x + (1/2)*x^2 + (1/2)*x^3 + (3/8)*x^4 + (13/40)*x^5 + (21/80)*x^6 + ...
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 39*x^5 + 189*x^6 + 1107*x^7 + ...
For n = 3: 123, 132, 231. For n = 4: 1234, 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
a(4)=9. The 9 plane (ordered) increasing unary-binary trees are
...................................................................
..4................................................................
..|................................................................
..3..........4...4...............4...4...............3...3.........
..|........./.....\............./.....\............./.....\........
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4...
..|.....\./.........\./.....\./.........\./.....\./.........\./....
..1......1...........1.......1...........1.......1...........1.....
...................................................................
..3...4...4...3....................................................
...\./.....\./.....................................................
....2.......2......................................................
....|.......|......................................................
....1.......1......................................................
...................................................................

Alois P. Heinz, Table of n, a(n) for n = 0..464
M. Aigner, Catalan and other numbers: a recurrent theme, Algebraic combinatorics and computer science, 347-390, Springer Italia, Milan, 2001.
Cyril Banderier et al., Explicit formulas for enumeration of lattice paths: basketball and the kernel method, arXiv:1609.06473 [math.CO], 2016. See series expansion p. 35.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, preprint.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
Miklós Bóna and István Mező, Limiting probabilities for vertices of a given rank in rooted trees, arXiv:1803.05033 [math.CO], 2018.
Miklós Bóna and István Mező, Limiting Probabilities for Vertices of a Given Rank in 1-2 Trees, The Electronic Journal of Combinatorics (2019) Vol. 26, No. 3, P#3.41.
Daniel Chen, Expected Number of Dice Rolls Until an Increasing Run of Three, arXiv:2308.14635 [math.CO], 2023. See p. 17.
G. Datolli, P. L. Ottaviani, A. Torre and L. Vazquez, Evolution operator equations: integration with algebraic and finite differences methods.[...], La Rivista del Nuovo Cimento 20,2 (1997) 1-133. eq. (I.2.18).
Colin Defant, Troupes, Cumulants, and Stack-Sorting, arXiv:2004.11367 [math.CO], 2020.
R. Ehrenborg, Cyclically consecutive permutation avoidance, arXiv:1312.2051 [math.CO], 2013.
S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125.
M. E. Jones and J. B. Remmel, Pattern matching in the cycle structures of permutations, Pure Math. Appl. (PU.M.A.), 22 (2011) 173-208.
Kaarel Hänni, Asymptotics of descent functions, MIT Mathematics UROP+ Papers (2020).
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.
Rupert Li, Vincular Pattern Avoidance on Cyclic Permutations, arXiv:2107.12353 [math.CO], 2021, p. 6.
Christopher Zhu, Enumerating Permutations and Rim Hooks Characterized by Double Descent Sets, arXiv:1910.12818 [math.CO], 2019.

Cf. A049774, A185415, A185416, A185422, A185423, A139605, A232864.
Cf. A000182, A002105, A234797.
Column k=0 of A162976.

a:= proc(n) if n < 2 then 1 else n! * sum((sqrt(3)/(2*Pi*(k+1/3)))^(n+1), k=-infinity..infinity) fi end: seq(a(n), n=0..30); # Richard Ehrenborg, Dec 09 2013
a := proc(n) option remember; local k; if n < 3 then 1 else
add(binomial(n-1, k)*a(k)*a(n-k-1), k = 0..n-2) fi end:
seq(a(n), n = 0..23); # Peter Luschny, May 24 2024

Table[n!, {n, 0, 40}]*CoefficientList[Series[ (1 + 1/Sqrt[3] Tan[Sqrt[3]/2 x])/(1 - 1/Sqrt[3] Tan[Sqrt[3]/2 x]), {x, 0, 40}], x]
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/2 + Sqrt[3]/2 Tan[ Pi/6 + Sqrt[3] x/2], {x, 0, n}]]; (* Michael Somos, May 22 2011 *)
Join[{1}, FullSimplify[Table[3^((n+1)/2) * n! * (Zeta[n+1, 1/3] - (-1)^n*Zeta[n+1, 2/3]) / (2*Pi)^(n+1), {n, 1, 20}]]] (* Vaclav Kotesovec, Aug 06 2021 *)

a(n):=if n=0 then 1 else sum((-3)^((n-k)/2)*((-1)^(n-k)+1)*sum(binomial(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^(j)*stirling2(n,j+k),j,0,n-k),k,1,n); /* Vladimir Kruchinin, Feb 13 2019 */

