A000182 -id:A000182 - cp's OEIS frontend

A000831 Expansion of e.g.f. (1 + tan(x))/(1 - tan(x)).

1, 2, 4, 16, 80, 512, 3904, 34816, 354560, 4063232, 51733504, 724566016, 11070525440, 183240753152, 3266330312704, 62382319599616, 1270842139934720, 27507470234550272, 630424777638805504, 15250953398036463616, 388362339077351014400, 10384044045105304174592

Offset: 0

N. J. A. Sloane

nonn

			(1+tan x)/(1-tan x) = 1 + 2*x/1! + 4*x^2/2! + 16*x^3/3! + 80*x^4/4! + 512*x^5/5! + ...

Alois P. Heinz, Table of n, a(n) for n = 0..432 (first 84 terms from R. J. Mathar)
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.
D. Dumont, Further triangles of Seidel-Arnold type and continued fractions related to Euler and Springer numbers, Adv. Appl. Math., 16 (1995), 275-296.
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.

Bisections: A002436 and A012393.

Cf. A000111, A000182, A155100, A258880, A258901, A258994.

m:=30; R:=PowerSeriesRing(Rationals(), m);
Coefficients(R!(Laplace( (1+Tan(x))/(1-Tan(x)) ))); // G. C. Greubel, Mar 21 2019; Apr 28 2023

A000831 := (1+tan(x))/(1-tan(x)) : for n from 0 to 200 do printf("%d %d ",n,n!*coeftayl(A000831,x=0,n)) ; end: # R. J. Mathar, Nov 19 2006
A000831 := n -> `if`(n=0,1,(-1)^((n^2-n)/2)*4^n*(euler(n,1/2)+euler( n,1))): # Peter Luschny, Nov 24 2010
# third Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, 2*add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> b(n, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Sep 02 2020

Range[0, 18]! CoefficientList[Series[(1+Tan[x])/(1-Tan[x]), {x,0,18}], x] (* Robert G. Wilson v, Apr 16 2011 *)
b[u_, o_] := b[u, o] = If[u+o == 0, 1, 2*Sum[b[o-1+j, u-j], {j, 1, u}]];
a[n_] := b[n, 0];
Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Dec 02 2023, after Alois P. Heinz *)

a(n):=sum(if evenp(n+k) then ((-1)^((n+k)/2)*sum(j!*stirling2(n,j)*2^(n-j+1)*(-1)^(j)*binomial(j-1,k-1),j,k,n)) else 0,k,1,n); /* Vladimir Kruchinin, Aug 19 2010 */

a(n) = if( n<1, n==0, n! * polcoeff( 1 + 2 / ( 1 / tan( x + x * O(x^n)) - 1), n)) /* Michael Somos, Apr 16 2011 */

a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( (cos(x + A) + sin(x + A)) / (cos(x + A) - sin(x + A)), n)) /* Michael Somos, Apr 16 2011 */

@CachedFunction
def sp(n,x) :
    if n == 0 : return 1
    return -add(2^(n-k)*sp(k,1/2)*binomial(n,k) for k in range(n)[::2])
A000831 = lambda n : abs(sp(n,x))
[A000831(n) for n in (0..21)]     # Peter Luschny, Jul 30 2012

def A000831_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( (1+tan(x))/(1-tan(x)) ).egf_to_ogf().list()
A000831_list(40) # G. C. Greubel, Mar 21 2019; Apr 28 2023

E.g.f.: tan(x+Pi/4).
a(n) = Sum_{k=1..n} (if even(n+k) ( (-1)^((n+k)/2)*Sum_{j=k..n} (j!*stirling2(n,j)*2^(n-j+1)*(-1)^(j)*binomial(j-1,k-1) ), n>0. - Vladimir Kruchinin, Aug 19 2010
a(n) = 4^n*(E_{n}(1/2) + E_{n}(1))*(-1)^((n^2-n)/2) for n > 0, where E_{n}(x) is an Euler polynomial. - Peter Luschny, Nov 24 2010
a(n) = 2^n * A000111(n). - Gerard P. Michon, Feb 24 2011
From Sergei N. Gladkovskii, Dec 01 2011 - Jan 24 2014: (Start)
Continued fractions:
E.g.f.: -1 + 2/(1-x*G(0)); G(k) = 1 - (x^2)/((x^2) - (2*k + 1)*(2*k + 3)/G(k+1)).
E.g.f.: 1 + 2*x/(U(0)-2*x) where U(k) = 4*k+1 + x/(1+x/ (4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1 + 2*x/(G(0)-x) where G(k) =  2*k+1 - x^2/G(k+1).
G.f.: 1 + 2*x/Q(0), where Q(k) = 1 - 2*x*(2*k+1) - 2*x^2*(2*k+1)*(2*k+2)/( 1 - 2*x*(2*k+2) - 2*x^2*(2*k+2)*(2*k+3)/Q(k+1)).
E.g.f.: tan(2*x) + sec(2*x) = (x-1)/(x+1) - 2*(2*x^2+3)/( T(0)*3*x*(1+x)- 2*x^2-3)/(x+1), where T(k) = 1 - x^4*(4*k-1)*(4*k+7)/( x^4*(4*k-1)*(4*k+7) - (4*k+1)*(4*k+5)*(16*k^2 + 8*k - 2*x^2 - 3)*(16*k^2 + 40*k - 2*x^2 + 21)/T(k+1)).
E.g.f.: 1 + 2*x/Q(0), where Q(k) = 4*k+1 -x/(1 - x/( 4*k+3 + x/(1 + x/Q(k+1)))).
E.g.f.: tan(2*x) + sec(2*x) = 2*R(0)-1, where R(k) = 1 + x/( 4*k+1 - x/(1 - x/( 4*k+3 + x/R(k+1)))).
G.f.: 1+ G(0)*2*x/(1-2*x), where G(k) = 1 - 2*x^2*(k+1)*(k+2)/(2*x^2*(k+1)*(k+2) - (1-2*x*(k+1))*(1-2*x*(k+2))/G(k+1)). (End)
a(n) ~ n! * (4/Pi)^(n+1). - Vaclav Kotesovec, Jun 15 2015
a(0) = 1; a(n) = 2 * Sum_{k=0..n-1} binomial(n-2,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jun 11 2020

