

A185896


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


9



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
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OFFSET

0,3


COMMENTS

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 <= n1, 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 nth 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).


LINKS

Table of n, a(n) for n=0..54.
K. Boyadzhiev, Derivative Polynomials for tanh, tan, sech and sec in Explicit Form
MP. Grosset and A. P. Veselov, Bernoulli numbers and solitons
A. Hodges and C. V. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. R. Soc. A (2007) 463,
24012414 doi:10.1098/rspa.2007.0001
Michael E. Hoffman, Derivative polynomials, Euler polynomials,and associated integer sequences
Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 2330.
M. JVerges, Enumeration of snakes and cyclealternating permutations


FORMULA

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/(12*t*z  2*(1+t^2)*z^2/(14*t*z ... n*(n+1)*(1+t^2)
*z^2/(12*n*(n+1)*t*z ....
RECURRENCE RELATION
(3)... T(n,k) = (k+1)*(T(n1,k1) + T(n1,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)... int((1t^2)*S(m,t)*S(n,t), t = 1..1) = (1)^((mn)/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,(tI)/(t+I))
with the inverse identity
(9)... A(n+1,t) = (I/2)^n*(1t)^n*R(n,I*(1+t)/(1t)),
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*t1/2 and T(n,x) is the nth 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

T(n,k)=(1)^(n+1)*(1)^((nk)/2)*sum_{j=k+1..n+1} j! *stirling2(n+1,j) *2^(n+1j) *(1)^(n+jk) *binomial(j1,k)), see A059419.
For n>1 and 0<i<=n
sum_{j=i+1..n+1}((1(1)^(ji))/(2*(ji))*(1)^((nj)/2)*T(n,j))=(n+1)*(1)^((n1i)/2)*T(n1,i).
[Leonid Bedratyuk, Aug 12 2012]
G.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 T. J. Stieltjes).  Sergei N. Gladkovskii, Dec 27 2013


EXAMPLE

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)
... int ((1t^2)*(16120*t^2+120*t^4)*(2723696*t^2+8400*t^45040*t^6),
t = 1..1) = 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.


MAPLE

R = proc(n) option remember;
if n=0 then RETURN(1);
else RETURN(expand(diff((u^2+1)*R(n1), u))); fi;
end proc;
for n from 0 to 12 do
t1 := series(R(n), u, 20);
lprint(seriestolist(t1));
od:


PROG

(PARI) {T(n, k)=if(n<0k<0k>n, 0, if(n==k, n!, (k+1)*(T(n1, k1)+T(n1,
k+1))))}


CROSSREFS

Cf. A000182, A000364, A000828 (row sums), A008292, A008293, A008294, A104035, A155100
Sequence in context: A221396 A221337 A157077 * A076256 A127467 A140333
Adjacent sequences: A185893 A185894 A185895 * A185897 A185898 A185899


KEYWORD

nonn,easy,tabl


AUTHOR

Peter Bala, Feb 07 2011


STATUS

approved