{a(n) = my(A); if( n<1, n==0, A = O(x); for( k=1, n, A = intformal( 1 + A + A^2)); n! * polcoeff( A, n))}; /* Michael Somos, Oct 04 2003 */

{a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(2*cosh(x +x*O(x^n)) - 1) ) )), n)}
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 22 2016

@CachedFunction
def c(n,k) :
    if n==k: return 1
    if k<1 or k>n: return 0
    return ((n-k)//2+1)*c(n-1,k-1)+2*k*c(n-1,k+1)
def A080635(n):
    return add(c(n,k) for k in (0..n))
[A080635(n) for n in (0..23)] # Peter Luschny, Jun 10 2014

E.g.f.: (1 + 1/sqrt(3) * tan(sqrt(3)/2 * x)) / (1 - 1/sqrt(3) * tan( sqrt(3)/2 * x)).
Recurrence: a(n+1) = (Sum_{k=0..n} binomial(n, k) * a(k) * a(n-k)) - a(n) + 0^n.
E.g.f.: A(x) satisfies A' = 1 - A + A^2. - Michael Somos, Oct 04 2003
E.g.f.: E(x) = (3*cos((1/2)*3^(1/2)*x) + (3^(1/2))*sin((1/2)*3^(1/2)*x))/(3*cos((1/2)*3^(1/2)*x) - (3^(1/2))* sin((1/2)*3^(1/2)*x)). See the Michael Somos comment. - Wolfdieter Lang, Sep 12 2005
O.g.f.: A(x) = 1+x/(1-x-2*x^2/(1-2*x-2*3*x^2/(1-3*x-3*4*x^2/(1-... -n*x-n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Peter Bala: (Start)
An alternative form of the e.g.f. for this sequence taken from [Bergeron et al.] is
(1)... (sqrt(3)/2)*tan((sqrt(3)/2)*x+Pi/6) [with constant term 1/2].
By comparing the egf for this sequence with the egf for the Eulerian numbers A008292 we can show that
(2)... a(n) = A(n,w)/(1+w)^(n-1) for n >= 1,
where w = exp(2*Pi*i/3) and {A(n,x),n>=1} = [1, 1+x, 1+4*x+x^2, 1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials. Equivalently,
(3)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k)*(-1/2 + sqrt(3)*i/6)^(k-1) for n >= 1, and
(4)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} (-1/2 + sqrt(3)*i/6)^(k-1)* Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n for n >= 1.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(5a)... a(p) == 1 (mod p) for prime p = 6*n+1,
(5b)... a(p) == -1 (mod p) for prime p = 6*n+5.
For the corresponding results for the case of non-plane unary-binary trees see A000111. For type B results see A001586. For a related sequence of polynomials see A185415. See also A185416 for a recursive method to compute this sequence. For forests of plane increasing unary binary trees see A185422 and A185423. (End)
O.g.f.: A(x) = x - (1/2)*x^2 + (1/2)*x^3 - (3/8)*x^4 + (13/40)*x^5 - (21/80)*x^6 + (123/560)*x^7 - (809/4480)*x^8 + (671/4480)*x^9 - (5541/44800)*x^10 + .... - Vladimir Kruchinin, Jan 18 2011
Let f(x) = 1+x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Oct 06 2011
From Sergei N. Gladkovskii, May 06 2013 - Dec 24 2013: (Start)
Continued fractions:
G.f.: 1 + 1/Q(0), where Q(k) = 1/(x*(k+1)) - 1 - 1/Q(k+1).
E.g.f.: 1 + 2*x/(W(0)-x), where W(k) = 4*k + 2 - 3*x^2/W(k+1).
G.f.: 1 + x/Q(0), m=1, where Q(k) = 1 - m*x*(2*k+1) - m*x^2*(2*k+1)*(2*k+2)/( 1 - m*x*(2*k+2) - m*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(k+1) - x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/( x^2*(k+1)*(k+2) - (1-x*(k+1))*(1-x*(k+2))/T(k+1) ).
G.f.: 1 + x/(G(0)-x), where G(k) = 1 + x*(k+1) - x*(k+1)/(1 - x*(k+2)/G(k+1) ). (End)
a(n) ~ 3^(3*(n+1)/2) * n^(n+1/2) / (exp(n)*(2*Pi)^(n+1/2)). - Vaclav Kotesovec, Oct 05 2013
a(n) = n! * Sum_{k=-oo..oo} (sqrt(3)/(2*Pi*(k+1/3)))^(n+1) for n >= 1. - Richard Ehrenborg, Dec 09 2013
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = (sqrt(3)/2)*tan((sqrt(3)/2)*x + Pi/6) satisfies the differential equation A"(x) = 2*A(x)*A'(x) with A(0) = 1/2 and A'(0) = 1, leading to the recurrence a(0) = 1/2, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the sequence [1/2, 1, 1, 3, 9, 39, 189, 1107, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 0 and a(1) = 1 produces A000182. Cf. A002105, A234797. (End)
E.g.f.: exp( Series_Reversion( Integral 1/(2*cosh(x) - 1) dx ) ). - Paul D. Hanna, Feb 22 2016
a(n) = Sum_{k=1..n} (-3)^((n-k)/2)*((-1)^(n-k)+1)*Sum_{j=0..n-k} C(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^j*Stirling2(n,j+k),n>0, a(0)=1. - Vladimir Kruchinin, Feb 13 2019
For n > 0, a(n) = 3^((n+1)/2) * n! * (zeta(n+1, 1/3) - (-1)^n*zeta(n+1, 2/3)) / (2*Pi)^(n+1). - Vaclav Kotesovec, Aug 06 2021