A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

Original entry on oeis.org

1, 2, 1, 4, 10, 1, 8, 60, 36, 1, 16, 296, 516, 116, 1, 32, 1328, 5168, 3508, 358, 1, 64, 5664, 42960, 64240, 21120, 1086, 1, 128, 23488, 320064, 900560, 660880, 118632, 3272, 1, 256, 95872, 2225728

Offset: 0

Johannes W. Meijer, Feb 20 2009, Jun 24 2009

Essentially the same as A185411. Row reverse of A185410. - Peter Bala, Jul 24 2012
The SF(z; n) formulas, see below, were discovered while studying certain properties of the Dirichlet eta function.
From Peter Bala, Apr 03 2011: (Start)
Let D be the differential operator 2*x*d/dx. The row polynomials of this table come from repeated application of the operator D to the function g(x) = 1/sqrt(1 - x). For example,
  D(g) = x*g^3
  D^2(g) = x*(2 + x)*g^5
  D^3(g) = x*(4 + 10*x + x^2)*g^7
  D^4(g) = x*(8 + 60*x + 36*x^2 + x^3)*g^9.
Thus this triangle is analogous to the triangle of Eulerian numbers A008292, whose row polynomials come from the  repeated application of the operator x*d/dx to the function 1/(1 - x). (End)

			The first few rows of the triangle are:
  [1]
  [2, 1]
  [4, 10, 1]
  [8, 60, 36, 1]
  [16, 296, 516, 116, 1]
The first few P(z;n) are:
  P(z; n=0) = 1
  P(z; n=1) = 2 + z
  P(z; n=2) = 4 + 10*z + z^2
  P(z; n=3) = 8 + 60*z + 36*z^2 + z^3
The first few SF(z;n) are:
  SF(z; n=0) = (1/2)*(1)/(1-z)^(3/2);
  SF(z; n=1) = (1/4)*(2+z)/(1-z)^(5/2);
  SF(z; n=2) = (1/8)*(4+10*z+z^2)/(1-z)^(7/2);
  SF(z; n=3) = (1/16)*(8+60*z+36*z^2+z^3)/(1-z)^(9/2);
In the Savage-Viswanathan paper, the coefficients appear as
  1;
  1,    2;
  1,   10,     4;
  1,   36,    60,     8;
  1,  116,   516,   296,    16;
  1,  358,  3508,  5168,  1328,   32;
  1, 1086, 21120, 64240, 42960, 5664, 64;
  ...

D. H. Lehmer, Interesting Series Involving the Central Binomial Coefficient, Am. Math. Monthly 92 (1985) 449-457, Polynomial V in eq (17). [R. J. Mathar, Feb 24 2009]
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, arXiv: 1204.4963v3 [math.CO], 2012.
Shi-Mei Ma, A family of two-variable derivative polynomials for tangent and secant, El J. Combinat. 20 (1) (2013) P11
S.-M. Ma and T. Mansour, The 1/k-Eulerian polynomials and k-Stirling permutations, arXiv preprint arXiv:1409.6525 [math.CO], 2014.
S.-M. Ma and Y.-N. Yeh, Stirling permutations, cycle structures of permutations and perfect matchings, arXiv preprint arXiv:1503.06601 [math.CO], 2015.
Carla D. Savage and Gopal Viswanathan, The 1/k-Eulerian polynomials, Elec. J. of Comb., Vol. 19, Issue 1, #P9 (2012).
Eric Weisstein's World of Mathematics, Dirichlet Eta Function

A142963 and this sequence can be mapped onto the A156920 triangle.
FP1 sequences A000340, A156922, A156923, A156924.
FP2 sequences A050488, A142965, A142966, A142968.
Appears in A162005, A000182, A162006 and A162007.
Cf. A186695, A187075.
Cf. A185410 (row reverse), A185411.

A156919 := proc(n,m) if n=m then 1; elif m=0 then 2^n ; elif m<0 or m>n then 0; else 2*(m+1)*procname(n-1,m)+(2*n-2*m+1)*procname(n-1,m-1) ; end if; end proc: seq(seq(A156919(n,m), m=0..n), n=0..7); # R. J. Mathar, Feb 03 2011

g[0] = 1/Sqrt[1-x]; g[n_] := g[n] = 2x*D[g[n-1], x]; p[n_] := g[n] / g[0]^(2n+1) // Cancel; row[n_] := CoefficientList[p[n], x] // Rest; Table[row[n], {n, 0, 9}] // Flatten (* Jean-François Alcover, Aug 09 2012, after Peter Bala *)
Flatten[Table[Rest[CoefficientList[Nest[2 x D[#, x] &, (1 - x)^(-1/2), k] (1 - x)^(k + 1/2), x]], {k, 10}]] (* Jan Mangaldan, Mar 15 2013 *)

SF(z; n) = Sum_{m >= 1} m^(n-1)*4^(-m)*z^(m-1)*Gamma(2*m+1)/(Gamma(m)^2) = P(z;n) / (2^(n+1)*(1-z)^((2*n+3)/2)) for n >= 0. The polynomials P(z;n) = Sum_{k = 0..n} a(k)*z^k generate the a(n) sequence.
If we write the sequence as a triangle the following relation holds: T(n,m) = (2*m+2)*T(n-1,m) + (2*n-2*m+1)*T(n-1,m-1) with T(n,m=0) = 2^n and T(n,n) = 1, n >= 0 and 0 <= m <= n.
G.f.: 1/(1-xy-2x/(1-3xy/(1-4x/(1-5xy/(1-6x/(1-7xy/(1-8x/(1-... (continued fraction). - Paul Barry, Jan 26 2011
From Peter Bala, Apr 03 2011 (Start)
E.g.f.: exp(z*(x + 2)) * (1 - x)/(exp(2*x*z) - x*exp(2*z))^(3/2) = Sum_{n >= 0} P(x,n)*z^n/n! = 1 + (2 + x)*z + (4 + 10*x + x^2)*z^2/2! + (8 + 60*x + 36*x^2 + x^3)*z^3/3! + ... .
Explicit formula for the row polynomials:
P(x,n-1) = Sum_{k = 1..n} 2^(n-2*k)*binomial(2k,k)*k!*Stirling2(n,k)*x^(k-1)*(1 - x)^(n-k).
The polynomials x*(1 + x)^n * P(x/(x + 1),n) are the row polynomials of A187075.
The polynomials x^(n+1) * P((x + 1)/x,n) are the row polynomials of A186695.
Row sums are A001147(n+1). (End)
Sum_{k = 0..n} (-1)^k*T(n,k) = (-1)^binomial(n,2)*A012259(n+1). - Johannes W. Meijer, Sep 27 2011

Minor edits from Johannes W. Meijer, Sep 27 2011

A060628 Triangle of coefficients in expansion of elliptic function sn(u) in powers of u and k.

