A002105 -id:A002105 - cp's OEIS frontend

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

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

A094665 Another version of triangular array in A083061: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 4, 15, 15, 0, 34, 147, 210, 105, 0, 496, 2370, 4095, 3150, 945, 0, 11056, 56958, 111705, 107415, 51975, 10395, 0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 0, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490, 18918900, 2027025

Offset: 0

Philippe Deléham, Jun 07 2004, Jun 12 2007

Diagonals: A000007, A002105; A001147, A001880.
Define polynomials P(n,x) = x(2x+1)P(n-1,x+1) - 2x^2P(n-1,x), P(0,x) = 1. Sequence gives triangle read by rows, defined by P(n,x) = Sum_{k = 0..n} T(n,k)*x^k. - Philippe Deléham, Jun 20 2004
From Johannes W. Meijer, May 24 2009: (Start)
In A160464 we defined the coefficients of the ES1 matrix by ES1[2*m-1,n=1] = 2*eta(2*m-1) and the recurrence relation ES1[2*m-1,n] = ((2*n-2)/(2*n-1))*(ES1[2*m-1,n-1] - ES1[2*m-3,n-1]/(n-1)^2) for m the positive and negative integers and n >= 1. As usual eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. It is well-known that ES1[1-2*m,n=1] = (4^m-1)*(-bernoulli(2*m))/m for m >= 1. and together with the recurrence relation this leads to ES1[-1,n] = 0.5 for n >= 1.
We discovered that the n-th term of the row coefficients ES1[1-2*m,n] for m >= 1, can be generated with the rather simple polynomials RES1(1-2*m,n) = (-1)^(m+1)*ECGP(1-2*m, n)/2^m. This discovery was enabled by the recurrence relation for the RES1(1-2*m,n) which we derived from the recurrence relation for the ES1[2*m-1,n] coefficients and the fact that RES1(-1,n) = 0.5. The coefficients of the ECGP(1-2*m,n) polynomials led to this triangle and subsequently to triangle A083061. (End)
From David Callan, Jan 03 2011: (Start)
T(n,k) is the number of increasing 0-2 trees (A002105) on 2n edges in which the minimal path from the root has length k.
Proof. The number a(n,k) of such trees satisfies the recurrence a(0,0)=1, a(1,1)=1 and, counting by size of the subtree rooted at the smaller child of the root,
a(n,k) = Sum_{j=1..n-1} C(2n-1,j)*a(j,k-1)*a(n-1-j)
for 2<=k<=n, where a(n) = Sum_{k>=0} a(n,k) is the reduced tangent number A002105 (indexed from 0). The recurrence translates into the differential equation
F_x(x,y) = y*F(x,y)*G(x)
for the GF F(x,y) = Sum_{n,k>=0} a(n,k)x^(2n)/(2n)!*y^k, where G(x):=Sum_{n>=0} a(n)x^(2n+1)/(2n+1)! is known to be sqrt(2)*tan(x/sqrt(2)). The differential equation has solution F(x,y) = sec(x/sqrt(2))^(2y). (End)

			Triangle begins:
.1;
.0, 1;
.0, 1, 3;
.0, 4, 15, 15;
.0, 34, 147, 210, 105;
.0, 496, 2370, 4095, 3150, 945;
.0, 11056, 56958, 111705, 107415, 51975, 10395;
.0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135;
From _Johannes W. Meijer_, May 24 2009: (Start)
The first few ECGP(1-2*m,n) polynomials are: ECGP(-1,n) = 1; ECGP(-3,n) = n; ECGP(-5,n) = n + 3*n^2; ECGP(-7,n) = 4*n + 15*n^2+ 15*n^3 .
The first few RES1(1-2*m,n) are: RES1(-1,n) = (1/2)*(1); RES1(-3,n) = (-1/4)*(n); RES1(-5,n) = (1/8)*(n+3*n^2); RES1(-7,n) = (-1/16)*(4*n+15*n^2+15*n^3).
(End)

Alois P. Heinz, Rows n = 0..140, flattened
H. J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, 40(2):175-180, May 2002.

Cf. A000364 A084938 A083061.
From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
Cf. A160464, A083061 and A160468.
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

nmax:=7; imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2 * x^2 * T1(i-1, x)): dx:=degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx) od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do T(n+1, k+1) := A083061(n, k) od: od: T(0, 0):=1: for n from 1 to nmax do T(n, 0):=0 od: seq(seq(T(n, k), k=0..n), n=0..nmax);
# Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012

nmax = 8;
T[n_, k_] := SeriesCoefficient[Sec[x/Sqrt[2]]^(2y), {x, 0, 2n}, {y, 0, k}]* (2n)!;
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Aug 10 2018 *)

Sum_{k = 0..n} T(n, k) = A002105(n+1).
Sum_{k = 0..n} T(n, k)*2^(n-k) = A000364(n); Euler numbers.
Sum_{k = 0..n} T(n, k)*(-2)^(n-k) = 1.
RES1(1-2*m,n) = n^2*RES1(3-2*m,n)-n*(2*n+1)*RES1(3-2*m,n+1)/2 for m >= 2, with RES1(-1,n) = 0.5 for n >= 1. - Johannes W. Meijer, May 24 2009
G.f.: Sum_{n,k>=0} T(n,k)x^n/n!*y^k = sec(x/sqrt(2))^(2y).

Term corrected by Johannes W. Meijer, Sep 23 2012

A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))(1 - exp(-3t)) + (1 - exp(-t))(1 - exp(-3t))(1 - exp(-5t)) + ... as a series in t.

