A080795 -id:A080795 - cp's OEIS frontend

A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

1, 3, 1, 10, 9, 1, 42, 67, 18, 1, 248, 510, 235, 30, 1, 1992, 4378, 2835, 605, 45, 1, 19600, 44268, 34888, 10605, 1295, 63, 1, 222288, 524748, 461748, 178913, 31080, 2450, 84, 1, 2851712, 7103088, 6728428, 3069612, 690753, 77112, 4242, 108, 1

Offset: 1

Peter Bala, Feb 07 2011

DEFINITION
Define a sequence of polynomials M(n,x) by means of the recurrence relation
(1)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)},
with starting value M(0,x) = 1. We call these the minimax polynomials.
The first few polynomials are
M(1,x) = x
M(2,x) = x*(x   +  3)
M(3,x) = x*(x^2 +  9*x   +  10)
M(4,x) = x*(x^3 + 18*x^2 +  67*x   +  42)
M(5,x) = x*(x^4 + 30*x^3 + 235*x^2 + 510*x + 248).
This triangle lists the coefficients of these polynomials (apart from M(0,x)) in ascending powers of x.
RELATION TO MINIMAX TREES
The value M(n,1) equals the number of minimax trees on n nodes - A080795(n). This result can be used to recursively calculate the entries of A080795 - see A185420.
In addition, the minimax polynomials M(n,x) occur in the formula for the number T(n,k) of forests of k minimax trees on n nodes. ... T(n,k) = 1/k!*sum {j = 0..k} (-1)^(k-j)*binomial(k,j)*M(n,j).
ANALOGIES WITH THE MONOMIALS
{M(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting M(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^[p]+...+(n-1)^[p] = -2*n^[p]+ 1/(p+1)*Sum_{k = 0..floor(p/2)} 8^k*binomial(p+1,2k)*B_(2k)*n^[p+1-2k], where {B_k}k>=0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For other polynomial sequences defined by recurrences similar to (1), and related to the zigzag numbers A000111 and the Springer numbers A001586, see A147309 and A185417, respectively. See also A185415.
The Bell transform of A143523(n). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
n\k|.....1......2......3......4......5......6......7
====================================================
..1|.....1
..2|.....3......1
..3|....10......9......1
..4|....42.....67.....18......1
..5|...248....510....235.....30......1
..6|..1992...4378...2835....605.....45......1
..7|.19600..44268..34888..10605...1295.....63......1
..
Example of the generalized Bernoulli summation formula:
The second row of the triangle gives x^[2] = 3*x+x^2.
Then 1^[2]+2^[2]+...+(n-1)^[2] = (n^3+3*n^2-4*n)/3 = 1/3*(MB(3,n)-MB(3,0)).
From _R. J. Mathar_, Mar 15 2013: (Start)
The matrix inverse starts
       1;
      -3,       1;
      17,      -9,        1;
    -147,      95,      -18,      1;
    1697,   -1245,      305,    -30,      1;
  -24483,   19687,    -5670,    745,    -45,    1;
  423857, -365757,   118237, -18690,   1540,  -63,   1;
-8560947, 7819287, -2761122, 498197, -50190, 2842, -84, 1; (End)

Cf. A080253 (coeffs. of delta operator), A080795 (row sums), A143523 (column 1), A147309, A185415, A185417, A185420.

#A185419
M := proc(n,x) option remember;
if n = 0 then
return 1
else return
x*(2*M(n-1,x+1)-M(n-1,x-1))
end if;
end proc:
with(PolynomialTools):
for n from 1 to 10 do
CoefficientList(M(n,x),x);
end do;

M[0, ] = 1; M[n, x_] := M[n, x] = x (2 M[n-1, x+1] - M[n-1, x-1]);
Table[CoefficientList[M[n, x], x] // Rest, {n, 1, 10}] (* Jean-François Alcover, Jun 26 2019 *)

# uses[bell_matrix from A264428]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A143523(n), 10) # Peter Luschny, Jan 18 2016