Original entry on oeis.org

1, 1, 1, 1, 14, 1, 1, 135, 135, 1, 1, 1228, 5478, 1228, 1, 1, 11069, 165826, 165826, 11069, 1, 1, 99642, 4494351, 13180268, 4494351, 99642, 1, 1, 896803, 116294673, 834687179, 834687179, 116294673, 896803, 1, 1, 8071256, 2949965020, 47152124264, 109645021894, 47152124264, 2949965020, 8071256, 1

Offset: 0

Vladeta Jovovic, Apr 13 2001

			sn u = u - (1 + k^2)*u^3/3! + (1 + 14*k^2 + k^4)*u^5/5! - (1 + 135*k^2 + 135*k^4 + k^6)*u^7/7! + ...
The triangle T(n, m) begins:
n\m 0      1         2         3         4         5      6  7
0:  1
2:  1      1
3:  1     14         1
4:  1    135       135         1
5:  1   1228      5478      1228         1
6:  1  11069    165826    165826     11069         1
7:  1  99642   4494351  13180268   4494351     99642      1
8:  1 896803 116294673 834687179 834687179 116294673 896803  1
... reformatted. - _Wolfdieter Lang_, Jul 05 2016

CRC Standard Mathematical Tables and Formulae, 30th ed. 1996, p. 526.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983, (5.2.24).

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, p. 575, 16.22.1.
A. Cayley, An Elementary Treatise on Elliptic Functions (page images), G. Bell and Sons, London, 1895, p. 56.
F. Clarke, The Taylor Series Coefficients of the Jacobi Elliptic Functions, slides. [broken link]
D. Dominici, Nested derivatives: A simple method for computing series expansions of inverse functions. arXiv:math/0501052v2 [math.CA], 2005.
A. Fransen, Conjectures on the Taylor series expansion coefficients of the Jacobian elliptic function sn(n,k), Math. Comp., 37 (1981), 475-497.
R. Fricke, Die elliptischen Funktionen und ihre Anwendungen, Erster Teil, p. 399 with p. 397.
C. L. Mallows, Letter to N. J. A. Sloane, May 16 1973
J. Tannery and J. Molk, Eléments de la Théorie des Fonctions Elliptiques (Vol. 4), Gauthier-Villars, Paris, 1902, p. 92.
G. Viennot, Une interprétation combinatoire des coefficients des développements en série entière des fonctions elliptiques de Jacobi, J. Combin. Theory, A 29 (1980), 121-133.
Eric Weisstein's World of Mathematics, Jacobi Elliptic Functions

Row sums give A000182. Diagonals: A004004, A004005, A002753, A032348, A032427, A032428, A032429, A032430, A032431, A032432, A032433.

Maple program from Rostislav Kollman (kollman(AT)dynasig.cz), Nov 05 2009: (Start) The program generates an "all in one" triangle of Taylor coefficients of the Jacobi SN,CN,DN functions.
"SN ", 1 "CN ", 1 "DN ", 1
"SN ", 1, 1 "CN ", 1, 4 "DN ", 4, 1
"SN ", 1, 14, 1 "CN ", 1, 44, 16 "DN ", 16, 44, 1
"SN ", 1, 135, 135, 1 "CN ", 1, 408, 912, 64 "DN ", 64, 912, 408, 1
"SN ", 1, 1228, 5478, 1228, 1 "CN ", 1, 3688, 30768, 15808, 256 "DN ", 256, 15808, 30768, 3688, 1
"SN ", 1, 11069, 165826, 165826, 11069, 1 "CN ", 1, 33212, 870640, 1538560, 259328, 1024 "DN ", 1024, 259328, 1538560, 870640, 33212, 1
#----------------------------------------------------------------
# Taylor series coefficients of Jacobi SN,CN,DN
#----------------------------------------------------------------
n := 6: g := x: for i from 1 to 2*n do g := simplify(y*z*diff(g,x) + x*z*diff(g,y) + x*y*diff(g,z)); if(type(i,odd))then SN := simplify(sort(subs(z = k,subs(y = 1,subs(x = 0,g)))) / k);
# lprint("SN ",SN); lprint("SN ",seq(coeff(SN, k, j),j=0..i-1,2)); else CN := simplify(sort(subs(z = 1,subs(y = 0,subs(x = k,g)))) / k); DN := simplify(sort(subs(z = 0,subs(y = k,subs(x = 1,g)))));
# lprint("CN ",CN); # lprint("DN ",DN); lprint("CN ",seq(coeff(CN, k, j),j=0..i-2,2)); lprint("DN ",seq(coeff(DN, k, j),j=2..i,2)); end; end: (End)
A060628 := proc(n,m) JacobiSN(z,k) ; coeftayl(%,z=0,2*n+1) ; (-1)^n*coeftayl(%,k=0,2*m)*(2*n+1)! ; end proc: # alternative program, R. J. Mathar, Jan 30 2011

maxn = 8; se = Series[ JacobiSN[u, m], {u, 0, 2*maxn + 1 }]; cc = Partition[ CoefficientList[se, u], 2][[All, 2]]; Flatten[ (CoefficientList[#, m] & /@ cc)* Table[(-1)^n*(2*n + 1)!, {n, 0, maxn}]] (* Jean-François Alcover, Sep 21 2011 *)