Several typos corrected by Olivier Gérard, Mar 26 2011

A147315 L-matrix for Euler numbers A000111(n+1).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 11, 6, 1, 16, 45, 35, 10, 1, 61, 211, 210, 85, 15, 1, 272, 1113, 1351, 700, 175, 21, 1, 1385, 6551, 9366, 5901, 1890, 322, 28, 1, 7936, 42585, 70055, 51870, 20181, 4410, 546, 36, 1, 50521, 303271, 563970, 479345, 218925, 58107, 9240, 870, 45, 1

Offset: 0

Paul Barry, Nov 05 2008

This is the inverse of the coefficient array for the orthogonal polynomials p(n,x) defined by: p(n,x)=if(n=-1,0,if(n=0,1,(x-n)p(n-1,x)-C(n,2)p(n-2,x))).
The Hankel array H for A000111(n+1) satisfies H=L*D*U with U the transpose of L.
Row sums are A000772(n+1) with e.g.f. dif(exp(-1)exp(sec(x)+tan(x)),x).
From Peter Bala, Jan 31 2011: (Start)
The following comments refer to the table with an offset of 1: i.e., both the row and column indexing starts at 1.
An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A000111(n) for n>=1 enumerates the number of increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2 (plane unary-binary trees in the notation of [Bergeron et al.])
The entry T(n,k) of the present table gives the number of forests of k increasing unordered trees on the vertex set {1,2,...,n} in which all outdegrees are <=2. See below for some examples.
For ordered forests of such trees see A185421. For forests of increasing ordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2, see A185422.
The Stirling number of the second kind Stirling2(n,k) is the number of partitions of the set [n] into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array, which counts increasing unary-binary trees, as generalized Stirling numbers of the second kind associated with A000111 or with the zigzag polynomials Z(n,x) of A147309 - see especially formulas (2) and (3) below.
See A145876 for generalized Eulerian numbers associated with A000111. (End)
The Bell transform of A000111(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
    1;
    1,    1;
    2,    3,    1;
    5,   11,    6,   1;
   16,   45,   35,  10,   1;
   61,  211,  210,  85,  15,  1;
  272, 1113, 1351, 700, 175, 21, 1;
  ...
The production array for L is the tridiagonal array
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  0,  0,  6,  4,  1;
  0,  0,  0, 10,  5,  1;
  0,  0,  0,  0, 15,  6,  1;
  0,  0,  0,  0,  0, 21,  7,  1;
  0,  0,  0,  0,  0,  0, 28,  8,  1,;
  0,  0,  0,  0,  0,  0,  0, 36,  9,  1;
From _Peter Bala_, Jan 31 2011: (Start)
Examples of forests:
The diagrams below are drawn so that the leftmost child of a binary node has the maximum label.
T(4,1) = 5. The 5 forests consisting of a single non-plane increasing unary-binary tree on 4 nodes are
...4... ........ .......... ........... ...........
...|... ........ .......... ........... ...........
...3... .4...3.. .4........ ........4.. ........3..
...|... ..\./... ..\....... ......./... ......./...
...2... ...2.... ...3...2.. ..3...2.... ..4...2....
...|... ...|.... ....\./... ...\./..... ...\./.....
...1... ...1.... .....1.... ....1...... ....1......
T(4,2) = 11. The 11 forests consisting of two non-plane increasing unary-binary trees on 4 nodes are
......... ...3.....
.3...2... ...|.....
..\./.... ...2.....
...1...4. ...|.....
......... ...1...4.
.
......... ...4.....
.4...2... ...|.....
..\./.... ...2.....
...1...3. ...|.....
......... ...1...3.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...1...2. ...|.....
......... ...1...2.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...2...1. ...|.....
......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... .......... (End)

