A185419 -id:A185419 - cp's OEIS frontend

A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 5, 0, 10, 0, 1, 0, 40, 0, 20, 0, 1, 61, 0, 175, 0, 35, 0, 1, 0, 768, 0, 560, 0, 56, 0, 1, 1385, 0, 4996, 0, 1470, 0, 84, 0, 1, 0, 24320, 0, 22720, 0, 3360, 0, 120, 0, 1

Offset: 0

Paul Barry, Nov 05 2008

Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Unsigned version of A147308. - N. J. A. Sloane, Nov 07 2008
From Peter Bala, Jan 26 2011: (Start)
Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
(1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
with starting condition Z(0,x) = 1. We call Z(n,x) the zigzag polynomial of degree n. This table lists the coefficients of these polynomials (for n >= 1) in ascending powers of x, row indices shifted by 1. The first few polynomials are
... Z(1,x) = x
... Z(2,x) = x^2
... Z(3,x) = x + x^3
... Z(4,x) = 4*x^2 + x^4
... Z(5,x) = 5*x + 10*x^3 + x^5.
The value Z(n,1) equals the zigzag number A000111(n). The polynomials Z(n,x) occur in formulas for the enumeration of permutations by alternating descents A145876 and in the enumeration of forests of non-plane unary binary labeled trees A147315.
{Z(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting Z(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[m]+...+(n-1)^[m] = (1/(m+1))*Sum_{k=0..m} (-1)^floor(k/2)*binomial(m+1,k)*B_k*n^[m+1-k],
where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
(End)
[gd(x)^(-1)]^m = Sum_{n>=m} Tg(n,m)*(m!/n!)*x^n, where gd(x) is Gudermannian function, Tg(n+1,m+1)=T(n,m). - Vladimir Kruchinin, Dec 18 2011
The Bell transform of abs(E(n)), E(n) the Euler numbers. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
   1;
   0,  1;
   1,  0,   1;
   0,  4,   0,  1;
   5,  0,  10,  0,  1;
   0, 40,   0, 20,  0, 1;
  61,  0, 175,  0, 35, 0, 1;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000111 (row sums), A000364, A012123, A012259, A145876, A147315, A185415, A185417, A185419, A185421, A185424. - Peter Bala, Jan 26 2011

Z := proc(n, x) option remember;
description 'zigzag polynomials Z(n, x)'
if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # Peter Bala, Jan 26 2011

t[n_, k_] := SeriesCoefficient[ 2^k*ArcTan[(E^x - 1)/(E^x + 1)]^k*n!/k!, {x, 0, n}]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten // Abs (* Jean-François Alcover, Jan 23 2015 *)

T(n, k)=local(X); if(k<1 || k>n, 0, X=x+x*O(x^n); n!*polcoeff(polcoeff((tan(X)+1/cos(X))^y, n), k)) \\ Paul D. Hanna, Feb 06 2011

R = PolynomialRing(QQ, 'x')
@CachedFunction
def zzp(n, x) :
    return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
def A147309_row(n) :
    x = R.gen()
    L = list(R(zzp(n, x)))
    del L[0]
    return L
for n in (1..10) : print(A147309_row(n)) # Peter Luschny, Jul 22 2012

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

From Peter Bala, Jan 26 2011: (Start)
GENERATING FUNCTION
The e.g.f., upon including a constant term of '1', is given by:
(1) F(x,t) = (tan(t) + sec(t))^x = Sum_{n>=0} Z(n,x)*t^n/n! = 1 + x*t + x^2*t^2/2! + (x+x^3)*t^3/3! + ....
Other forms include
(2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
(3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
ROW GENERATING POLYNOMIALS
One easily checks from (1) that
d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
and so the row generating polynomials Z(n,x) satisfy the recurrence relation
(4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
The e.g.f. for the odd-indexed row polynomials is
(5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
The e.g.f. for the even-indexed row polynomials is
(6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
(7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
The zeros of Z(n,x) lie on the imaginary axis (use (4) and adapt the proof given in A185417 for the zeros of the polynomial S(n,x)).
BINOMIAL EXPANSION
The form of the e.g.f. shows that {Z(n,x)} n >= 0 is a sequence of polynomials of binomial type. In particular, we have the expansion
(8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
The delta operator D* associated with this binomial type sequence is
(9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
the relation
(10) tan(D*)+sec(D*) = exp(D).
The delta operator D* acts as a lowering operator on the zigzag polynomials:
(11) (D*)Z(n,x) = n*Z(n-1,x).
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
More generally, for k = 1,2,3,...
(13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
RELATIONS WITH OTHER SEQUENCES
Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
(End)
T(n,m) = Sum(i=0..n-m, s(i)/(n-i)!*Sum(k=m..n-i, A147315(n-i,k)*Stirling1(k,m))), m>0, T(n,0) = s(n), s(n)=[1,0,1,0,5,0,61,0,1385,0,50521,...] (see A000364). - Vladimir Kruchinin, Mar 10 2011