Sum_{n>=0} Sum_{k=0..n} (-1)^n*T(n, k)*y^(2*k)*x^(2*n+1)/(2*n+1)! = JacobiSN(x, y).
JacobiSN(x, y) = 1*x + (-1/6 - (1/6)*y^2)*x^3 + (1/120 + (7/60)*y^2 + (1/120)*y^4)*x^5 + (-1/5040 - (3/112)*y^4 - (3/112)*y^2 - (1/5040)*y^6)*x^7 + (1/362880 + (307/90720)*y^6 + (913/60480)*y^4 + (307/90720)*y^2 + (1/362880)*y^8)*x^9 + O(x^11).
From Peter Bala, Aug 23 2011: (Start)
Let f(x) = sqrt((1-x^2)*(1-k^2*x^2)).
Define the nested derivative D^n[f](x) by means of the recursion D^0[f](x) = 1 and D^(n+1)[f](x) = d/dx(f(x)*D^n[f](x)) for n >= 0.
Then the coefficient polynomial R(2*n+1,k) of u^(2*n+1)/(2*n+1)! is given by R(2*n+1,k) = D^(2*n)[f](0) - apply [Dominici, Theorem 4.1].
See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x).
(End)
sn(u|k^2) = Sum_{n>=0} a_n(k^2)*u^(2*n+1)/(2*n+1)!. For the recurrence of the row polynomials a_n(k^2) = Sum_{m=0..n} (-1)^n*T(n, m)*k^(2*m), see the Fricke reference. - Wolfdieter Lang, Jul 05 2016

A080635 Number of permutations on n letters without double falls and without initial falls.

Original entry on oeis.org

1, 1, 1, 3, 9, 39, 189, 1107, 7281, 54351, 448821, 4085883, 40533129, 435847959, 5045745069, 62594829027, 828229153761, 11644113200031, 173331882039141, 2723549731505163, 45047085512477049, 782326996336904679, 14233537708408467549, 270733989894887810547

Offset: 0

Emanuele Munarini, Feb 28 2003

A permutation w has a double fall at k if w(k) > w(k+1) > w(k+2) and has an initial fall if w(1) > w(2).
exp(x*(1-y+y^2)*D_y)*f(y)|_{y=0} = f(1-E(-x)) for any function f with a Taylor series. D_y means differentiation with respect to y and E(x) is the e.g.f. given below. For a proof of exp(x*g(y)*D_y)*f(y) = f(F^{-1}(x+F(y))) with the compositional inverse F^{-1} of F(y)=int(1/g(y),y) with F(0)=0 see, e.g., the Datolli et al. reference.
Number of increasing ordered trees on vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <= 2. - David Callan, Mar 30 2007
Number of increasing colored 1-2 trees of order n with choice of two colors for the right branches of the vertices of outdegree 2. - Wenjin Woan, May 21 2011

			E.g.f. = 1 + x + (1/2)*x^2 + (1/2)*x^3 + (3/8)*x^4 + (13/40)*x^5 + (21/80)*x^6 + ...
G.f. = 1 + x + x^2 + 3*x^3 + 9*x^4 + 39*x^5 + 189*x^6 + 1107*x^7 + ...
For n = 3: 123, 132, 231. For n = 4: 1234, 1243, 1324, 1342, 1423, 2314, 2341, 2413, 3412.
a(4)=9. The 9 plane (ordered) increasing unary-binary trees are
...................................................................
..4................................................................
..|................................................................
..3..........4...4...............4...4...............3...3.........
..|........./.....\............./.....\............./.....\........
..2....2...3.......3...2...3...2.......2...3...4...2.......2...4...
..|.....\./.........\./.....\./.........\./.....\./.........\./....
..1......1...........1.......1...........1.......1...........1.....
...................................................................
..3...4...4...3....................................................
...\./.....\./.....................................................
....2.......2......................................................
....|.......|......................................................
....1.......1......................................................
...................................................................

Alois P. Heinz, Table of n, a(n) for n = 0..464
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Cyril Banderier et al., Explicit formulas for enumeration of lattice paths: basketball and the kernel method, arXiv:1609.06473 [math.CO], 2016. See series expansion p. 35.
Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, preprint.
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Rupert Li, Vincular Pattern Avoidance on Cyclic Permutations, arXiv:2107.12353 [math.CO], 2021, p. 6.
Christopher Zhu, Enumerating Permutations and Rim Hooks Characterized by Double Descent Sets, arXiv:1910.12818 [math.CO], 2019.

Cf. A049774, A185415, A185416, A185422, A185423, A139605, A232864.
Cf. A000182, A002105, A234797.
Column k=0 of A162976.

a:= proc(n) if n < 2 then 1 else n! * sum((sqrt(3)/(2*Pi*(k+1/3)))^(n+1), k=-infinity..infinity) fi end: seq(a(n), n=0..30); # Richard Ehrenborg, Dec 09 2013
a := proc(n) option remember; local k; if n < 3 then 1 else
add(binomial(n-1, k)*a(k)*a(n-k-1), k = 0..n-2) fi end:
seq(a(n), n = 0..23); # Peter Luschny, May 24 2024

Table[n!, {n, 0, 40}]*CoefficientList[Series[ (1 + 1/Sqrt[3] Tan[Sqrt[3]/2 x])/(1 - 1/Sqrt[3] Tan[Sqrt[3]/2 x]), {x, 0, 40}], x]
a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ 1/2 + Sqrt[3]/2 Tan[ Pi/6 + Sqrt[3] x/2], {x, 0, n}]]; (* Michael Somos, May 22 2011 *)
Join[{1}, FullSimplify[Table[3^((n+1)/2) * n! * (Zeta[n+1, 1/3] - (-1)^n*Zeta[n+1, 2/3]) / (2*Pi)^(n+1), {n, 1, 20}]]] (* Vaclav Kotesovec, Aug 06 2021 *)

a(n):=if n=0 then 1 else sum((-3)^((n-k)/2)*((-1)^(n-k)+1)*sum(binomial(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^(j)*stirling2(n,j+k),j,0,n-k),k,1,n); /* Vladimir Kruchinin, Feb 13 2019 */

{a(n) = my(A); if( n<1, n==0, A = O(x); for( k=1, n, A = intformal( 1 + A + A^2)); n! * polcoeff( A, n))}; /* Michael Somos, Oct 04 2003 */