Alois P. Heinz, Rows n = 0..140, flattened
Paul Barry, A Note on Three Families of Orthogonal Polynomials defined by Circular Functions, and Their Moment Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.7.2.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
Tom Copeland, Mathemagical Forests
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A145876, A147309, A185421, A185422, A185424.

A147315 := proc(n,k) n!*exp(x*(sec(t)+tan(t)-1)) - 1: coeftayl(%,t=0,n) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A147315(n,k),k=1..n),n=0..12) ; # R. J. Mathar, Mar 04 2011
# second Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
g:= proc(n) option remember; expand(`if`(n=0, 1, add(
      g(n-j)*x*binomial(n-1, j-1)*b(j, 0), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n+1))(g(n+1)):
seq(T(n), n=0..10);  # Alois P. Heinz, May 19 2021

t[n_, k_] := t[n, k] = t[n-1, k-1] + (k+1)*t[n-1, k] + 1/2*(k+1)*(k+2)*t[n-1, k+1]; t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[0, 0] = t[1, 0] = 1; Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 21 2011, after PARI prog. *)

Co(n,k):=sum(binomial(k,j)*(if oddp(n-k+j) then 0 else if (n-k+j)/2A147315(n,m):=1/m!*sum((if oddp(n-k) then 0 else 2^(1-k)*sum((-1)^(floor((n+k)/2)-i)*binomial(k,i)*(2*i-k)^n,i,0,floor(k/2)))*(sum(Co(i,m)*binomial(k-i+m-1,m-1),i,1,k)),k,m,n); /* Vladimir Kruchinin, Feb 17 2011 */

T(n,m):=(sum(binomial(k+m,m)*((-1)^(n-k-m)+1)*sum(binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)*stirling2(n+1,j+k+m+1), j,0,n-k-m), k,0,n-m))/(m+1)!; /* Vladimir Kruchinin, May 17 2011 */

{T(n,k)=if(k<0||k>n,0,if(n==0,1,T(n-1,k-1)+(k+1)*T(n-1,k)+(k+1)*(k+2)/2*T(n-1,k+1)))} /* offset=0 */

{T(n,k)=local(X=x+x*O(x^(n+2)));(n+1)!*polcoeff(polcoeff(exp(y*((1+sin(X))/cos(X)-1))-1,n+1,x),k+1,y)} /* offset=0 */

/* Generate from the production matrix P: */
{T(n,k)=local(P=matrix(n,n,r,c,if(r==c-1,1,if(r==c,c,if(r==c+1,c*(c+1)/2)))));if(k<0||k>n,0,if(n==k,1,(P^n)[1,k+1]))}

# uses[bell_matrix from A264428, A000111]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A000111(n+1), 10) # Peter Luschny, Jan 18 2016

