A008294 -id:A008294 - cp's OEIS frontend

A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = kT(n-1,k-1) + (k+1)T(n-1,k+1).

1, 0, 1, 1, 0, 2, 0, 5, 0, 6, 5, 0, 28, 0, 24, 0, 61, 0, 180, 0, 120, 61, 0, 662, 0, 1320, 0, 720, 0, 1385, 0, 7266, 0, 10920, 0, 5040, 1385, 0, 24568, 0, 83664, 0, 100800, 0, 40320, 0, 50521, 0, 408360, 0, 1023120, 0, 1028160, 0, 362880, 50521, 0, 1326122, 0, 6749040

Offset: 0

Philippe Deléham, Apr 06 2005

Or, triangle of coefficients (with exponents in increasing order) in polynomials Q_n(u) defined by d^n sec x / dx^n = Q_n(tan x)*sec x.
Interpolates between factorials and Euler (or secant) numbers. Related to Springer numbers.
Companion triangles are A155100 (derivative polynomials of tangent function) and A185896 (derivative polynomials of squared secant function).
A combinatorial interpretation for the polynomial Q_n(u) 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. Adjoin x_0 = 0 to the front of the permutation and x_(n+1) = (-1)^n*(n+1) to the end to form x_0,x_1,...,x_n,x_(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n;0) when x_0 < x_1 > x_2 < ... x_(n+1). For example, 0 3 -1  2 -4 is a snake of type S(3;0).
Let sc be the number of sign changes through a snake ... sc = #{i, 0 <= i <= n, x_i*x_(i+1) < 0}. For example, the snake 0 3 -1 2 -4 has sc = 3. The polynomial Q_n(u) is the generating function for the sign change statistic on snakes of type S(n;0): ... Q_n(u) = sum {snakes in S(n;0)} u^sc. See the example section below for the cases n = 2 and n = 3.
PRODUCTION MATRIX
Let D = subdiag(1,2,3,...) be the array with the indicated sequence on the first subdiagonal and zeros elsewhere and let C = transpose(D). The production matrix for this triangle is C+D: the first row of (C+D)^n is the n-th row of this triangle. D represents the derivative operator d/dx and C represents the operator p(x) -> x*d/dx(x*p(x)) acting on the basis monomials {x^n}n>=0. See Formula (1) below.

			The polynomials Q_0(u) through Q_6(u) (with exponents in decreasing order) are:
  1
  u
  2*u^2 + 1
  6*u^3 + 5*u
  24*u^4 + 28*u^2 + 5
  120*u^5 + 180*u^3 + 61*u
  720*u^6 + 1320*u^4 + 662*u^2 + 61
Triangle begins:
  1
  0 1
  1 0 2
  0 5 0 6
  5 0 28 0 24
  0 61 0 180 0 120
  61 0 662 0 1320 0 720
  0 1385 0 7266 0 10920 0 5040
  1385 0 24568 0 83664 0 100800 0 40320
  0 50521 0 408360 0 1023120 0 1028160 0 362880
  50521 0 1326122 0 6749040 0 13335840 0 11491200 0 3628800
  0 2702765 0 30974526 0 113760240 0 185280480 0 139708800 0 39916800
  2702765 0 98329108 0 692699304 0 1979524800 0 2739623040 0 1836172800 0 479001600
Examples of sign change statistic sc on snakes of type S(n;0)
= = = = = = = = = = = = = = = = = = = = = =
.....Snakes....# sign changes sc.......u^sc
= = = = = = = = = = = = = = = = = = = = = =
n=2
...0 1 -2 3...........2.................u^2
...0 2  1 3...........0.................1
...0 2 -1 3...........2.................u^2
yields Q_2(u) = 2*u^2 + 1.
n=3
...0 1 -2  3 -4.......3.................u^3
...0 1 -3  2 -4.......3.................u^3
...0 1 -3 -2 -4.......1.................u
...0 2  1  3 -4.......1.................u
...0 2 -1  3 -4.......3.................u^3
...0 2 -3  1 -4.......3.................u^3
...0 2 -3 -2 -4.......1.................u
...0 3  1  2 -4.......1.................u
...0 3 -1  2 -4.......3.................u^3
...0 3 -2  1 -4.......3.................u^3
...0 3 -2 -1 -4.......1.................u
yields Q_3(u) = 6*u^3 + 5*u.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see pp. 445 and 469.

Reinhard Zumkeller, Rows n = 0..125 of triangle, 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.
Gordon Haigh, A "natural" approach to Pick's theorem, Math. Gaz. 64 (1980), no. 429, 173-180.
Michael E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly, 102 (1995), 23-30.
Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, Volume 6.1 (1999): Research paper R21, 13 p.
M. Josuat-Vergès, Enumeration of snakes and cycle-alternating permutations, arXiv:1011.0929 [math.CO], 2010.
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.