Original entry on oeis.org

1, 1, 5, 55, 1073, 32671, 1431665, 85363615, 6646603073, 654896692351, 79656194515025, 11722538113191775, 2052949879753739873, 421931472111868912831, 100568330857984368195185

Offset: 0

Peter Bala, Mar 24 2009

We appear to get the same sequence by expanding 1 - (1 - exp(t)) + (1 - exp(t))*(1 - exp(2*t)) - (1 - exp(t))*(1 - exp(2*t))*(1 - exp(3*t)) + ... as a series in t. Compare with A079144. For other sequences with generating functions of a similar type see A000364, A000464, A002105 and A002439.
From Peter Bala, Mar 13 2017: (Start)
It appears that the g.f. has two other forms: either F(exp(-t)) where F(q) = Sum_{n >= 0} q^(n+1)*Product_{k = 1..n} 1 - q^(2*k) = q + q^2 + q^3 - q^7 - q^8 - q^10 - q^11 - ... is a g.f. for A003475 or 1/2*G(exp(t)) where G(q) = 1 + Sum_{n >= 0} (-1)^n*q^(n+1)*Product_{k = 1..n} 1 - q^k = 1 + q - q^2 + 2*q^3 - 2*q^4 + q^5 + q^7 - 2*q^8 + ... is a g.f. for A003406. See Zagier, Example 1. (End)
From Peter Bala, Dec 18 2021: (Start)
Conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 16 begins [1, 1, 5, 7, 1, 15, 1, 15, 1, 15, 1, 15, ...] with an apparent pre-period of length 4 and a period of length 2.
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
If true, then for each i the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. For example, the expansion of exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + x + 3*x^2 + 21*x^3 + 291*x^4 + 6861*x^5 + 246171*x^6 + 12458901*x^7 + 843915891*x^8 + 73640674461*x^9 + 8041227405771*x^10 + ... appears to have integer coefficients. (End)

			G.f. A(x) = 1 + x + 5*x^2 + 55*x^3 + 1073*x^4 + 32671*x^5 + 1431665*x^6 + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..170
Hsien-Kuei Hwang, and Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
D. Zagier, Quantum modular forms, Quanta of Maths: Conference in honor of Alain Connes, Clay Mathematics Proceedings 11, AMS and Clay Mathematics Institute 2010, 659-675

Cf. A000364, A000464, A002105, A002439, A079144, A158691, A209832, A215066, A003406, A003475.

max = 14; se = Series[1 + Sum[ Product[1 - E^(-(2*k - 1)*t), {k, 1, n}], {n, 1, max}], {t, 0, max}]; CoefficientList[se, t]*Range[0, max]! (* Jean-François Alcover, Mar 06 2013 *)

{a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, 1-exp(-(2*k-1)*x+x*O(x^n)))), n)} \\ Paul D. Hanna, Aug 01 2012

{a(n)=n!*polcoeff(sum(m=0, n, prod(k=1, m, exp(k*x+x*O(x^n))-1)), n)} \\ Paul D. Hanna, Aug 01 2012

Basic hypergeometric generating function: 1 + Sum_{n >= 0} Product_{k = 1..n} (1 - exp(2*k-1)*t) = 1 + t + 5*t^2/2! + 55*t^3/3! + ....
a(n) ~ 6*sqrt(2) * 12^n * (n!)^2 / Pi^(2*n+2). - Vaclav Kotesovec, May 04 2014
Conjectural g.f.: G(exp(t)) as a formal power series in t, where G(q) := Sum_{n >= 0} q^(2*n+1) * Product_{k = 1..2*n} (1 - q^k). - Peter Bala, May 16 2017
E.g.f.: Sum_{n>=0} exp(n*(n+1)/2*x) / Product_{k=0..n} (1 + exp(k*x)). - Paul D. Hanna, Oct 14 2020

A079144 Number of labeled interval orders on n elements: (2+2)-free posets.

Original entry on oeis.org

1, 1, 3, 19, 207, 3451, 81663, 2602699, 107477247, 5581680571, 356046745023, 27365431508779, 2494237642655487, 266005087863259291, 32815976815540917183, 4636895313201764853259, 743988605732990946684927

Offset: 0

Detlef Pauly (dettodet(AT)yahoo.de), Dec 27 2002

From Peter Bala, Dec 26 2021: (Start)
We make the following conjectures:
1) Taking the sequence modulo an integer k gives an eventually periodic sequence with period dividing phi(k). For example, the sequence taken modulo 8 begins [1, 1, 3, 3, 7, 3, 7, 3, 7,  ...] and appears to have a pre-period of length 3 and a period of length 2 = (1/2)*phi(8).
2) Let i >= 0 and define a_i(n) = a(n+i). Then for each i the Gauss congruences a_i(n*p^k) == a_i(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k. If true, then for each i the expansion of exp( Sum_{n >= 1} a_i(n)*x^n/n ) has integer coefficients. (End)

			1 + x + 3*x^2 + 19*x^3 + 207*x^4 + 3451*x^5 + 81663*x^6 + 2602699*x^7 + ...

Alois P. Heinz, Table of n, a(n) for n = 0..260
Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Graham Brightwell and Mitchel T. Keller, Asymptotic Enumeration of Labelled Interval Orders, arXiv:1111.6766 [math.CO], 2011.
Anders Claesson, Mark Dukes and Martina Kubitzke, Partition and composition matrices, arXiv:1006.1312 [math.CO], 2010-2011.
Hsien-Kuei Hwang and Emma Yu Jin, Asymptotics and statistics on Fishburn matrices and their generalizations, arXiv:1911.06690 [math.CO], 2019.
Don Zagier, Vassiliev invariants and a strange identity related to the Dedekind eta-function, Topology, vol.40, pp.945-960 (2001); see p. 952.
Yan X Zhang, Four Variations on Graded Posets, arXiv preprint arXiv:1508.00318 [math.CO], 2015.