{a(n) = n! * polcoeff( exp( serreverse( intformal( 1/(2*cosh(x +x*O(x^n)) - 1) ) )), n)}
for(n=0, 30, print1(a(n), ", ")) \\ Paul D. Hanna, Feb 22 2016

@CachedFunction
def c(n,k) :
    if n==k: return 1
    if k<1 or k>n: return 0
    return ((n-k)//2+1)*c(n-1,k-1)+2*k*c(n-1,k+1)
def A080635(n):
    return add(c(n,k) for k in (0..n))
[A080635(n) for n in (0..23)] # Peter Luschny, Jun 10 2014

E.g.f.: (1 + 1/sqrt(3) * tan(sqrt(3)/2 * x)) / (1 - 1/sqrt(3) * tan( sqrt(3)/2 * x)).
Recurrence: a(n+1) = (Sum_{k=0..n} binomial(n, k) * a(k) * a(n-k)) - a(n) + 0^n.
E.g.f.: A(x) satisfies A' = 1 - A + A^2. - Michael Somos, Oct 04 2003
E.g.f.: E(x) = (3*cos((1/2)*3^(1/2)*x) + (3^(1/2))*sin((1/2)*3^(1/2)*x))/(3*cos((1/2)*3^(1/2)*x) - (3^(1/2))* sin((1/2)*3^(1/2)*x)). See the Michael Somos comment. - Wolfdieter Lang, Sep 12 2005
O.g.f.: A(x) = 1+x/(1-x-2*x^2/(1-2*x-2*3*x^2/(1-3*x-3*4*x^2/(1-... -n*x-n*(n+1)*x^2/(1- ...))))) (continued fraction). - Paul D. Hanna, Jan 17 2006
From Peter Bala: (Start)
An alternative form of the e.g.f. for this sequence taken from [Bergeron et al.] is
(1)... (sqrt(3)/2)*tan((sqrt(3)/2)*x+Pi/6) [with constant term 1/2].
By comparing the egf for this sequence with the egf for the Eulerian numbers A008292 we can show that
(2)... a(n) = A(n,w)/(1+w)^(n-1) for n >= 1,
where w = exp(2*Pi*i/3) and {A(n,x),n>=1} = [1, 1+x, 1+4*x+x^2, 1+11*x+11*x^2+x^3,...] denotes the sequence of Eulerian polynomials. Equivalently,
(3)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} k!*Stirling2(n,k)*(-1/2 + sqrt(3)*i/6)^(k-1) for n >= 1, and
(4)... a(n) = (-i*sqrt(3))^(n-1)*Sum_{k=1..n} (-1/2 + sqrt(3)*i/6)^(k-1)* Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*j^n for n >= 1.
This explicit formula for a(n) may be used to obtain various congruence results. For example,
(5a)... a(p) == 1 (mod p) for prime p = 6*n+1,
(5b)... a(p) == -1 (mod p) for prime p = 6*n+5.
For the corresponding results for the case of non-plane unary-binary trees see A000111. For type B results see A001586. For a related sequence of polynomials see A185415. See also A185416 for a recursive method to compute this sequence. For forests of plane increasing unary binary trees see A185422 and A185423. (End)
O.g.f.: A(x) = x - (1/2)*x^2 + (1/2)*x^3 - (3/8)*x^4 + (13/40)*x^5 - (21/80)*x^6 + (123/560)*x^7 - (809/4480)*x^8 + (671/4480)*x^9 - (5541/44800)*x^10 + .... - Vladimir Kruchinin, Jan 18 2011
Let f(x) = 1+x+x^2. Then a(n+1) = (f(x)*d/dx)^n f(x) evaluated at x = 0. - Peter Bala, Oct 06 2011
From Sergei N. Gladkovskii, May 06 2013 - Dec 24 2013: (Start)
Continued fractions:
G.f.: 1 + 1/Q(0), where Q(k) = 1/(x*(k+1)) - 1 - 1/Q(k+1).
E.g.f.: 1 + 2*x/(W(0)-x), where W(k) = 4*k + 2 - 3*x^2/W(k+1).
G.f.: 1 + x/Q(0), m=1, where Q(k) = 1 - m*x*(2*k+1) - m*x^2*(2*k+1)*(2*k+2)/( 1 - m*x*(2*k+2) - m*x^2*(2*k+2)*(2*k+3)/Q(k+1) ).
G.f.: 1 + x/Q(0), where Q(k) = 1 - x*(k+1) - x^2*(k+1)*(k+2)/Q(k+1).
G.f.: 1 + T(0)*x/(1-x), where T(k) = 1 - x^2*(k+1)*(k+2)/( x^2*(k+1)*(k+2) - (1-x*(k+1))*(1-x*(k+2))/T(k+1) ).
G.f.: 1 + x/(G(0)-x), where G(k) = 1 + x*(k+1) - x*(k+1)/(1 - x*(k+2)/G(k+1) ). (End)
a(n) ~ 3^(3*(n+1)/2) * n^(n+1/2) / (exp(n)*(2*Pi)^(n+1/2)). - Vaclav Kotesovec, Oct 05 2013
a(n) = n! * Sum_{k=-oo..oo} (sqrt(3)/(2*Pi*(k+1/3)))^(n+1) for n >= 1. - Richard Ehrenborg, Dec 09 2013
From Peter Bala, Sep 11 2015: (Start)
The e.g.f. A(x) = (sqrt(3)/2)*tan((sqrt(3)/2)*x + Pi/6) satisfies the differential equation A"(x) = 2*A(x)*A'(x) with A(0) = 1/2 and A'(0) = 1, leading to the recurrence a(0) = 1/2, a(1) = 1, else a(n) = 2*Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the sequence [1/2, 1, 1, 3, 9, 39, 189, 1107, ...].
Note, the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence n! and with a(0) = 0 and a(1) = 1 produces A000182. Cf. A002105, A234797. (End)
E.g.f.: exp( Series_Reversion( Integral 1/(2*cosh(x) - 1) dx ) ). - Paul D. Hanna, Feb 22 2016
a(n) = Sum_{k=1..n} (-3)^((n-k)/2)*((-1)^(n-k)+1)*Sum_{j=0..n-k} C(j+k-1,j)*(j+k)!*2^(-j-k)*(-1)^j*Stirling2(n,j+k),n>0, a(0)=1. - Vladimir Kruchinin, Feb 13 2019
For n > 0, a(n) = 3^((n+1)/2) * n! * (zeta(n+1, 1/3) - (-1)^n*zeta(n+1, 2/3)) / (2*Pi)^(n+1). - Vaclav Kotesovec, Aug 06 2021

Several typos corrected by Olivier Gérard, Mar 26 2011

A008303 Triangle read by rows: T(n,k) (n >= 1, 0 <= k <= ceiling(n/2)-1) = number of permutations of [n] with k peaks.

