A185896 - OEIS

%I #64 Apr 21 2021 07:58:57

%S 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,

%T 8400,0,5040,0,7936,0,48384,0,80640,0,40320,7936,0,168960,0,645120,0,

%U 846720,0,362880,0,353792,0,3256320,0,8951040,0,9676800,0,3628800

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

%C DEFINITION

%C Define polynomials R(n,t) with t = tan(x) by

%C ... (d/dx)^n sec^2(x) = R(n,tan(x))*sec^2(x).

%C The first few are

%C ... R(0,t) = 1

%C ... R(1,t) = 2*t

%C ... R(2,t) = 2 + 6*t^2

%C ... R(3,t) = 16*t + 24*t^3.

%C This triangle shows the coefficients of R(n,t) in ascending powers of t called the tangent number triangle in [Hodges and Sukumar].

%C 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.

%C COMBINATORIAL INTERPRETATION

%C 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].

%C 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.

%C Let x_1,...,x_n be a signed permutation.

%C Then 0,x_1,...,x_n,0 is a snake of type S(n;0,0) when 0 < x_1 > x_2 < ... 0.

%C For example, 0 4 -3 -1 -2 0 is a snake of type S(4;0,0).

%C Let sc be the number of sign changes through a snake

%C ... sc = #{i, 1 <= i <= n-1, x_i*x_(i+1) < 0}.

%C 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):

%C ... R(n,t) = sum {snakes in S(n+1;0,0)} t^sc.

%C See the example section below for the cases n=1 and n=2.

%C PRODUCTION MATRIX

%C Define three arrays R, L, and S as

%C ... R = superdiag[2,3,4,...]

%C ... L = subdiag[1,2,3,...]

%C ... S = diag[2,4,6,...]

%C 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.

%C 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.

%C The three arrays satisfy the commutation relations

%C ... [R,L] = S, [R,S] = 2*R, [L,S] = -2*L

%C and hence give a representation of the Lie algebra sl(2).

%H G. C. Greubel, <a href="/A185896/b185896.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H K. Boyadzhiev, <a href="http://arxiv.org/abs/0903.0117">Derivative Polynomials for tanh, tan, sech and sec in Explicit Form</a>, arXiv:0903.0117 [math.CA], 2009-2010.

%H M-P. Grosset and A. P. Veselov, <a href="http://arxiv.org/abs/math/0503175">Bernoulli numbers and solitons</a>, arXiv:math/0503175 [math.GM], 2005.

%H A. Hodges and C. V. Sukumar, <a href="http://dx.doi.org/10.1098/rspa.2007.0001">Bernoulli, Euler, permutations and quantum algebras</a>, Proc. R. Soc. A (2007) 463, 2401-2414 doi:10.1098/rspa.2007.0001

%H Michael E. Hoffman, <a href="http://www.combinatorics.org/ojs/index.php/eljc/article/view/v6i1r21">Derivative polynomials, Euler polynomials,and associated integer sequences</a>, Electronic Journal of Combinatorics, Volume 6 (1999), Research Paper #R21.

%H Michael E. Hoffman, <a href="http://www.jstor.org/stable/2974853">Derivative polynomials for tangent and secant</a>, Amer. Math. Monthly, 102 (1995), 23-30.

%H M. Josuat-Verges, <a href="http://arxiv.org/abs/1011.0929">Enumeration of snakes and cycle-alternating permutations</a>, arXiv:1011.0929 [math.CO], 2010.

%H Shi-Mei Ma, Qi Fang, Toufik Mansour, Yeong-Nan Yeh, <a href="https://arxiv.org/abs/2104.09374">Alternating Eulerian polynomials and left peak polynomials</a>, arXiv:2104.09374, 2021

%F GENERATING FUNCTION

%F E.g.f.:

%F (1)... F(t,z) = 1/(cos(z)-t*sin(z))^2 = sum {n>=0} R(n,t)*z^n/n!

%F = 1 +  (2*t)*z + (2+6*t^2)*z^2/2! + (16*t+24*t^3)*z^3/3! + ....

%F The e.g.f. equals the square of the e.g.f. of A104035.

%F Continued fraction representation for the o.g.f:

%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)

%F *z^2/(1-2*n*(n+1)*t*z -....

%F RECURRENCE RELATION

%F (3)... T(n,k) = (k+1)*(T(n-1,k-1) + T(n-1,k+1)).

%F ROW POLYNOMIALS

%F The polynomials R(n,t) satisfy the recurrence relation

%F (4)... R(n+1,t) = d/dt{(1+t^2)*R(n,t)} with R(0,t) = 1.