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).

A080795 Number of minimax trees on n nodes.

Original entry on oeis.org

1, 1, 4, 20, 128, 1024, 9856, 110720, 1421312, 20525056, 329334784, 5812797440, 111923560448, 2334639652864, 52444850814976, 1262260748288000, 32405895451246592, 883950436237705216, 25530268718794276864

Offset: 0

Emanuele Munarini, Mar 14 2003

A minimax tree is (i) rooted, (ii) binary (i.e., each node has at most two sons), (iii) topological (i.e., the left son is different from the right son), (iv) labeled (i.e., there is a bijection between the nodes and a finite totally ordered set). Moreover it has the following property: (v) the label of each node x is the minimum or the maximum of all the labels of the nodes of the subtree whose root is x.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA]
Dominique Foata & Guo-Niu Han, Arbres minimax et polynomes d'André , Advances in Appl. Math., 27, 2001, p. 367-389.
Dominique Foata and Guo-Niu Han, Arbres minimax et polynomes d'André. Special issue in honor of Dominique Foata's 65th birthday (Philadelphia, PA, 2000). Adv. in Appl. Math. 27 (2001), no. 2-3, 367-389.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411, 2013

Cf. A185419, A185420. A008292, A185896.

w := (sqrt(2) - 1)/2:
seq(simplify((2*sqrt(2))^(n-1)*add(k!*Stirling2(n, k)*w^(k-1), k = 1..n)), n = 1..20); # Peter Bala, Oct 31 2024

Range[0, 18]! CoefficientList[ Series[ Tanh[ ArcTanh[ Sqrt[2]] - Sqrt[2] x]/Sqrt[2], {x, 0, 18}], x] (* Robert G. Wilson v *)

{Stirling2(n,k)=(1/k!)*sum(j=0,k,(-1)^j*binomial(k,j)*(k-j)^n)}
/* Finite sum given by Peter Bala: */
{a(n)=local(w=(sqrt(2)-1)/2);if(n==0,1,round((2*sqrt(2))^(n-1)*sum(k=1,n,k!*Stirling2(n,k)*w^(k-1))))}