Original entry on oeis.org

1, 2, 4, 2, 8, 16, 16, 88, 16, 32, 416, 272, 64, 1824, 2880, 272, 128, 7680, 24576, 7936, 256, 31616, 185856, 137216, 7936, 512, 128512, 1304832, 1841152, 353792, 1024, 518656, 8728576, 21253376, 9061376, 353792, 2048, 2084864, 56520704, 222398464, 175627264, 22368256

Offset: 1

N. J. A. Sloane

From Petros Hadjicostas, Aug 06 2019: (Start)
André (1895) first defined these numbers. In his notation, T(n, k) = Q(n+1, 2*(k+1)) for n >= 1 and 0 <= k <= ceiling(n/2)-1.
His triangle is as follows (p. 148):
  Q_{2,2}
  Q_{3,2}
  Q_{4,2} Q_{4,4}
  Q_{5,2} Q_{5,4}
  Q_{6,2} Q_{6,4} Q_{6,6}
  Q_{7,2} Q_{7,4} Q_{7,6}
  ...
He has Q(n, s) = 0 when either s is odd, or n <= 1, or s > n. Also, Q_{n,2} = 2^(n-2) for n >= 2.
His recurrence is Q(n, s) = s * Q(n-1, s) + (n - s + 1) * Q(n-1, s-2) for n >= 3 and s >= 2. (Obviously, for s odd, we get Q(n, s) = 0 + 0 = 0.)
In terms of the current array, André's (1895) recurrence becomes T(n, k) = (2*k + 2) * T(n-1, k) + (n - 2*k) * T(n-1, k-1) for n >= 2 and 1 <= k <= n with T(n, 0) = 2^(n-1) for n >= 1. In this recurrence, we assume T(n, k) = 0 for k >= ceiling(n/2) or k < 0. (End)
From Petros Hadjicostas, Aug 07 2019: (Start)
We clarify further the quantity Q(n, s) defined by André (1895). In his paper, André considers circular permutations of [n] and deals with maxima, minima, and so-called "séquences" in a permutation.
The term "séquence" in a permutation, as used by André in several of his papers in the 19th century, means a list of consecutive numbers in the permutation that go from a maximum to a minimum, or vice versa, and do not contain any interior minima or maxima. This terminology is also repeated in Ex. 13 (pp. 260-261) by Comtet (even though he refers to the corresponding indices rather than the numbers in the permutation itself).
Some authors call these so-called "séquences" (defined by André and Comtet) "alternate runs" (or just "runs"). Here we are actually dealing with "circular runs" if we read these so-called "séquences" in ascending order in one of the two directions on a circle.
Q(n, s) is the number of circular permutations of [n] (out of the (n-1)! in total) that have exactly s of these so-called "séquences" ("alternate runs").
André (1895) proves that, in a circular permutation of [n], the number of maxima equals the number of minima and that the number of his so-called "séquences" ("alternate runs") is always even (i.e., Q(n, s) = 0 for s odd).
He also shows that, if v = floor(n/2), then the only possible values for the length of a so-called "séquence" ("alternate run") in a circular permutation of [n] are 2, 4, ..., 2*v. That is why Q(n, s) = 0 when either s is odd, or n <= 1, or s > n.
Note that Sum_{t = 1..floor(n/2)} Q_{n, 2*t} = Sum_{t = 1..floor(n/2)} T(n-1, t-1) = (n-1)! = total number of circular permutations of [n].
Since T(n, k) = Q(n+1, 2*(k+1)) for n >= 1 and 0 <= k <= ceiling(n/2)-1, we conclude that the number of (linear) permutations of [n] with k peaks equals the number of circular permutations of [n+1] with exactly 2*(k+1) of these so-called "séquences" ("alternate runs"). (End)
From Petros Hadjicostas, Aug 08 2019: (Start)
The author of this array indirectly assumes that a "peak" of a (linear) permutation of [n] is an interior maximum of the permutation; i.e., we ignore maxima at the endpoints of a permutation.
Similarly, a "valley" of a (linear) permutation of [n] is an interior minimum of the permutation; i.e., we ignore minima at the endpoints of the permutation.
Since the complement of a permutation a_1 a_2 ... a_n (using one-line notation, not cycle notation) is (n+1-a_1) (n+1-a_2) ... (n+1-a_n), it follows that, for n >= 2 and 0 <= k <= ceiling(n/2) - 1, T(n, k) is also the number of (linear) permutations of [n] with exactly k valleys. (End)

			Triangle T(n,k) (with rows n >= 1 and columns k >= 0) starts as follows:
  [ 1]    1;
  [ 2]    2;
  [ 3]    4,       2;
  [ 4]    8,      16;
  [ 5]   16,      88,      16;
  [ 6]   32,     416,     272;
  [ 7]   64,    1824,    2880,     272;
  [ 8]  128,    7680,   24576,    7936;
  [ 9]  256,   31616,  185856,  137216,    7936;
  [10]  512,  128512, 1304832, 1841152,  353792;
    A000079, A000431, A000487, A000517, A179708, ...
T(3,1) = 2 because we have 132 and 231.
From _Petros Hadjicostas_, Aug 07 2019: (Start)
In terms of André's (1895) notation (see the comments above), we have Q(4, 2) = T(3, 0) = 4 and Q(4, 4) = T(3, 1) = 2.
Out of the (4-1)! = 6 circular permutations of [4], each of the permutations 1324 and 1423 has exactly 4 so-called "séquences" ("alternate runs"), while each of the rest (1234, 1243, 1342, and 1432) has exactly 2 so-called "séquences" ("alternate runs").
In detail, we list the so-called "séquences" ("alternate runs") of the above circular permutations:
  1234 --> 1234 and 41 (maximum 4 and minimum 1).
  1243 --> 124 and 431 (maximum 4 and minimum 1).
  1324 --> 13, 32, 24, and 41 (maxima 3, 4, and minima 1, 2).
  1342 --> 134 and 421 (maximum 4 and minimum 1).
  1423 --> 14, 42, 23, and 31 (maxima 3, 4 and minima 1, 2),
  1432 --> 14 and 4321 (maximum 4 and minimum 1).
(End)