GENERATING FUNCTION
Let a = 3-2*sqrt(2). Let f(t) = (1/2)*sqrt(2)*((1+a*exp(2*sqrt(2)*t))/ (1-a*exp(2*sqrt(2)*t))) = 1 + t + 4*t^2/2! + 20*t^3/3! + ... be the e.g.f. for A080795. Then the e.g.f. for the current table, including a constant 1, is
(1)... F(x,t) = f(t)^x = Sum_{n>=0} M(n,x)*t^n/n! = 1 + x*t + (3*x+x^2)*t^2/2! + (10*x+9*x^2+x^3)*t^3/3! + ....
ROW POLYNOMIALS
One easily checks that d/dt(F(x,t)) = x*(2*F(x+1,t)-F(x-1,t)) and hence the row generating polynomials M(n,x) satisfy the recurrence relation
(2)... M(n+1,x) = x*{2*M(n,x+1)-M(n,x-1)}.
The form of the e.g.f shows that the row polynomials are a polynomial sequence of binomial type. The associated delta operator D* is given by
(3)... D* = sqrt(2)/4*log((3+2*sqrt(2))*(sqrt(2)*exp(D)-1)/(sqrt(2)*exp(D)+1)),
where D is the derivative operator d/dx. This expands to
(4)... D* = D - 3*D^2/2! + 17*D^3/3! - 147*D^4/4! + ....
The sequence of coefficients [1,3,17,147,...] is A080253.
The delta operator D* acts as a lowering operator for the minimax polynomials
(5)...(D*) M(n,x) = n*M(n-1,x).
In what follows it will be convenient to denote M(n,x) by x^[n].
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(6)... x^[p] = (-1)^((p^2-1)/8)*x (mod p).
More generally, for k = 1,2,...
(7)... x^[p+k-1] = (-1)^((p^2-1)/8)*x^[k] (mod p).
GENERALIZED BERNOULLI POLYNOMIALS ASSOCIATED WITH THE MINIMAX POLYNOMIALS
The generalized Bernoulli polynomial MB(k,x) associated with the minimax polynomial x^[k] (= M(k,x)) may be defined as the result of applying the differential operator D*/(exp(D)-1) to the polynomial x^[k]:
(8)... MB(k,x) := {D*/(exp(D)-1)} x^[k].
The first few generalized Bernoulli polynomials are
MB(0,x) = 1,
MB(1,x) = x   -  2,
MB(2,x) = x^2 -    x   + 4/3,
MB(3,x) = x^3 +  3*x^2 - 4*x,
MB(4,x) = x^4 + 10*x^3 + 3*x^2 - 14*x - 32/15.
Since exp(D)-1 is the forward difference operator it follows from (5) and (8) that
(9)... MB(k,x+1) - MB(k,x) = k*x^[k-1].
Summing (9) from x = 1 to x = n-1 and telescoping we find a closed form expression for the finite sums
(10)... 1^[p]+2^[p]+...+(n-1)^[p] = 1/(p+1)*{MB(p+1,n)-MB(p+1,1)}.
The generalized Bernoulli polynomials can be expanded in terms of the minimax polynomials x^[k]. Use (3) to express exp(D)-1 in terms of D*.
Substitute the resulting expression in (8) and expand as a power series in D* to arrive at the expansion:
(11)... MB(k,x) = -2*k*x^[k-1] + Sum_{j=0..floor(k/2)} 2^(3*j) * binomial(k,2j)*B_(2j)*x^[k-2j], where {B_j}j>=0 = [1,-1/2,1/6,0,-1/30,...] denotes the Bernoulli number sequence.
RELATION WITH OTHER SEQUENCES
Column 1 [1, 3, 10, 42, 248, ...] = A143523 with an offset of 1.
Row sums [1, 1, 4, 20, 128, 1024, ...] = A080795.

A185896 Triangle of coefficients of (1/sec^2(x))*D^n(sec^2(x)) in powers of t = tan(x), where D = d/dx.