E.g.f.: ( tanh(arctanh(sqrt(2)) - sqrt(2)*x) )/sqrt(2) = sqrt(2)/2* (1 + (3-2*sqrt(2))* exp(2*sqrt(2)*x) )/( 1 - (3-2*sqrt(2))* exp(2*sqrt(2)*x) ).
Recurrence: a(n+1) = 2*(Sum_{k=0..n} binomial(n,k)*a(k)*a(n-k)) - 0^n.
a(2*n) = 2^n * A006154(2*n), n>0 (conjectured). - Ralf Stephan, Apr 29 2004
For n>0, a(n) = sqrt(2)^(3*n+1)*Sum_{k>=0} k^n/(1+sqrt(2))^(2*k). - Benoit Cloitre, Jan 12 2005
From Peter Bala, Jan 30 2011: (Start)
A finite sum equivalent to the previous formula of Benoit Cloitre is
a(n) = (2*sqrt(2))^(n-1)*Sum_{k = 1..n} k!*Stirling2(n,k)*w^(k-1), for n >= 1, with w = (sqrt(2) - 1)/2.
This formula can be used to prove congruences for a(n). For example, a(p) == (-1)^((p^2-1)/8) (mod p) for odd prime p.
For similar formulas for labeled plane and non-plane unary-binary trees see A080635 and A000111 respectively.
For a sequence of related polynomials see A185419. For a recursive table to calculate a(n) see A185420.
The e.g.f. A(x) satisfies the autonomous differential equation d/dx (A(x)) = 2*A(x)^2 - 1. (End)
From Peter Bala, Aug 26 2011: (Start)
The inverse function A(x)^(-1) of the generating function A(x) satisfies A(x)^(-1) = Integral_{t = 1..x} 1/(2*t^2 - 1) dt.
Let f(x) = 2*x^2 - 1. Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0 (see A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x)). Then by [Dominici, Theorem 4.1] we have a(n+1) = D^n[f](1).
For n >= 1 we have a(n) = (2 + sqrt(2))^(n-1)*A(n, 3 - 2*sqrt(2)), where {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 (see A008292).
a(n+1) = (-1)^n*(sqrt(-2))^n * R(n, sqrt(-2)) where R(n, x) are the polynomials defined in A185896 (derivative polynomials associated with the function sec^2(x)). (End)
G.f.: 1 + x/G(0) where G(k) =  1 - 4*x*(k+1) - 2*x^2*(k+1)*(k+2)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 11 2013
G.f.: 1 + x/(G(0) -x), where G(k) = 1 - x*(k+1) - 2*x*(k+1)/(1 - x*(k+2)/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 24 2013
E.g.f.: sqrt(2)*( -1/2 + (3+2*sqrt(2))/(4 + 2*sqrt(2)- E(0) )), where E(k) = 2 + 2*sqrt(2)*x/( 2*k+1 - 2*sqrt(2)*x/E(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Dec 27 2013
a(n) ~ n! * 2^((3*n+1)/2) / (log(3+2*sqrt(2)))^(n+1). - Vaclav Kotesovec, Feb 25 2014

A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

Original entry on oeis.org

1, 1, 2, 3, 4, 4, 11, 26, 12, 8, 57, 120, 136, 32, 16, 361, 970, 760, 560, 80, 32, 2763, 7052, 8860, 3680, 2000, 192, 64, 24611, 72530, 72884, 58520, 15120, 6496, 448, 128, 250737, 716528, 976464, 538048, 314720, 55552, 19712, 1024, 256

Offset: 1

Peter Bala, Jan 28 2011

Define a polynomial sequence S(n,x) recursively by
(1)... S(n+1,x) = x*S(n,x-1)+(x+1)*S(n,x+1) with S(0,x) = 1.
This table lists the coefficients of these polynomials (for n>=1) in ascending powers of x.
The first few polynomials are
S(0,x) = 1
S(1,x) = 2*x+1
S(2,x) = 4*x^2+4*x+3
S(3,x) = 8*x^3+12*x^2+26*x+11.
The sequence [1,1,3,11,57,...] of constant terms of the polynomials is the sequence of Springer numbers A001586. The zeros of the polynomials S(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.
Compare the recurrence (1) with the recurrence relation satisfied by the coefficients T(n,k) of the polynomials of A104035, namely
(2)... T(n+1,k) = k*T(n,k-1)+(k+1)*T(n,k+1).

			Table begin
n\k|.....0.....1.....2.....3.....4.....5......6
===============================================
0..|.....1
1..|.....1.....2
2..|.....3.....4.....4
3..|....11....26....12.....8
4..|....57...120...136....32...16
5..|...361...970...760...560...80.....32
6..|..2763..7052..8860..3680..2000...192....64
...