Cf. A022493 (unlabeled interval orders).
Cf. A002439 (Glaisher's T numbers), A002114 (Glaisher's H numbers).
Cf. A001318, A138265.

A002439 := proc(n) option remember; if n = 0 then 1; else (-4)^n-add((-9)^k*binomial(2*n+1,2*k)*procname(n-k),k=1..n+1) ; end if;end proc:
seq(1/(24^n)*add(binomial(n,k)*A002439(k), k = 0..n), n = 0..20); # Peter Bala, Dec 26 2021

nmax=20; rk=Rest[CoefficientList[Series[Sum[Product[1-1/(1+x)^j,{j,1,n}],{n,0,nmax}],{x,0,nmax}],x]]; Flatten[{1,Table[Sum[rk[[k]] * k! * StirlingS2[n,k],{k,1,n}],{n,1,nmax}]}] (* Vaclav Kotesovec, May 03 2014 *)

{a(n) = if( n<0, 0, n! * polcoeff( subst( sum( i=0, n, prod( j=1, i, 1 - (1 - x + O(x^(n - i + 2)))^j )), x, 1 - exp( -x + x * O(x^n))), n))} /* Michael Somos, Apr 01 2012 */

a(n) = (1/(24^n))*Sum_{k=0..n} binomial(n, k)*A002439(k). Zagier 2001, p. 954.
G.f.: Sum(Product(1-exp(-k*x),k = 1 .. n),n = 0 .. infinity). a(n) = Sum_{k=0..n} k!*Stirling2(n,k)*A138265(k). - Vladeta Jovovic, Mar 11 2008
From Peter Bala, Mar 19 2009: (Start)
Conjectural form for the o.g.f. as a continued fraction:
1/(1-x/(1-2*x/(1-5*x/(1-7*x/(1-12*x/(1-15*x/(1- ...))))))) = 1 + x + 3*x^2 + 19*x^3 + 207*x^4 + ..., where the sequence [1,2,5,7,12,15,..] is the sequence of generalized pentagonal numbers A001318. Compare with the continued fraction form of the o.g.f. of A002105. (End)
E.g.f.: 1+(exp(x)-1)/(G(0)+1-exp(x)), where G(k)= 2*exp(x*(k+1))-1-exp(x*(k+1))*(exp(x*(k+2))-1)/G(k+1); (continued fraction, Euler's kind, 1-step). - Sergei N. Gladkovskii, Jan 06 2012
Asymptotics (Brightwell and Keller, 2011): a(n) ~ 12*sqrt(3)/Pi^(5/2) * (n!)^2 * sqrt(n) * (6/Pi^2)^n. - Vaclav Kotesovec, May 03 2014
From Peter Bala, May 11 2017: (Start)
For a proof of above conjectural continued fraction representation of the o.g.f. see the Bala link.
G.f.: 1/(1 + x - 2*x/(1 - 1*x/(1 + x - 7*x/(1 - 5*x/(1 + x - 15*x/(1 - 12*x/(1 + x - 26*x/(1 - 22*x/(1 + x - ...))))))))), where the sequence of unsigned partial numerators [2, 1, 7, 5, 15, 12, ...] is obtained from A001318 by swapping adjacent terms.
E.g.f.: F(q) = Sum_{n >= 0} q^(n+1)*Product_{i = 1..n} (1 - q^i)^2 at q = exp(t). Note that F(q) at q = 1/(1 - t) is a g.f. for unlabeled interval orders A022493, and at q = 1 - t gives a g.f. for A138265. (End)
From Peter Bala, Dec 26 2021: (Start)
a(6*n + 5) == 0 (mod 7); a(10*n + 7) == 0 (mod 11);
a(12*n + 11) == 0 (mod 13); a(16*n + 5) == a(16*n + 8) == 0 (mod 17);
a(18*n + 3) == 0 (mod 19); a(22*n + 4) == 0 (mod 23). (End)

A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

Original entry on oeis.org

1, 1, 3, 4, 15, 15, 34, 147, 210, 105, 496, 2370, 4095, 3150, 945, 11056, 56958, 111705, 107415, 51975, 10395, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135, 14873104, 85389132, 197722980, 244909665, 178378200, 77567490

Offset: 0

Hans J. H. Tuenter, Apr 19 2003

This polynomial arises in the setting of a symmetric Bernoulli random walk and occurs in an expression for the even moments of the absolute distance from the origin after an even number of timesteps. The Gandhi polynomial, sequence A036970, occurs in an expression for the odd moments.
When formatted as a square array, first row is A002105, first column is A001147, second column is A001880.
Another version of the triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, 9, ...] = 1; 0, 1; 0, 1, 3; 0, 4, 15, 15; 0, 34, 147, 210, 105; ... where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 07 2004
In A160464 we defined the coefficients of the ES1 matrix. Our discovery that the n-th term of the row coefficients ES1[1-2*m,n] for m>=1, can be generated with rather simple polynomials led to triangle A094665 and subsequently to this one. - Johannes W. Meijer, May 24 2009
Related to polynomials defined in A160485 by a shift of +-1/2 and scaling by a power of 2. - Richard P. Brent, Jul 15 2014

			Triangle starts (with an additional first column 1,0,0,...):
[1]
[0,      1]
[0,      1,       3]
[0,      4,      15,      15]
[0,     34,     147,     210,     105]
[0,    496,    2370,    4095,    3150,     945]
[0,  11056,   56958,  111705,  107415,   51975,  10395]
[0, 349504, 1911000, 4114110, 4579575, 2837835, 945945, 135135]

R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
Marc Joye, Pascal Paillier and Berry Schoenmakers, On Second-Order Differential Power Analysis, in Cryptographic Hardware and Embedded Systems-CHES 2005, editors: Josyula R. Rao and Berk Sunar, Lecture Notes in Computer Science 3659 (2005) 293-308, Springer-Verlag.
H. J. H. Tuenter, Walking into an absolute sum, The Fibonacci Quarterly, 40 (2002), 175-180.