Original entry on oeis.org

1, 0, 2, 2, 0, 6, 0, 16, 0, 24, 16, 0, 120, 0, 120, 0, 272, 0, 960, 0, 720, 272, 0, 3696, 0, 8400, 0, 5040, 0, 7936, 0, 48384, 0, 80640, 0, 40320, 7936, 0, 168960, 0, 645120, 0, 846720, 0, 362880, 0, 353792, 0, 3256320, 0, 8951040, 0, 9676800, 0, 3628800

Offset: 0

Peter Bala, Feb 07 2011

DEFINITION
Define polynomials R(n,t) with t = tan(x) by
... (d/dx)^n sec^2(x) = R(n,tan(x))*sec^2(x).
The first few are
... R(0,t) = 1
... R(1,t) = 2*t
... R(2,t) = 2 + 6*t^2
... R(3,t) = 16*t + 24*t^3.
This triangle shows the coefficients of R(n,t) in ascending powers of t called the tangent number triangle in [Hodges and Sukumar].
The polynomials R(n,t) form a companion polynomial sequence to Hoffman's two polynomial sequences - P(n,t) (A155100), the derivative polynomials of the tangent and Q(n,t) (A104035), the derivative polynomials of the secant. See also A008293 and A008294.
COMBINATORIAL INTERPRETATION
A combinatorial interpretation for the polynomial R(n,t) as the generating function for a sign change statistic on certain types of signed permutation can be found in [Verges].
A signed permutation is a sequence (x_1,x_2,...,x_n) of integers such that {|x_1|,|x_2|,...|x_n|} = {1,2...,n}. They form a group, the hyperoctahedral group of order 2^n*n! = A000165(n), isomorphic to the group of symmetries of the n dimensional cube.
Let x_1,...,x_n be a signed permutation.
Then 0,x_1,...,x_n,0 is a snake of type S(n;0,0) when 0 < x_1 > x_2 < ... 0.
For example, 0 4 -3 -1 -2 0 is a snake of type S(4;0,0).
Let sc be the number of sign changes through a snake
... sc = #{i, 1 <= i <= n-1, x_i*x_(i+1) < 0}.
For example, the snake 0 4 -3 -1 -2 0  has sc = 1. The polynomial R(n,t) is the generating function for the sign change statistic on snakes of type S(n+1;0,0):
... R(n,t) = sum {snakes in S(n+1;0,0)} t^sc.
See the example section below for the cases n=1 and n=2.
PRODUCTION MATRIX
Define three arrays R, L, and S as
... R = superdiag[2,3,4,...]
... L = subdiag[1,2,3,...]
... S = diag[2,4,6,...]
with the indicated sequences on the main superdiagonal, the main subdiagonal and main diagonal, respectively, and 0's elsewhere. The array R+L is the production array for this triangle: the first row of (R+L)^n produces the n-th row of the triangle.
On the vector space of complex polynomials the array R, the raising operator, represents the operator p(x) - > d/dx (x^2*p(x)), and the array L, the lowering operator, represents the differential operator d/dx - see Formula (4) below.
The three arrays satisfy the commutation relations
... [R,L] = S, [R,S] = 2*R, [L,S] = -2*L
and hence give a representation of the Lie algebra sl(2).

			Table begins
  n\k|.....0.....1.....2.....3.....4.....5.....6
  ==============================================
  0..|.....1
  1..|.....0.....2
  2..|.....2.....0.....6
  3..|.....0....16.....0....24
  4..|....16.....0...120.....0...120
  5..|.....0...272.....0...960.....0...720
  6..|...272.....0..3696.....0..8400.....0..5040
Examples of recurrence relation
  T(4,2) = 3*(T(3,1) + T(3,3)) = 3*(16 + 24) = 120;
  T(6,4) = 5*(T(5,3) + T(5,5)) = 5*(960 + 720) = 8400.
Example of integral formula (6)
... Integral_{t = -1..1} (1-t^2)*(16-120*t^2+120*t^4)*(272-3696*t^2+8400*t^4-5040*t^6) dt = 2830336/1365 = -2^13*Bernoulli(12).
Examples of sign change statistic sc on snakes of type (0,0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......t^sc
= = = = = = = = = = = = = = = = = = = = = =
n=1
...0 1 -2 0...........1................t
...0 2 -1 0...........1................t
yields R(1,t) = 2*t;
n=2
...0 1 -2 3 0.........2................t^2
...0 1 -3 2 0.........2................t^2
...0 2 1 3 0..........0................1
...0 2 -1 3 0.........2................t^2
...0 2 -3 1 0.........2................t^2
...0 3 1 2 0..........0................1
...0 3 -1 2 0.........2................t^2
...0 3 -2 1 0.........2................t^2
yields
R(2,t) = 2 + 6*t^2.