See A008294 for another version of this triangle.
Setting u=0,1,2,3,4 gives A000364, A001586, A156129, A156131, A156132.
Setting u=sqrt(2) gives A156134 and A156138; u=sqrt(3) gives A002437 and A002439.
Cf. A060187, A155100, A185896, A186492.

a104035 n k = a104035_tabl !! n !! k
a104035_row n = a104035_tabl !! n
a104035_tabl = iterate f [1] where
   f xs = zipWith (+)
     (zipWith (*) [1..] (tail xs) ++ [0,0]) ([0] ++ zipWith (*) [1..] xs)
-- Reinhard Zumkeller, Apr 27 2014

nmax = 10; t[n_, k_] := t[n, k] = k*t[n-1, k-1] + (k+1)*t[n-1, k+1]; t[0, 0] = 1; t[0, ] = 0; Flatten[ Table[t[n, k], {n, 0, nmax}, {k, 0, n}]] (* _Jean-François Alcover, Nov 14 2011 *)

T(n, n) = n!; T(n, 0) = 0 if n = 2m + 1; T(n, 0) = A000364(m) if n = 2m.
Sum_{k>=0} T(m, k)*T(n, k) = T(m+n, 0).
Sum_{k>=0} T(n, k) = A001586(n): Springer numbers.
G.f.: Sum_{n >= 0} Q_n(u)*t^n/n! = 1/(cos t - u sin t).
From Peter Bala: (Start)
RECURRENCE RELATION
For n>=0,
(1)... Q_(n+1)(u) = d/du Q_n(u) + u*d/du(u*Q_n(u))
... = (1+u^2)*d/du Q_n(u) + u*Q_n(u),
with starting condition Q_0(u) = 1. Compare with Formula (4) of A186492.
RELATION WITH TYPE B EULERIAN NUMBERS
(2)... Q_n(u) = ((u+i)/2)^n*B(n,(u-i)/(u+i)), where i = sqrt(-1) and
[B(n,u)]n>=0 = [1,1+u,1+6*u+u^2,1+23*u+23*u^2+u^3,...] is the sequence of type B Eulerian polynomials (with a factor of u removed) - see A060187.
(End)
T(n,0) = abs(A122045(n)). - Reinhard Zumkeller, Apr 27 2014

Entry revised by N. J. A. Sloane, Nov 06 2009

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

A008293 Triangle of coefficients in expansion of D^n (tan x) in powers of tan x.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 8, 6, 16, 40, 24, 16, 136, 240, 120, 272, 1232, 1680, 720, 272, 3968, 12096, 13440, 5040, 7936, 56320, 129024, 120960, 40320, 7936, 176896, 814080, 1491840, 1209600, 362880, 353792, 3610112, 12207360, 18627840, 13305600, 3628800

Offset: 0

N. J. A. Sloane

James Gregory calculated the first eight rows of this table (with some numerical errors) in 1671. See Roy, p. 299. - Peter Bala, Sep 06 2016

			From _Peter Bala_, Sep 06 2016: (Start)
Table begins
    1
    1     1
    2     2
    2     8      6
   16    40     24
   16   136    240    120
  272  1232   1680    720
  272  3968  12096  13440  5040
  ...
D(tan(x)) = 1 + tan(x)^2.
D^2(tan(x)) = 2*tan(x) + 2*tan(x)^3.
D^3(tan(x)) = 2 + 8*tan(x)^2 + 6*tan(x)^4.
D^4(tan(x)) = 16*tan(x) + 40*tan(x)^3 + 24*tan(x)^5. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..990
William Y. C. Chen and Amy M. Fu, The Dumont Ansatz for the Eulerian Polynomials, Peak Polynomials and Derivative Polynomials, arXiv:2204.01497 [math.CO], 2022.
M.-P. Grosset and A. P. Veselov, Bernoulli numbers and solitons, arXiv:math/0503175 [math.GM], 2005.
Gordon Haigh, A "natural" approach to Pick's theorem, Math. Gaz. 64 (1980), no. 429, 173-180.
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.
Shi-Mei Ma, Qi Fang, Toufik Mansour, and Yeong-Nan Yeh, Alternating Eulerian polynomials and left peak polynomials, arXiv:2104.09374, 2021
R. Roy, The Discovery of the Series Formula for Pi by Leibniz, Gregory and Nilakantha, Mathematics Magazine Vol. 63, No. 5 (Dec., 1990), 291-306.
M. S. Tokmachev, Correlations Between Elements and Sequences in a Numerical Prism, Bulletin of the South Ural State University, Ser. Mathematics. Mechanics. Physics, 2019, Vol. 11, No. 1, 24-33.