Cf. A036970, A160485.
From Johannes W. Meijer, May 24 2009 and Jun 27 2009: (Start)
Cf. A160464, A094665 and A160468.
A002105 equals the row sums (n>=2) and the first left hand column (n>=1).
A001147, A001880, A160470, A160471 and A160472 are the first five right hand columns.
Appears in A162005, A162006 and A162007.
(End)

imax := 6;
T1(0, x) := 1:
T1(0, x+1) := 1:
for i from 1 to imax do
    T1(i, x) := expand((2*x+1) * (x+1) * T1(i-1, x+1) - 2*x^2*T1(i-1, x)):
    dx := degree(T1(i, x)):
    for k from 0 to dx do
        c(k) := coeff(T1(i, x), x, k)
    od:
    T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx):
od:
for i from 0 to imax do
    for j from 0 to i do
        a(i, j) := coeff(T1(i, x), x, j)
    od:
od:
seq(seq(a(i, j), j = 0..i), i = 0..imax);
# Johannes W. Meijer, Jun 27 2009, revised Sep 23 2012

b[0, 0] = 1;
b[n_, k_] := b[n, k] = Sum[2^j*(Binomial[k + j, 1 + j] + Binomial[k + j + 1, 1 + j])*b[n - 1, k - 1 + j], {j, Max[0, 1 - k], n - k}];
a[0, 0] = 1;
a[n_, k_] := b[n, k]/2^(n - k);
Table[a[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 19 2018, after Philippe Deléham *)

# uses[fr2_row from A088874]
A083061_row = lambda n: [(-1)^(n-k)*m*2^(-n+k) for k,m in enumerate(fr2_row(n))]
for n in (0..7): print(A083061_row(n)) # Peter Luschny, Sep 19 2017

Let T(i, x)=(2x+1)(x+1)T(i-1, x+1)-2x^2T(i-1, x), T(0, x)=1; so that T(1, x)=1+3x; T(2, x)=4+15x+15x^2; T(3, x)=34+147x+210x^2+105x^3, etc. Then the (i, j)-th entry in the table is the coefficient of x^j in T(i, x).

a(n, k)*2^(n-k) = A085734(n, k). - Philippe Deléham, Feb 27 2005

A008301 Poupard's triangle: triangle of numbers arising in enumeration of binary trees.

Original entry on oeis.org

1, 1, 2, 1, 4, 8, 10, 8, 4, 34, 68, 94, 104, 94, 68, 34, 496, 992, 1420, 1712, 1816, 1712, 1420, 992, 496, 11056, 22112, 32176, 40256, 45496, 47312, 45496, 40256, 32176, 22112, 11056, 349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000

Offset: 0

N. J. A. Sloane

The doubloon polynomials evaluated at q=1. [Note the error in (D1) of the Foata-Han article in the Ramanujan journal which should read d_{1,j}(q) = delta_{2,j}.] - R. J. Mathar, Jan 27 2011

T(n,k), 0 <= k <= 2n-2, is the number of increasing 0-2 trees on vertices [0,2n] in which the parent of 2n is k (Poupard). A little more generally, for fixed m in [k+1,2n], T(n,k) is the number of trees in which m is a leaf with parent k. (The case m=2n is Poupard's result.) T(n,k) is the number of increasing 0-2 trees on vertices [0,2n] in which the minimal path from the root ends at k+1 (Poupard). - David Callan, Aug 23 2011

			[1],
[1, 2, 1],
[4, 8, 10, 8, 4],
[34, 68, 94, 104, 94, 68, 34],
[496, 992, 1420, 1712, 1816, 1712, 1420, 992, 496],
[11056, 22112, 32176, 40256, 45496, 47312, 45496, 40256, 32176, 22112, 11056],
[349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000, 1666688, 1528384, 1309568, 1026400, 699008, 349504], ...

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
Neil J. Y. Fan, Liao He, The Complete cd-Index of Boolean Lattices, Electron. J. Combin., 22 (2015), #P2.45.
D. Foata, G.-N. Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126.
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432.
D. Foata and G.-N. Han, Tree Calculus for Bivariable Difference Equations, 2012. - From _N. J. A. Sloane_, Feb 02 2013
D. Foata and G.-N. Han, Tree Calculus for Bivariable Difference Equations, arXiv:1304.2484 [math.CO], 2013.
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory, Ser. A, 115 (2008), pp. 293-303.
C. Poupard, Deux propriétés des arbres binaires ordonnés stricts, European J. Combin., 10 (1989), 369-374.

Cf. A107652. Leading diagonal and row sums = A002105.