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form, arXiv:0903.0117 [math.CA], 2009-2010.
M-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
A. Hodges and C. V. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. R. Soc. A (2007) 463, 2401-2414 doi:10.1098/rspa.2007.0001
Michael E. Hoffman, Derivative polynomials, Euler polynomials,and associated integer sequences, Electronic Journal of Combinatorics, Volume 6 (1999), Research Paper #R21.
Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 23-30.
M. Josuat-Verges, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
Shi-Mei Ma, Qi Fang, Toufik Mansour, Yeong-Nan Yeh, Alternating Eulerian polynomials and left peak polynomials, arXiv:2104.09374, 2021

Cf. A000182, A000364, A000828 (row sums), A008292, A008293, A008294, A104035, A155100.

R = proc(n) option remember;
if n=0 then RETURN(1);
else RETURN(expand(diff((u^2+1)*R(n-1), u))); fi;
end proc;
for n from 0 to 12 do
t1 := series(R(n), u, 20);
lprint(seriestolist(t1));
od:

Table[(-1)^(n + 1)*(-1)^((n - k)/2)*Sum[j!*StirlingS2[n + 1, j]*2^(n + 1 - j)*(-1)^(n + j - k)*Binomial[j - 1, k], {j, k + 1, n + 1}], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Jul 22 2017 *)

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

{T(n, k) = my(A); if( n<0 || k>n, 0, A=1; for(i=1, n, A = ((1 + x^2) * A)'); polcoeff(A, k))}; /* Michael Somos, Jun 24 2017 */

GENERATING FUNCTION
E.g.f.:
(1)... F(t,z) = 1/(cos(z)-t*sin(z))^2 = Sum_{n>=0} R(n,t)*z^n/n! = 1 + (2*t)*z + (2+6*t^2)*z^2/2! + (16*t+24*t^3)*z^3/3! + ....
The e.g.f. equals the square of the e.g.f. of A104035.
Continued fraction representation for the o.g.f:
(2)... F(t,z) = 1/(1-2*t*z - 2*(1+t^2)*z^2/(1-4*t*z -...- n*(n+1)*(1+t^2)*z^2/(1-2*n*(n+1)*t*z -....
RECURRENCE RELATION
(3)... T(n,k) = (k+1)*(T(n-1,k-1) + T(n-1,k+1)).
ROW POLYNOMIALS
The polynomials R(n,t) satisfy the recurrence relation
(4)... R(n+1,t) = d/dt{(1+t^2)*R(n,t)} with R(0,t) = 1.
Let D be the derivative operator d/dt and U = t, the shift operator.
(5)... R(n,t) = (D + DUU)^n 1
RELATION WITH OTHER SEQUENCES
A) Derivative Polynomials A155100
The polynomials (1+t^2)*R(n,t) are the polynomials P_(n+2)(t) of A155100.
B) Bernoulli Numbers A000367 and A002445
Put S(n,t) = R(n,i*t), where i = sqrt(-1). We have the definite integral evaluation
(6)... Integral_{t = -1..1} (1-t^2)*S(m,t)*S(n,t) dt = (-1)^((m-n)/2)*2^(m+n+3)*Bernoulli(m+n+2).
The case m = n is equivalent to the result of [Grosset and Veselov]. The methods used there extend to the general case.
C) Zigzag Numbers A000111
(7)... R_n(1) = A000828(n+1) = 2^n*A000111(n+1).
D) Eulerian Numbers A008292
The polynomials R(n,t) are related to the Eulerian polynomials A(n,t) via
(8)... R(n,t) = (t+i)^n*A(n+1,(t-i)/(t+i))
with the inverse identity
(9)... A(n+1,t) = (-i/2)^n*(1-t)^n*R(n,i*(1+t)/(1-t)),
where {A(n,t)}n>=1 = [1,1+t,1+4*t+t^2,1+11*t+11*t^2+t^3,...] is the sequence of Eulerian polynomials and i = sqrt(-1).
E) Ordered set partitions A019538
(10)... R(n,t) = (-2*i)^n*T(n+1,x)/x,
where x = i/2*t - 1/2 and T(n,x) is the n-th row po1ynomial of A019538;
F) Miscellaneous
Column 1 is the sequence of tangent numbers - see A000182.
A000670(n+1) = (-i/2)^n*R(n,3*i).
A004123(n+2) = 2*(-i/2)^n*R(n,5*i).
A080795(n+1) =(-1)^n*(sqrt(-2))^n*R(n,sqrt(-2)). - Peter Bala, Aug 26 2011
From Leonid Bedratyuk, Aug 12 2012: (Start)
T(n,k) = (-1)^(n+1)*(-1)^((n-k)/2)*Sum_{j=k+1..n+1} j! *stirling2(n+1,j) *2^(n+1-j) *(-1)^(n+j-k) *binomial(j-1,k), see A059419.
Sum_{j=i+1..n+1}((1-(-1)^(j-i))/(2*(j-i))*(-1)^((n-j)/2)*T(n,j))=(n+1)*(-1)^((n-1-i)/2)*T(n-1,i), for n>1 and  0G.f.:  1/G(0,t,x), where G(k,t,x) = 1 - 2*t*x - 2*k*t*x - (1+t^2)*(k+2)*(k+1)*x^2/G(k+1,t,x); (continued fraction due to T. J. Stieltjes). - Sergei N. Gladkovskii, Dec 27 2013