From Peter Bala, Jan 31 2011: (Start)
The following formulas refer to the table with an offset of 1: i.e., both the row n and column k indexing start at 1.
GENERATING FUNCTION
E.g.f.:
(1)... exp(x*(sec(t)+tan(t)-1)) - 1 = Sum_{n>=1} R(n,x)*t^n/n!
= x*t + (x+x^2)*t^2/2! + (2*x+3*x^2+x^3)*t^3/3! + ....
TABLE ENTRIES
(2)... T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j),
where Z(n,x) denotes the zigzag polynomials as described in A147309.
Compare (2) with the formula for the Stirling numbers of the second kind
(3)... Stirling2(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
Recurrence relation
(4)... T(n+1,k) = T(n,k-1) + k*T(n,k) + (1/2)*k(k+1)*T(n,k+1).
ROW POLYNOMIALS
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x+x^2
R(3,x) = 2*x+3*x^2+x^3
They satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x) + (1/2)*R''(n,x)},
where ' indicates differentiation with respect to x. This should be compared with the recurrence satisfied by the Bell polynomials Bell(n,x)
(6)... Bell(n+1,x) = x*(Bell(n,x) + Bell'(n,x)). (End)
From Vladimir Kruchinin, Feb 17 2011: (Start)
Sum_{m=1..n} T(n,m) = A000772(n).
Sum_{m=1..2n-1} T(2n-1,m)* Stirling1(m,1) = A000364(n).
Let Co(n,k) = Sum_{j=1..k} binomial(k,j)*(if (n-k+j) is odd then 0 else if (n-k+j)/2T(n,m) = m!* Sum_{k=m..n} (if n-k is odd then 0 else 2^(1-k)) * Sum_{i=0..floor(k/2)} (-1)^(floor((n+k)/2)-i) * binomial(k,i) * (2*i-k)^n)))) * Sum_{i=1..k} Co(i,m) * binomial(k-i+m-1,m-1), n>0.
(End)
T(n,m) = Sum_{k = 0..n-m} binomial(k+m,m)*((-1)^(n-k-m)+1)*Sum_{j=0..n-k-m} binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)* Stirling2(n+1,j+k+m+1)/(m+1)!. - Vladimir Kruchinin, May 17 2011
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2/2!)*d/dt. Cf. A008277 and A094198. See also A185422. - Peter Bala, Nov 25 2011

More terms from Michel Marcus, Mar 01 2014

A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 0, 8, 0, 1, 18, 0, 20, 0, 1, 0, 148, 0, 40, 0, 1, 378, 0, 658, 0, 70, 0, 1, 0, 5040, 0, 2128, 0, 112, 0, 1, 14562, 0, 33992, 0, 5628, 0, 168, 0, 1, 0, 277164, 0, 158480, 0, 12936, 0, 240, 0, 1

Offset: 1

Peter Bala, Jan 27 2011

Define a sequence of polynomials P(n,x) by means of the recurrence relation
(1)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}
with starting value P(0,x) = 1. The first few polynomials are
  P(1,x) = x
  P(2,x) = x^2
  P(3,x) = x*(x^2+2),
  P(4,x) = x^2*(x^2+8),
  P(5,x) = x*(x^4+20*x^2+18).
This triangle lists the coefficients of these polynomials in ascending powers of x. The triangle has links with A080635, which gives the number of ordered increasing 0-1-2 trees on n nodes (plane unary-binary trees in the notation of [BERGERON et al.]). The number of forests of k such trees on n nodes is given by the formula
... 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
See A185422.
We also have A080635(n) = P(n,1), which can be used to calculate the terms of A080635 - see A185416.
For similarly defined polynomial sequences for other families of trees see A147309 and A185419. See also A185417.
Exponential Riordan array [(3/2)*(1-sqrt(3)*tan((Pi+3*sqrt(3)*x)/6))/(-1+2*sin((Pi-6*sqrt(3)*x)/6)), log((1/2)*(1+sqrt(3)*tan(sqrt(3)*x/2+Pi/6)))]. Production matrix is the exponential Riordan array [2*cosh(x)-1,x] beheaded. A185422*A008277^{-1}.

			Triangle begins:
  n\k|....1......2......3......4......5......6......7......8
  ==========================================================
  ..1|....1
  ..2|....0......1
  ..3|....2......0......1
  ..4|....0......8......0......1
  ..5|...18......0.....20......0......1
  ..6|....0....148......0.....40......0......1..
  ..7|..378......0....658......0.....70......0......1
  ..8|....0...5040......0...2128......0....112......0......1

F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of Increasing Trees, in Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1922, pp. 24-48.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees

Cf. A080635, A147309, A185417, A185419, A185422, A185423.