%F Let D be the derivative operator d/dt and U = t, the shift operator.

%F (5)... R(n,t) = (D + DUU)^n 1

%F RELATION WITH OTHER SEQUENCES

%F A) Derivative Polynomials A155100

%F The polynomials (1+t^2)*R(n,t) are the  polynomials P_(n+2)(t) of

%F A155100.

%F B) Bernoulli Numbers A000367 and A002445

%F Put S(n,t) = R(n,i*t), where i = sqrt(-1). We have the definite integral

%F evaluation

%F (6)... int((1-t^2)*S(m,t)*S(n,t), t = -1..1) = (-1)^((m-n)/2)*2^(m+n+3)

%F *Bernoulli(m+n+2).

%F The case  m = n is equivalent to the result of [Grosset and Veselov]. The

%F methods used there extend to the general case.

%F C) Zigzag Numbers A000111

%F (7)... R_n(1) = A000828(n+1) = 2^n*A000111(n+1).

%F D) Eulerian Numbers A008292

%F The polynomials R(n,t) are related to the Eulerian polynomials A(n,t) via

%F (8)... R(n,t) = (t+i)^n*A(n+1,(t-i)/(t+i))

%F with the inverse identity

%F (9)... A(n+1,t) = (-i/2)^n*(1-t)^n*R(n,i*(1+t)/(1-t)),

%F where {A(n,t)}n>=1 = [1,1+t,1+4*t+t^2,1+11*t+11*t^2+t^3,...] is the

%F sequence of Eulerian polynomials and i = sqrt(-1).

%F E) Ordered set partitions A019538

%F (10)... R(n,t) = (-2*i)^n*T(n+1,x)/x,

%F where x = i/2*t - 1/2 and T(n,x) is the n-th row po1ynomial of A019538;

%F F) Miscellaneous

%F Column 1 is the sequence of tangent numbers - see A000182.

%F A000670(n+1) = (-i/2)^n*R(n,3*i).

%F A004123(n+2) = 2*(-i/2)^n*R(n,5*i).

%F A080795(n+1) =(-1)^n*(sqrt(-2))^n*R(n,sqrt(-2)). - Peter Bala, Aug 26 2011

%F From _Leonid Bedratyuk_, Aug 12 2012: (Start)

%F 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.

%F 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)

%F 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

%e Table begins

%e n\k|.....0.....1.....2.....3.....4.....5.....6

%e ==============================================

%e 0..|.....1

%e 1..|.....0.....2

%e 2..|.....2.....0.....6

%e 3..|.....0....16.....0....24

%e 4..|....16.....0...120.....0...120

%e 5..|.....0...272.....0...960.....0...720

%e 6..|...272.....0..3696.....0..8400.....0..5040

%e .

%e Examples of recurrence relation

%e T(4,2) = 3*(T(3,1) + T(3,3)) = 3*(16 + 24) = 120;

%e T(6,4) = 5*(T(5,3) + T(5,5)) = 5*(960 + 720) = 8400.

%e Example of integral formula (6)

%e ... int ((1-t^2)*(16-120*t^2+120*t^4)*(272-3696*t^2+8400*t^4-5040*t^6),

%e t = -1..1) =  2830336/1365 = -2^13*Bernoulli(12).

%e Examples of sign change statistic sc on snakes of type (0,0)

%e = = = = = = = = = = = = = = = = = = = = = =

%e .....Snakes....# sign changes sc.......t^sc

%e = = = = = = = = = = = = = = = = = = = = = =

%e n=1

%e ...0 1 -2 0...........1................t

%e ...0 2 -1 0...........1................t

%e yields R(1,t) = 2*t;

%e n=2

%e ...0 1 -2 3 0.........2................t^2

%e ...0 1 -3 2 0.........2................t^2

%e ...0 2 1 3 0..........0................1

%e ...0 2 -1 3 0.........2................t^2

%e ...0 2 -3 1 0.........2................t^2

%e ...0 3 1 2 0..........0................1

%e ...0 3 -1 2 0.........2................t^2

%e ...0 3 -2 1 0.........2................t^2

%e yields

%e R(2,t) = 2 + 6*t^2.

%p R = proc(n) option remember;

%p if n=0 then RETURN(1);

%p else RETURN(expand(diff((u^2+1)*R(n-1), u))); fi;

%p end proc;

%p for n from 0 to 12 do

%p t1 := series(R(n), u, 20);

%p lprint(seriestolist(t1));

%p od:

%t 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 *)

%o (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))))};

%o (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 */

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

%K nonn,easy,tabl

%O 0,3

%A _Peter Bala_, Feb 07 2011