Florence Nightingale David and D. E. Barton, Combinatorial Chance, Charles Griffin, 1962; see Table 10.6, p. 163. [They use the notation T_{N,t^*}^{**}, where N is the length of the permutation and t^* is the number of peaks in the permutation. They also give André's recurrence. So, here n = N and k = t^*. - Petros Hadjicostas, Aug 09 2019]
Florence Nightingale David, Maurice George Kendall, and D. E. Barton, Symmetric Functions and Allied Tables, Cambridge, 1966, p. 261, Table 7.3.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, John Wiley and Sons, N.Y., 1983, Ex. 3.3.46. - Emeric Deutsch, Jul 26 2009
T. K. Petersen, Eulerian Numbers, Birkhäuser, 2015, Chapter 4.

Alois P. Heinz, Rows n = 1..200, flattened (first 30 rows from Vincenzo Librandi)
Max A. Alekseyev, On the number of permutations with bounded run lengths, arXiv preprint arXiv:1205.4581 [math.CO], 2012-2013. - From _N. J. A. Sloane_, Oct 23 2012
Désiré André, Mémoire sur les séquences des permutations circulaires, Bulletin de la S. M. F., tome 23 (1895), pp. 122-184.
T. Austin, R. Fagen, T. Lehrer, and W. Penney, The distribution of the number of locally maximal elements in a random sample, Ann. Math. Statist. 28 (1957), 786-790. - _Ira M. Gessel_, Aug 06 2014
Jean-Luc Baril and José L. Ramírez, Some distributions in increasing and flattened permutations, arXiv:2410.15434 [math.CO], 2024. See p. 3.
S. Billey, K. Burdzy, and B. E. Sagan, Permutations with given peak set, arXiv: 1209.0693 [math.CO], 2012.
S. Billey, K. Burdzy, and B. E. Sagan, Permutations with given peak set, J. Int. Seq. 16 (2013), #13.6.1.
C.-O. Chow and S.-M. Ma, Counting signed permutations by their alternating runs, Discrete Mathematics, 323 (2014), 49-57.
C.-O. Chow, S.-M. Ma, T. Mansour, and M. Shattuck, Counting permutations by cyclic peaks and valleys, Annales Mathematicae et Informaticae 43 (2014), 43-54.
Kieran Clenaghan, In Praise of Sequence (Co-)Algebra and its implementation in Haskell, arXiv:1812.05878 [math.CO], 2019. See page 36.
Colin Defant, Troupes, Cumulants, and Stack-Sorting, arXiv:2004.11367 [math.CO], 2020.
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. 13.
S. Elizalde and M. Noy, Consecutive patterns in permutations, Adv. Appl. Math. 30 (2003), 110-125; see Section 5.
R. C. Entringer, Enumeration of permutations of (1,...,n) by number of maxima, Duke Math. J. 36 (1969), 575-579. - _Ira M. Gessel_, Oct 23 2013
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.
FindStat - Combinatorial Statistic Finder, The number of inner peaks of a permutation, The number of peaks of a permutation, The number of valleys of a permutation.
W. O. Kermack and A. G. McKendrick, Some distributions associated with a randomly arranged set of numbers, Proc. Royal Soc. of Edinburgh 67 (1937), 332-376.
W. O. Kermack and A. G. McKendrick, Some properties of points arranged on a Möbius surface, Mathematical Gazette 22 (1938), 66-72.
Shi-Mei Ma, Derivative polynomials and permutations by numbers of interior peaks and left peaks, arXiv preprint arXiv:1106.5781 [math.CO], 2011.
Shi-Mei Ma, Enumeration of permutations by number of alternating runs, Discrete Math., 313 (2013), 1816-1822.
S.-M. Ma, T. Mansour, and D. G. L. Wang, Combinatorics of Dumont differential system on the Jacobi elliptic functions, arXiv preprint arXiv:1403.0233 [math.CO], 2014.
Shi-Mei Ma, Toufik Mansour, David G.L. Wang, and Yeong-Nan Yeh, Several variants of the Dumont differential system and permutation statistics, Science China Mathematics 60 (2018).
Shi-Mei Ma, Jun Ma, Jean Yeh, and Yeong-Nan Yeh, The 1/k-Eulerian polynomials of type B, arXiv:2001.07833 [math.CO], 2020.
A. Mendes and J. B. Remmel, Permutations and words counted by consecutive patterns, Adv. Appl. Math. 37(4) (2006), 443-480.
Tom Roberts and Thomas Prellberg, Improving Convergence of Generalised Rosenbluth Sampling for Branched Polymer Models by Uniform Sampling, arXiv:2401.12201 [cond-mat.stat-mech], 2024. See p. 14.
Louis W. Shapiro, Wen-Jin Woan, and Seyoum Getu, Runs, slides and moments, SIAM J. Algebraic and Discrete Methods 4 (1983), 459-466; see p. 461.
Bao-Xuan Zhu, Stieltjes moment properties and continued fractions from combinatorial triangles, arXiv:2007.14924 [math.CO], 2020, see p. 27.
Yan Zhuang, Monoid networks and counting permutations by runs, arXiv preprint arXiv:1505.02308 [math.CO], 2015.
Yan Zhuang, Counting permutations by runs, J. Comb. Theory Ser. A 142 (2016), pp. 147-176.