#A185415
P := proc(n,x)
description 'polynomial sequence P(n,x)'
if n = 0
return 1
else return
x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
end proc:
with(PolynomialTools):
for n from 1 to 10
CoefficientList(P(n,x),x);
end do;

p[0][x_] = 1; p[n_][x_] := p[n][x] = x*(p[n-1][x-1] - p[n-1][x] + p[n-1][x+1]) // Expand; row[n_] := CoefficientList[ p[n][x], x]; Table[row[n] // Rest, {n, 1, 10}] // Flatten (* Jean-François Alcover, Sep 11 2012 *)

GENERATING FUNCTION
The e.g.f. is
(1)... F(x, t) = E(t)^x = Sum_{n >= 0} P(n, x) * t^n/n!,
where
E(t) = 1/2+sqrt(3)/2*tan[sqrt(3)/2*t+Pi/6] = 1 + t + t^2/2! + 3*t^3/3! + 9*t^4/4! + ... is the e.g.f. for A080635.
ROW POLYNOMIALS
One easily checks that
... d/dt(F(x,t)) = x*(F(x-1,t)-F(x,t)+F(x+1,t))
and hence the row generating polynomials P(n,x) satisfy the recurrence relation
(2)... P(n+1,x) = x*{P(n,x-1)-P(n,x)+P(n,x+1)}.
RELATIONS WITH OTHER SEQUENCES
A080635(n) = P(n,1).
A185422(n,k) = 1/k!*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).
A185423(n,k) = Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*P(n,j).

A185423 Ordered forests of k increasing plane unary-binary trees with n nodes.

Original entry on oeis.org

1, 1, 2, 3, 6, 6, 9, 30, 36, 24, 39, 150, 270, 240, 120, 189, 918, 1980, 2520, 1800, 720, 1107, 6174, 16254, 25200, 25200, 15120, 5040, 7281, 47070, 142884, 266616, 327600, 272160, 141120, 40320, 54351, 394470, 1369710, 2948400, 4331880

Offset: 1

Peter Bala, Jan 28 2011

An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A080635 enumerates increasing plane (ordered) unary-binary trees with n nodes with labels drawn from the set {1,2,...,n}. The entry T(n,k) of the present table counts ordered forests of k increasing plane unary-binary trees with n nodes. See below for an example. See A185421 for the corresponding table in the non-plane case. See A185422 for the table for unordered forests of increasing plane unary-binary trees.

			Triangle begins
n\k|....1......2......3......4......5......6......7
===================================================
..1|....1
..2|....1......2
..3|....3......6......6
..4|....9.....30.....36.....24
..5|...39....150....270....240....120
..6|..189....918...1980...2520...1800....720
..7|.1107...6174..16254..25200..25200..15120...5040
..
Examples of recurrence relation:
T(5,3) = 270 = 3*(T(4,2) + T(4,3) + T(4,4)) = 3*(30+36+24);
T(6,1) = 1*(T(5,0) + T(5,1) + T(5,2)) = 39+150.
Examples of ordered forests:
T(4,2) = 30. The 15 unordered forests consisting of two plane increasing unary-binary trees on 4 nodes are shown below. We can order the trees in a forest in two ways giving rise to 30 ordered forests.
......... ......... ...3.....
.2...3... .3...2... ...|.....
..\./.... ..\./.... ...2.....
...1...4. ...1...4. ...|.....
......... ......... ...1...4.
.
......... ......... ...4.....
.2...4... .4...2... ...|.....
..\./.... ..\./.... ...2.....
...1...3. ...1...3. ...|.....
......... ......... ...1...3.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...1...2. ...1...2. ...|.....
......... ......... ...1...2.
.
......... ......... ...4.....
.3...4... .4...3... ...|.....
..\./.... ..\./.... ...3.....
...2...1. ...2...1. ...|.....
......... ......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... ..........

Cf. A019538, A080635, A185421, A185422.

#A185423
P := proc(n,x)
description 'polynomial sequence P(n,x) A185415'
if n = 0 return 1
else
return x*(P(n-1,x-1)-P(n-1,x)+P(n-1,x+1))
end proc:
with combinat:
T:=(n,k) -> add ((-1)^(k-j)*binomial(k,j)*P(n,j),j = 0..k):
for n from 1 to 10 do
seq(T(n,k),k = 1..n);
end do;

{T(n,k)=if(n<1||k<1||k>n,0,if(n==1,if(k==1,1,0),k*(T(n-1,k-1)+T(n-1,k)+T(n-1,k+1))))}