Peter Bala, The zeros of the row polynomials of A185417

Cf A001586 (1st column and row sums), A104035, A126156, A147309, A185415, A185418, A185419

#A185417
S := proc(n,x) option remember;
description 'polynomials S(n,x)'
if n = 0 return 1 else return x*S(n-1,x-1)+(x+1)*S(n-1,x+1)
end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(S(n,x),x); end do;

S[0, ] = 1; S[n, x_] := S[n, x] = x*S[n-1, x-1] + (x+1)*S[n-1, x+1]; Table[ CoefficientList[S[n, x], x], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 15 2015 *)

E.g.f: F(x,t) = 1/(cos(t)-sin(t))*(tan(2*t)+sec(2*t))^x
= (cos(t)+sin(t))^x/(cos(t)-sin(t))^(x+1)
= 1 + (2*x+1)*t + (4*x^2+4*x+3)*t^2/2! + ....
Note that (tan(t)+sec(t))^x is the e.g.f for table A147309.
ROW POLYNOMIALS
The easily checked identity d/dt F(x,t) = x*F(x-1,t)+(x+1)*F(x+1,t) shows that the row generating polynomials of this table are the polynomials S(n,x) described in the Comments section above.
The polynomials S(n,-x) satisfy a Riemann hypothesis: that is, the zeros of S(n,-x) lie on the vertical line Re(x) = 1/2 in the complex plane - see the link.
RELATION WITH OTHER SEQUENCES
1st column [1,1,3,11,57,...] is A001586.
Row sums sequence [1,3,11,57,...] is also A001586.
For n>=1, the values 1/2^n*P(2*n,-1/2) = [1,7,139,5473,...] appear to be A126156.

A185420 Square array, read by antidiagonals, used to recursively calculate the number of minimax trees A080795.

Original entry on oeis.org

1, 4, 1, 20, 5, 1, 128, 32, 6, 1, 1024, 256, 46, 7, 1, 9856, 2464, 432, 62, 8, 1, 110720, 27680, 4784, 662, 80, 9, 1, 1421312, 355328, 60864, 8224, 952, 100, 10, 1, 20525056, 5131264, 873664, 116128, 13048, 1308, 122, 11, 1

Offset: 1

Peter Bala, Jan 30 2011

The table entries T(n,k), for n,k>=1, are defined by means of the recurrence relation
(1)... T(n+1,k) = (2*k+2)*T(n,k+1)-(k-1)*T(n,k-1),
with boundary condition T(1,k) = 1.
The first column of the table gives A080795.
For similarly defined tables used to calculate the zigzag numbers A000111 and the Springer numbers A001586 see A185414 and A185418, respectively.
See also A185416.

			Square array begins
n\k|......1.......2.......3........4.......5.........6
======================================================
..1|......1.......1.......1........1........1........1
..2|......4.......5.......6........7........8........9
..3|.....20......32......46.......62.......80......100
..4|....128.....256.....432......662......952.....1308
..5|...1024....2464....4784.....8224....13048....19544
..6|...9856...27680...60864...116128...201632...327096
..7|.110720..355328..873664..1833728..3460640..6046720
..
Examples of recurrence relation:
T(4,3) = 432 = 8*T(3,4) - 2*T(3,2) = 8*62 - 2*32;
T(6,2) = 27680 = 6*T(5,3) - 1*T(5,1) = 6*4784 - 1*1024.

Cf. A080795, A185414, A185416, A185418, A185419.

#A185420
M := proc(n,x) option remember;
description 'minimax polynomials M(n,x)'
if n = 0
return 1
else return
x*(2*M(n-1,x+1)-M(n-1,x-1))
end proc:
for n from 1 to 10 do
seq(M(n,k)/k, k = 1..10);
end do;

M[n_, x_] := M[n, x] = If[n == 0, 1, x (2 M[n - 1, x + 1] - M[n - 1, x - 1])];
T[n_, k_] := M[n, k]/k;
Table[T[d - k + 1, k], {d, 1, 9}, {k, 1, d}] // Flatten (* Jean-François Alcover, Sep 24 2022 *)

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

(1)... T(n,k) = M(n,k)/k with M(n,x) the polynomials described in A185419.
(2)... First column: T(n,1) = A080795(n).
(3)... Second column: T(n,2) = (1/4)*A080795(n+1).

cp's OEIS Frontend

A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

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

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A080795 Number of minimax trees on n nodes.

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A185417 Table of coefficients of a polynomial sequence related to the Springer numbers.

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A185420 Square array, read by antidiagonals, used to recursively calculate the number of minimax trees A080795.

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