Cf. A210108 (left half).

a008301 n k = a008301_tabf !! n !! k
a008301_row n = a008301_tabf !! n
a008301_tabf = iterate f [1] where
   f zs = zs' ++ tail (reverse zs') where
     zs' = (sum zs) : h (0 : take (length zs `div` 2) zs) (sum zs) 0
     h []        = []
     h (x:xs) y' y = y'' : h xs y'' y' where y'' = 2*y' - 2*x - y
-- Reinhard Zumkeller, Mar 17 2012

doubloon := proc(n,j,q) option remember; if n = 1 then if j=2 then 1; else 0; end if; elif j >= 2*n+1 or ( n>=1 and j<=1 ) then 0 ; elif j=2 and n>=1 then add(q^(k-1)*procname(n-1,k,q),k=1..2*n-2) ; elif n>=2 and 3<=j and j<=2*n then 2*procname(n,j-1,q)-procname(n,j-2,q)-(1-q)*add( q^(n+i+1-j)*procname(n-1,i,q),i=1..j-3) - (1+q^(n-1))*procname(n-1,j-2,q)+(1-q)*add(q^(i-j+1)*procname(n-1,i,q),i=j-1..2*n-1) ; else error; end if; expand(%) ; end proc:
A008301 := proc(n,k) doubloon(n+1,k+2,1) ; end proc:
seq(seq(A008301(n,k),k=0..2*n),n=0..12) ; # R. J. Mathar, Jan 27 2011
# Second program based on the Poupard numbers g_n(k) (A236934).
T := proc(n,k) option remember; local j;
  if n = 1 then 1
elif k = 1 then 0
elif k = 2 then 2*add(T(n-1, j), j=1..2*n-3)
elif k > n then T(n, 2*n-k)
else 2*T(n, k-1)-T(n, k-2)-4*T(n-1, k-2)
  fi end:
A008301 := (n,k) -> T(n+1,k+1)/2^n;
seq(print(seq(A008301(n,k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 12 2014

doubloon[1, 2, q_] = 1; doubloon[1, j_, q_] = 0; doubloon[n_, j_, q_] /; j >= 2n+1 || n >= 1 && j <= 1 = 0; doubloon[n_ /; n >= 1, 2, q_] := doubloon[n, 2, q] = Sum[ q^(k-1)*doubloon[n-1, k, q], {k, 1, 2n-2}]; doubloon[n_, j_, q_] /; n >= 2 <= j && j <= 2n := doubloon[n, j, q] = 2*doubloon[n, j-1, q] - doubloon[n, j-2, q] - (1-q)*Sum[ q^(n+i+1-j)*doubloon[n-1, i, q], {i, 1, j-3}] - (1 + q^(n-1))*doubloon[n-1, j-2, q] + (1-q)* Sum[ q^(i-j+1)*doubloon[n-1, i, q], {i, j-1, 2n-1}]; A008301[n_,k_] := doubloon[n+1, k+2, 1]; Flatten[ Table[ A008301[n, k], {n, 0, 6}, {k, 0, 2n}]] (* Jean-François Alcover, Jan 23 2012, after R. J. Mathar *)
T[n_, k_] := T[n, k] = Which[n==1, 1, k==1, 0, k==2, 2*Sum[T[n-1, j], {j, 1, 2*n-3}], k>n, T[n, 2*n-k], True, 2*T[n, k-1] - T[n, k-2] - 4*T[n-1, k - 2]]; A008301[n_, k_] := T[n+1, k+1]/2^n; Table[A008301[n, k], {n, 1, 6}, {k, 1, 2*n-1}] // Flatten (* Jean-François Alcover, Nov 28 2015, after Peter Luschny *)

Recurrence relations are given on p. 370 of the Poupard paper; however, in line -5 the summation index should be k and in line -4 the expression 2_h^{k-1} should be replaced by 2d_h^(k-1). - Emeric Deutsch, May 03 2004
If we write the triangle like this:
                 0,    1,   0
            0,   1,    2,   1,   0
       0,   4,   8,   10,   8,   4,   0
  0,  34,  68,  94,  104,  94,  68,  34,  0
then the first nonzero term is the sum of the previous row and the remaining terms in each row are obtained by the rule illustrated by 104 = 2*94 - 2*8 - 1*68. - N. J. A. Sloane, Jun 10 2005
Continuing Sloane's remark: If we also set the line "... 1 ..." on the top of the pyramid, then we obtain T(n,k) = A236934(n+1,k+1)/2^n for n >= 1 and 1 <= k <= 2n-1 (see the second Maple program). - Peter Luschny, May 12 2014

More terms from Emeric Deutsch, May 03 2004

A177385 E.g.f.: Sum_{n>=0} Product_{k=1..n} sinh(k*x).

Original entry on oeis.org

1, 1, 4, 37, 616, 16081, 605164, 31011457, 2076192976, 175951716481, 18411425885524, 2331339303739777, 351341718484191736, 62144180030978834881, 12748469150999320273084, 3002313213700366145858497

Offset: 0

Paul D. Hanna, May 15 2010

nonn

Compare to the e.g.f. for A002105, the reduced tangent numbers:
. Sum_{n>=0} A002105(n+1)*x^n/n! = Sum_{n>=0} Product_{k=1..n} tanh(k*x).
Limit n->infinity n!*A177386(n) / (2^n*A177385(n)) = 1. - Vaclav Kotesovec, Nov 06 2014

			E.g.f: A(x) = 1 + x + 4*x^2/2! + 37*x^3/3! + 616*x^4/4! +...
A(x) = 1 + sinh(x) + sinh(x)*sinh(2x) + sinh(x)*sinh(2x)*sinh(3x) + ...

Vaclav Kotesovec, Table of n, a(n) for n = 0..250

Cf. A249698, A177386, A249564, A249748.

