|
| |
|
|
A185896
|
|
Triangle of coefficients of (1/sec^2(x))*D^n(sec^2(x)) in powers of t = tan(x), where D = d/dx.
|
|
10
|
|
|
|
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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
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 <= 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).
|
|
|
LINKS
|
|
|
|
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/(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
The polynomials (1+t^2)*R(n,t) are the polynomials P_(n+2)(t) of
Put S(n,t) = R(n,i*t), where i = sqrt(-1). We have the definite integral
evaluation
(6)... int((1-t^2)*S(m,t)*S(n,t), t = -1..1) = (-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.
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).
(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.
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)^((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 0<i<=n. (End)
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 to 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 ((1-t^2)*(16-120*t^2+120*t^4)*(272-3696*t^2+8400*t^4-5040*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(n-1), u))); fi;
end proc;
for n from 0 to 12 do
t1 := series(R(n), u, 20);
lprint(seriestolist(t1));
od:
|
|
|
MATHEMATICA
|
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 *)
|
|
|
PROG
|
(PARI) {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))))};
(PARI) {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 */
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