TABLE ENTRIES
(1)... T(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*P(n,j),
where P(n,x) are the polynomials described in A185415.
Compare (1) with the formula for the number of ordered set partitions
(2)... A019538(n,k) = Sum_{j = 0..k} (-1)^j*binomial(k,j)*j^n.
RECURRENCE RELATION
(3)... T(n+1,k) = k*(T(n,k-1) + T(n,k) + T(n,k+1))
GENERATING FUNCTION
Let E(t) = 1/2 + (sqrt(3)/2)*tan(sqrt(3)/2*t + Pi/6) be the e.g.f. for A080635.
The e.g.f. for the present triangle is
(4)... 1/(1 + x*(1 - E(t))) = Sum {n >= 0} R(n,x)*t^n/n!
= 1 + x*t + (x + 2*x^2)*t^2/2! + (3*x + 6*x^2 + 6*x^3)*t^3/3! + ....
ROW POLYNOMIALS
The ordered Bell polynomials OB(n,x) are the row polynomials of A019538 given by
(5)... OB(n,x) = Sum_{k = 1..n} k!*Stirling2(n,k)*x^k.
By comparing the e.g.f.s for A019538 and the present table we obtain the surprising identity
(6)... (-i*sqrt(3))^(n-1)*OB(n,x)/x = R(n,y)/y,
where i = sqrt(-1) and x = (sqrt(3)*i/3)*y + (-1/2+sqrt(3)*i/6).
RELATIONS WITH OTHER SEQUENCES
Column 1: A080635.
A185422(n,k) = T(n,k)/k!.
Setting y = 0 in (6) gives
(7)... A080635(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k) * (-1/2+sqrt(3)*i/6)^(k-1).
The row polynomials have their zeros on the vertical line Re(z) = -1/2 in the complex plane (this follows from the above connection with the ordered Bell polynomials, which have negative real zeros). - Peter Bala, Oct 23 2023

A000772 E.g.f. exp(tan(x) + sec(x) - 1).

Original entry on oeis.org

1, 1, 2, 6, 23, 107, 583, 3633, 25444, 197620, 1684295, 15618141, 156453857, 1683050189, 19344093070, 236497985706, 3063827565763, 41916787157011, 603799270943519, 9132945141812301, 144708157060239704, 2396568154933265024, 41403636316192616995

Offset: 0

N. J. A. Sloane

nonn

The number of elevated increasing binary trees. There is no restriction on the outdegree at the root. - Wenjin Woan, Jan 09 2008

Alois P. Heinz, Table of n, a(n) for n = 0..478 (first 101 terms from T. D. Noe)
Letong Hong and Rupert Li, Length-Four Pattern Avoidance in Inversion Sequences, arXiv:2112.15081 [math.CO], 2021.

Cf. A000110, A094198, A147315, A185422.

m:=25; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(Tan(x) + Sec(x) - 1))); [Factorial(n-1)*b[n]: n in [1..m]]; // Vincenzo Librandi, Jan 30 2020

b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= proc(n) option remember; `if`(n=0, 1, add(
      a(n-j)*binomial(n-1,j-1)*b(j, 0), j=1..n))
    end:
seq(a(n), n=0..23);  # Alois P. Heinz, May 19 2021

nn = 25; Range[0, nn]! CoefficientList[Series[Exp[Tan[x] + Sec[x] - 1], {x, 0, nn}], x] (* T. D. Noe, Jun 20 2012 *)

a(n) = Sum_{k=1..n} A147315(n-1,k-1), n>0, a(0)=1. - Vladimir Kruchinin, Mar 10 2011
a(n) = D^n(exp(x)) evaluated at x = 0, where D is the operator (1+x+x^2/2!)*d/dx. Cf. A000110 and A094198. See also A185422. - Peter Bala, Nov 25 2011
a(n) ~ 2^n * exp(2/Pi - 1 + 4*sqrt(n/Pi) - n) * n^(n - 1/4) / Pi^(n + 1/4). - Vaclav Kotesovec, Jan 27 2020

Showing 1-6 of 6 results.

cp's OEIS Frontend

A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

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A080635 Number of permutations on n letters without double falls and without initial falls.

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A147315 L-matrix for Euler numbers A000111(n+1).

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A185415 Table of coefficients of a polynomial sequence of binomial type related to A080635.

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A185423 Ordered forests of k increasing plane unary-binary trees with n nodes.

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