Table[n!*SeriesCoefficient[Sum[Product[Sinh[k*x],{k,1,j}],{j,0,n}],{x,0,n}], {n,0,20}] (* Vaclav Kotesovec, Nov 01 2014 *)
nn=20; tab = ConstantArray[0,nn]; tab[[1]] = Series[Sinh[x],{x,0,nn}]; Do[tab[[k]] = Series[tab[[k-1]]*Sinh[k*x],{x,0,nn}],{k,2,nn}]; Flatten[{1,Rest[CoefficientList[Sum[tab[[k]],{k,1,nn}],x] * Range[0,nn]!]}] (* Vaclav Kotesovec, Nov 04 2014 (more efficient) *)

{a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,sinh(k*X)));n!*polcoeff(Egf,n)}

a(n) ~ c * d^n * (n!)^2, where d = A249748 = 1.04689919262595424111342518325311817976789046475647184115584744582777576864..., c = 0.880333778211172907563073031129920597506533414605109200048966773434616066... . - Vaclav Kotesovec, Nov 04 2014

A210108 Left half of Poupard's triangle, A008301.

Original entry on oeis.org

1, 1, 2, 4, 8, 10, 34, 68, 94, 104, 496, 992, 1420, 1712, 1816, 11056, 22112, 32176, 40256, 45496, 47312, 349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000, 14873104, 29746208, 43920304, 56696384, 67419664, 75523808, 80571184, 82285184, 819786496

Offset: 0

Reinhard Zumkeller, Mar 17 2012

Reinhard Zumkeller, Rows n=0..120 of triangle, flattened

Cf. A002105 (left edge), A005799 (right edge); A210111.

a210108 n k = a210108_tabl !! n !! k
a210108_row n = a210108_tabl !! n
a210108_tabl = zipWith take [1..] a008301_tabf

T(n,k) = A008301(n,k), 0 <= k <= n.

A365673 Array A(n, k) read by ascending antidiagonals. Polygonal number weighted generalized Catalan sequences.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 4, 1, 1, 1, 4, 15, 8, 1, 1, 1, 5, 34, 105, 16, 1, 1, 1, 6, 61, 496, 945, 32, 1, 1, 1, 7, 96, 1385, 11056, 10395, 64, 1, 1, 1, 8, 139, 2976, 50521, 349504, 135135, 128, 1, 1, 1, 9, 190, 5473, 151416, 2702765, 14873104, 2027025, 256, 1

Offset: 0

Peter Luschny, Sep 30 2023

Using polygonal numbers as weights, a recursion for triangles is defined, whose main diagonals represents a family of sequences, which include, among others, the powers of 2, the double factorial of odd numbers, the reduced tangent numbers, and the Euler numbers.
Apart from the edge cases k = 0 and k = n the recursion is T(n, k) = w(n, k) * T(n, k - 1) + T(n - 1, k). T(n, 0) = 1 and T(n, n) = T(n, n-1) if n > 0.
The weights w(n, k) identical to 1 yield the recursion of the Catalan triangle A009766 (with main diagonal the Catalan numbers). Here the polygonal numbers are used as weights in the form w(n, k) = p(s, n - k + 1), where the parameter s is the number of sides of the polygon and p(s, n) = ((s-2) * n^2 - (s-4) * n) / 2, see A317302.

			Array A(n, k) starts:                            (polygon|diagonal|triangle)
[0] 1, 1, 1,   1,     1,       1,         1, ...  A258837  A000012
[1] 1, 1, 2,   4,     8,      16,        32, ...  A080956  A011782
[2] 1, 1, 3,  15,   105,     945,     10395, ...  A001477  A001147  A001498
[3] 1, 1, 4,  34,   496,   11056,    349504, ...  A000217  A002105  A365674
[4] 1, 1, 5,  61,  1385,   50521,   2702765, ...  A000290  A000364  A060058
[5] 1, 1, 6,  96,  2976,  151416,  11449296, ...  A000326  A126151  A366138
[6] 1, 1, 7, 139,  5473,  357721,  34988647, ...  A000384  A126156  A365672
[7] 1, 1, 8, 190,  9080,  725320,  87067520, ...  A000566  A366150  A366149
[8] 1, 1, 9, 249, 14001, 1322001, 188106489, ...  A000567
           A054556                         A366137

Wikipedia, Polygonal number.

Poly weights: A258837, A080956, A001477, A000217, A000290, A000326, A000384.
Rows: A000012, A011782, A001147, A002105, A000364, A126151, A126156, A366150.
Triangles: A001498, A365674, A060058, A366138, A365672, A366149.
Cf. A009766, A366137 (central diagonal), A317302 (table of polygonal numbers).
Cf. A112934, A303943, A305532, A305533.

poly := (s, n) -> ((s - 2) * n^2 - (s - 4) * n) / 2:
T := proc(s, n, k) option remember; if k = 0 then 1 else if k = n then T(s, n, k-1) else poly(s, n - k + 1) * T(s, n, k - 1) + T(s, n - 1, k) fi fi end:
for n from 0 to 8 do A := (n, k) -> T(n, k, k): seq(A(n, k), k = 0..9) od;
# Alternative, using continued fractions:
A := proc(p, L) local CF, poly, k, m, P, ser;
   poly := (s, n) -> ((s - 2)*n^2 - (s - 4)*n)/2;
   CF := 1 + x;
   for k from 1 to L do
       m := L - k + 1;
       P := poly(p, m);
       CF := 1/(1 - P*x*CF)
   od;
   ser := series(CF, x, L);
   seq(coeff(ser, x, m), m = 0..L-1)
end:
for p from 0 to 8 do lprint(A(p, 8)) od;

poly[s_, n_] := ((s - 2) * n^2 - (s - 4) * n) / 2;
T[s_, n_, k_] := T[s, n, k] = If[k == 0, 1, If[k == n, T[s, n, k - 1], poly[s, n - k + 1] * T[s, n, k - 1] + T[s, n - 1, k]]];
A[n_, k_] := T[n, k, k];
Table[A[n - k, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 27 2023, from first Maple program *)