Cf. A008294. Other versions of same triangle: A101343, A155100.
T(n,ceiling(n/2)) gives A000142.
Bisection of column k=0 gives A000182.
Row sums give A000831.

row[n_] := CoefficientList[ D[Tan[x], {x, n}] /. Tan -> Identity /. Sec -> Function[Sqrt[1 + #^2]], x] // DeleteCases[#, 0]&; Table[row[n], {n, 0, 10}] // Flatten // Prepend[#, 1] & (* Jean-François Alcover, Apr 05 2013 *)
T[ n_, k_] := If[n<1, Boole[n==0 && k==1], (k-1)*T[n-1, k-1] + (k+1)*T[n-1, k+1]]; (* Michael Somos, Jul 08 2024 *)

{T(n, k) = if(n<1, n==0 && k==1, (k-1)*T(n-1, k-1) + (k+1)*T(n-1, k+1))}; /* Michael Somos, Jul 08 2024 */

T(0, k) = delta(1, k), T(n, k) = (k-1)*T(n-1, k-1) + (k+1)*T(n-1, k+1).

A151775 Triangle read by rows: T(n,k) = value of (d^2n/dx^2n) (tan^(2k)(x)/cos(x)) at the point x = 0.

Original entry on oeis.org

1, 1, 2, 5, 28, 24, 61, 662, 1320, 720, 1385, 24568, 83664, 100800, 40320, 50521, 1326122, 6749040, 13335840, 11491200, 3628800, 2702765, 98329108, 692699304, 1979524800, 2739623040, 1836172800, 479001600, 199360981, 9596075582

Offset: 0

N. J. A. Sloane, Jun 24 2009, at the suggestion of Alexander R. Povolotsky

From Emeric Deutsch, Jun 27 2009: (Start)
T(n,0) = A000364(n), the Euler (or secant) numbers.
Sum of entries in row n = A000281(n).
(End)

			Triangle begins:
      1;
      1,       2;
      5,      28,      24;
     61,     662,    1320,      720;
   1385,   24568,   83664,   100800,    40320;
  50521, 1326122, 6749040, 13335840, 11491200, 3628800;

Alois P. Heinz, Rows n = 0..100, flattened
C. Radoux, The Hankel Determinant of Exponential Polynomials: A Very Short Proof and a New Result Concerning Euler Numbers, Amer. Math. Monthly, 109 (2002), 277-278.

A subtriangle of A008294.

Cf. A000364, A000281. [Emeric Deutsch, Jun 27 2009]

A151775 := proc(n,k) if n= 0 then 1 ; else taylor( (tan(x))^(2*k)/cos(x),x=0,2*n+1) ; diff(%,x$(2*n)) ; coeftayl(%,x=0,0) ; fi; end: for n from 0 to 10 do for k from 0 to n do printf("%d ", A151775(n,k)) ; od: printf("\n") ; od: # R. J. Mathar, Jun 24 2009
T := proc (n, k) if n = 0 and k = 0 then 1 elif n = 0 then 0 else simplify(subs(x = 0, diff(tan(x)^(2*k)/cos(x), `$`(x, 2*n)))) end if end proc: for n from 0 to 7 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form; Emeric Deutsch, Jun 27 2009
# alternative Maple program:
T:= (n, k)-> (2*n)!*coeff(series(tan(x)^(2*k)/cos(x), x, 2*n+1), x, 2*n):
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Aug 06 2017

T[n_, k_] := (2n)! SeriesCoefficient[Tan[x]^(2k)/Cos[x], {x, 0, 2n}];
T[0, 0] = 1;
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 10 2019, after Alois P. Heinz *)

More values from R. J. Mathar and Emeric Deutsch, Jun 24 2009

Showing 1-4 of 4 results.

cp's OEIS Frontend

A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = k*T(n-1,k-1) + (k+1)*T(n-1,k+1).

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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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A008293 Triangle of coefficients in expansion of D^n (tan x) in powers of tan x.

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A151775 Triangle read by rows: T(n,k) = value of (d^2n/dx^2n) (tan^(2k)(x)/cos(x)) at the point x = 0.

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A104035 Triangle T(n,k), 0 <= k <= n, read by rows, defined by T(0,0) = 1; T(0,k) = 0 if k>0 or if k<0; T(n,k) = kT(n-1,k-1) + (k+1)T(n-1,k+1).