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

A143523 a(n) = n-fold Dumont operator of x evaluated at x=y=1, z=3.

Original entry on oeis.org

1, 3, 10, 42, 248, 1992, 19600, 222288, 2851712, 41075328, 658359040, 11621260032, 223832419328, 4669549335552, 104894256056320, 2524539033397248, 64811332658757632, 1767891945806266368, 51060500413513400320

Offset: 0

Paul D. Hanna, Aug 23 2008

nonn

The Dumont operator: D = y*z*dx + z*x*dy + x*y*dz is used to generate expansions for the Jacobi elliptic functions sn, cn and dn.

			Given the Dumont operator: D = y*z*dx + z*x*dy + x*y*dz,
illustrate a(n) = D^n x evaluated at x=1, y=1, z=3:
D^0 x = x --> a(0) = 1;
D^1 x = y*z --> a(1) = 3;
D^2 x = (y^2 + z^2)*x --> a(2) = 10;
D^3 x = 4*z*y*x^2 + (z*y^3 + z^3*y) --> a(3) = 42;
D^4 x = (4*y^2 + 4*z^2)*x^3 + (y^4 + 14*z^2*y^2 + z^4)*x --> a(4) = 248;
D^5 x = 16*z*y*x^4 + (44*z*y^3 + 44*z^3*y)*x^2 + (z*y^5 + 14*z^3*y^3 + z^5*y) --> a(5) = 1992.

Cf. A143522, A080795.

{a(n)=local(F=x);if(n>=0,for(i=1,n,F=y*z*deriv(F,x)+z*x*deriv(F,y)+x*y*deriv(F,z)));subst(subst(subst(F,x,1),y,1),z,3)}

{a(n)=local(r=2*sqrt(2)+x*O(x^n));round(n!*polcoeff(2*r*(3-r)*exp(r*x)/(1-(3-r)^2*exp(2*r*x)),n))}

E.g.f.: 2*r*(3-r)*exp(r*x)/(1 - (3-r)^2*exp(2*r*x)) where r=2*sqrt(2).
E.g.f.: G'(x)/G(x) where G(x) is the e.g.f. of A080795 (number of minimax trees on n nodes).
G.f.: 1/Q(0), where Q(k) = 1 - 3*x*(2*k+1) - x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Sep 28 2013
a(n) ~ n! * 2^(5*(n+1)/2) / log(17+12*sqrt(2))^(n+1). - Vaclav Kotesovec, Oct 08 2013

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A185419 Table of coefficients of a polynomial sequence of binomial type related to the enumeration of minimax trees A080795.

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

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A003704 Expansion of log(1+sinh(x)).

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A185896 Triangle of coefficients of (1/sec^2(x))*D^n(sec^2(x)) in powers of t = tan(x), where D = d/dx.

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

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A143523 a(n) = n-fold Dumont operator of x evaluated at x=y=1, z=3.

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