A(p, n) = {
       my(CF = 1 + x,
           poly(s, n) = ((s - 2)*n^2 - (s - 4)*n)/2,
           m, P
       );
       for(k = 1, n,
           m = n - k + 1;
           P = poly(p, m);
           CF = 1/(1 - P*x*CF)
        );
        Vec(CF + O(x^(n)))
}
for(p = 0, 8, print(A(p, 8)))
\\  Michel Marcus and Peter Luschny, Oct 02 2023

from functools import cache
@cache
def T(s, n, k):
    if k == 0: return 1
    if k == n: return T(s, n, k - 1)
    p = (n - k + 1) * ((s - 2) * (n - k + 1) - (s - 4)) // 2
    return p * T(s, n, k - 1) + T(s, n - 1, k)
def A(n, k): return T(n, k, k)
for n in range(9): print([A(n, k) for k in range(9)])

A094503 Triangle read by rows: coefficients d(n,k) of Andre polynomials D(x,n) = Sum_{k>0} d(n,k)*x^k.

Original entry on oeis.org

1, 1, 1, 1, 1, 4, 1, 11, 4, 1, 26, 34, 1, 57, 180, 34, 1, 120, 768, 496, 1, 247, 2904, 4288, 496, 1, 502, 10194, 28768, 11056, 1, 1013, 34096, 166042, 141584, 11056, 1, 2036, 110392, 868744, 1372088, 349504, 1, 4083, 349500, 4247720, 11204160, 6213288

Offset: 1

Philippe Deléham, Jun 09 2004

a(n,k) is the number of increasing 0-1-2 trees on [n] with k leaves. An increasing 0-1-2 tree on [n] is an unordered tree on [n], rooted at 1, in which each vertex has <= 2 children and the labels increase along each path from the root. Example: a(4,2)=4 counts the trees with edges as follows, {1->2->3,1->4}, {1->2->4,1->3}, {1->2->4,2->3}, {1->3->4,1->2}. - David Callan, Oct 24 2004

			Triangle begins:
  1
  1
  1  1
  1  4
  1 11   4
  1 26  34
  1 57 180 34
  ...
From _Peter Bala_, Jun 26 2012: (Start)
Recurrence equation: T(6,3) = 3*T(5,3) + 2*T(5,2) = 3*4 + 2*11 = 34.
n = 4: the 5 weighted non-plane increasing 0-1-2 trees on 4 vertices are
.........................................................
..4......................................................
..|......................................................
..3............4............4.............3.......3...4..
..|.........../............/............./.........\./...
..2......2...3........3...2.........4...2........(t)2....
..|.......\./..........\./...........\./............|....
..1.....(t)1.........(t)1..........(t)1.............1....
.........................................................
Hence row polynomial R(4,t) = (1 + 4*t)*t.
(End)

Alois P. Heinz, Rows n = 1..200, flattened
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
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.
David Callan, A note on downup permutations and increasing 0-1-2 trees
Chak-On Chow and Wai Chee Shiu, Counting Simsun Permutations by Descents, Ann. Comb. 15, 625-635 (2011). See p. 627 and comments section in A113897.
Ming-Jian Ding and Bao-Xuan Zhu, Some results related to Hurwitz stability of combinatorial polynomials, Advances in Applied Mathematics, Volume 152, (2024), 102591. See p. 35.
Filippo Disanto and Thomas Wiehe, Exact enumeration of cherries and pitchforks in ranked trees under the coalescent model, arXiv preprint arXiv:1112.1295 [math.CO], 2011-2012, see Y(x,z).
Filippo Disanto and Thomas Wiehe, Exact enumeration of cherries and pitchforks in ranked trees under the coalescent model, Math. Biosci. 242 (2013), no. 2, 195-200.
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
Dominique Foata and Guoniu Han, Arbres minimax et polynomes d'André .
D. Foata and Guo-Niu Han, Arbres minimax et polynomes d'André, Adv. in Appl. Math., 27 (2001), no.2-3, 367-389.
Guo-Niu Han and Shi-Mei Ma, Derivatives, Eulerian polynomials and the g-indexes of Young tableaux, arXiv:2006.14064 [math.CO], 2020.
Shi-Mei Ma, Enumeration of permutations by number of alternating runs, Discrete Math., 313 (2013), 1816-1822.
Shi-Mei Ma, Qi Fang, Toufik Mansour, and Yeong-Nan Yeh, Alternating Eulerian polynomials and left peak polynomials, arXiv:2104.09374, 2021
Shi-Mei Ma and Yeong-Nan Yeh, The Peak Statistics on Simsun Permutations, Elect. J. Combin., 23 (2016), P2.14; arXiv preprint, arXiv:1601.06505 [math.CO], 2016.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411 [math.RA], 2013.

Cf. A000012, A000295, A008292, A101280, A113897.

A094503:=proc(n,k) options remember: if(n=1 and k=1) then RETURN(1) elif(1<=k and k<=floor((n+1)/2) and n>=1) then RETURN(k*A094503(n-1,k)+(n+2-2*k)*A094503(n-1,k-1)) else RETURN(0) fi: end; seq(seq(A094503(n,k),k=1..floor((n+1)/2)),n=1..14);

t[1, 1] = 1; t[n_, k_] /; Not[1 <= k <= (n+1)/2] = 0; t[n_, k_] := t[n, k] = k*t[n-1, k] + (n+2-2*k)*t[n-1, k-1]; Table[t[n, k], {n, 0, 13}, {k, 1, (n + 1)/2}] // Flatten (* Jean-François Alcover, Nov 22 2012, after Maple *)

def p(n) :
    s = var('s'); u = sqrt(s^2-2)
    egf = u*x-2*ln((exp(u*x)*(1-s/u)+s/u+1)/2)
    return factorial(n+2)*egf.series(x,n+4).coefficient(x,n+2)
def A094503_row(n) : return [p(n).coefficient(s,n-2*i) for i in (0..n//2)]
for n in (0..6): print(A094503_row(n)) # Peter Luschny, Jul 01 2012

D(1, n) = A000111(n), Euler or up/down numbers. D(1/2, n) = A000142(n)*(1/2)^n. D(1/4, n) = A080795(n)*(1/4)^n.
From Peter Bala, Jun 26 2012: (Start):
Recurrence equation: T(n,k) = k*T(n-1,k) + (n+2-2*k)*T(n-1,k-1) for n >= 1 and 1 <= k <= floor((n+1)/2).
Let r(t) = sqrt(1-2*t) and w(t) = (1-r(t))/(1+r(t)). The e.g.f. is F(t,z) = r(t)*(1 + w(t)*exp(r(t)*z))/(1 - w(t)*exp(r(t)*z)) = 1 + t*z + t*z^2/2! + (t+t^2)*z^3/3! + (t+4*t^2)*z^4/4! + ... (Foata and Han, 2001, section 7).
Note that (F(2*t,z) - 1)/(2*t) is the e.g.f. for A101280.
The modified e.g.f. A(t,z) := (F(t,z) - 1)/t = z + z^2/2! + (1+t)*z^3/3! + (1+4*t)*z^4/4! + ... satisfies the autonomous partial differential equation dA/dz = 1 + A + 1/2*t*A^2 with A(t,0) = 1. It follows that the inverse function A(t,z)^(-1) may be expressed as an integral: A(t,z)^(-1) = Integral_{x = 0..z} 1/(1+x+1/2*t*x^2) dx.
Applying [Dominici, Theorem 4.1] to invert the integral gives the following method for calculating the row polynomials R(n,t) of the table: let f(t,x) = 1+x+1/2*t*x^2 and let D be the operator f(t,x)*d/dx. Then R(n+1,t) = t*D^n(f(t,x)) evaluated at x = 0.
By Bergeron et al., Theorem 1, the shifted row polynomial 1/t*R(n,t) is the generating function for rooted non-plane increasing 0-1-2 trees on n vertices, where the vertices of outdegree 2 have weight t and all other vertices have weight 1. An example is given below.
1/(2*t)*(1+t)^(n+1)*R(n,2*t/(1+t)^2) = the n-th Eulerian polynomial of A008292. For example, n = 5 gives 1/(2*t)*(1+t)^6*R(5,2*t/(1+t)^2) = 1 + 26*t + 66*t^2 + 26*t^3 + t^4.
A000142(n) = 2^n*R(n,1/2); A080795(n) = 4^n*R(n,1/4);
A000670(n) = 3/4*3^n*R(n,4/9); A004123(n+1) = 5/6*5^n*R(n,12/25).
(End)
There is a second family of polynomials which also matches the data and is different from the André polynomials as defined by Foata and Han (2001), formula 3.5. Let u = sqrt(s^2-2) and F(s,x) = u*x-2*log((exp(u*x)*(1-s/u)+s/u+1)/2), then for n>=0 the sequence of polynomials p_{n}(s) = (n+2)!*[x^(n+2)]F(s,x) starts 1, s, s^2+1, s^3+4*s, s^4+11*s^2+4, s^5+26*s^3+34*s, s^6+57*s^4+180*s^2+34, ... and the nonzero coefficients of these polynomials in descending order coincide with the sequence a(n). p_{n}(0) is an aerated version of the reduced tangent numbers, p_{2*n}(0) = A002105(n+1) for n>=0. In contrast, the André polynomials vanish at t=0 except for n=0. - Peter Luschny, Jul 01 2012
T(n,k) = A008303(n,k)/2^(n-k). - Ammar Khatab, Aug 17 2024

cp's OEIS Frontend

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A094665 Another version of triangular array in A083061: triangle T(n,k), 0<=k<=n, read by rows; given by [0, 1, 3, 6, 10, 15, 21, 28, ...] DELTA [1, 2, 3, 4, 5, 6, 7, 8, ...] where DELTA is the operator defined in A084938.

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A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))*(1 - exp(-3*t)) + (1 - exp(-t))*(1 - exp(-3*t))*(1 - exp(-5*t)) + ... as a series in t.

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A079144 Number of labeled interval orders on n elements: (2+2)-free posets.

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A083061 Triangle of coefficients of a companion polynomial to the Gandhi polynomial.

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A008301 Poupard's triangle: triangle of numbers arising in enumeration of binary trees.

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A158690 Expansion of the basic hypergeometric series 1 + (1 - exp(-t)) + (1 - exp(-t))(1 - exp(-3t)) + (1 - exp(-t))(1 - exp(-3t))(1 - exp(-5t)) + ... as a series in t.