From Emeric Deutsch, Jul 26 2009: (Start)
Sum of entries in row n is n! = A000142(n).
T(n,0) = 2^(n-1) = A000079(n-1).
T(n,1) = A000431(n).
T(n,2) = A000487(n).
T(n,3) = A000517(n).
T(2n, n-1) = T(2n+1, n) = A000182(n+1) (the tangent numbers). (End)
Columns k = 0-6 give: A011782, A000431, A000487, A000517, A179708, A179709, A179710.
Cf. A000111, A000182, A090672, A101280, A185896, A242783, A242784.

# The Maple program yields (by straightforward counting) the generating polynomial of the row n specified in the program.
n := 8: with(combinat): P := permute(n): st := proc (p) local ct, j: ct := 0: for j from 2 to nops(p)-1 do if p[j-1] < p[j] and p[j+1] < p[j] then ct := ct+1 else end if end do: ct end proc: sort(add(t^st(P[j]), j = 1 .. factorial(n))); # Emeric Deutsch, Jul 26 2009
# Second Maple program:
a := 1+sqrt(1-t): b := 1-sqrt(1-t): G := (exp(b*z)-exp(a*z))/(b*exp(a*z)-a*exp(b*z)): Gser := simplify(series(G, z = 0, 15)): for n to 12 do P[n] := sort(expand(factorial(n)*coeff(Gser, z, n))) end do: for n to 12 do seq(coeff(P[n], t, j), j = 0 .. ceil((1/2)*n)-1) end do; # yields sequence in triangular form - Emeric Deutsch, Jul 26 2009
# Third Maple program:
b:= proc(u, o, t) option remember; expand(`if`(u+o=0, 1,
      add(b(u-j, o+j-1, 0)*x^t, j=1..u)+
      add(b(u+j-1, o-j, 1), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0$2)):
seq(T(n), n=1..15);  # Alois P. Heinz, May 22 2014
# Recurrence of D. André (1895).
T := proc(n, k) option remember;
if n < 1 or 2*k > (n-1) then return 0 fi;
if k = 0 then return 2^(n-1) fi;
(2*k + 2)*T(n-1, k) + (n - 2*k)*T(n-1, k-1) end:
seq(seq(T(n, k), k=0..(n-1)/2), n=1..12); # Peter Luschny, Aug 06 2019

From Luc Roy, Jul 08 2010: (Start)
It appears that one-half of the sequence A008303 can be obtained with this Mathematica program:
Expand[CoefficientList[Simplify[InverseSeries[Integrate[
Series[(1 + m Sinh[x]^2)^(-1), {x, 0, 15}, {m, 0, 15}], x]]], x]
Denominator[CoefficientList[Series[Exp[x], {x, 0, 15}], x]]]
(* Mathematica Output of Luc Roy's program *)
{0, 1, 0, 2 m, 0, 8 m + 16 m^2, 0, 32 m + 416 m^2 + 272 m^3, 0, 128 m + 7680 m^2 + 24576 m^3 + 7936 m^4, 0, 512 m + 128512 m^2 + 1304832 m^3 + 1841152 m^4 + 353792 m^5, 0, 2048 m + 2084864 m^2 + 56520704 m^3 + 222398464 m^4 + 175627264 m^5 + 22368256 m^6, 0, 8192 m + 33497088 m^2 + 2230947840 m^3 + 20261765120 m^4 + 41731645440 m^5 + 21016670208 m^6 + 1903757312 m^7}
(End)
(* Another Mathematica program *)
m = 14; a = 1 + Sqrt[1 - t]; b = 1 - Sqrt[1 - t];
g[z_] = (E^(b*z) - E^(a*z))/(b*E^(a*z) - a*E^(b*z));
gser = Series[g[z], {z, 0, m}];
Do[p[n]=n!*Coefficient[gser, z, n]//Simplify, {n, 0, m}];
Flatten[ Table[ Coefficient[p[n], t, j], {n, 0, m}, {j, 0, Ceiling[n/2] - 1}]]
(* Jean-François Alcover, May 27 2011, after Emeric Deutsch *)
(* To get the triangle from Jean-François Alcover's Mathematica program *)
FormTable[Table[Coefficient[p[n], t, j], {n, 0, m}, {j, 0, Ceiling[n/2] - 1}]]
(* Petros Hadjicostas, Aug 06 2019 *)
gf := Sqrt[x - 1] Cot[y Sqrt[x - 1]] - 1; ser := Series[1/gf, {y, 0, 16}];
cy[n_] := n! Coefficient[ser, y, n]; row[n_] := CoefficientList[cy[n], x];
Table[row[n], {n, 1, 12}] // Flatten (* Peter Luschny, Aug 06 2019 *)

{T(n, k) = if(n<1, 0, my(z = sqrt(1 - y + y*O(y^(n\2)))); n!*polcoef(polcoef(z/(z - tanh(x*z)), n, x), k))}; /* Michael Somos, May 24 2023 */

From Emeric Deutsch, Jul 26 2009: (Start)
E.g.f.: G(t,z)=[exp(bz)-exp(az)]/[b*exp(az)-a*exp(bz)], where a+b=2 and ab=t, i.e., a=1+sqrt(1-t), b=1-sqrt(1-t) (see the Goulden-Jackson reference). -
Sum_{k>=0} k*T(n,k) = n!*(n-2)/3 = A090672(n-1).
Row n has ceiling(n/2) terms. (End)
E.g.f.: tan(t*sqrt(x-1))/(sqrt(x-1)-tan(t*sqrt(x-1))) = Sum_{n>=0} P(n,x)*t^n/n! = t + 2*t^2/2! + (4+2*x)*t^3/3! + (8+16*x)*t^4/4! + .... The row generating polynomials P(n,x) satisfy x^(n-1)*P(n,1+1/x^2) = R(n-1,x), where R(n,x) are the row polynomials of A185896. A000670(n) = (3/2)^(n-1)*P(n,8/9). - Peter Bala, Oct 14 2011
From Jinyuan Wang, Dec 28 2020: (Start)
T(n, k) = (n - 2*k + 2)*T(n-1, k-1) + 2*k*T(n-1, k) for n > 1 and k > 1; T(n, 1) = 2^(n - 1); T(1, k) = 0 for k > 1.
T(2*k-1, k) = A000182(k). (End)
From Ammar Khatab, Aug 17 2024: (Start)
T(2*n,k) = 4^(n-k+1)* Sum_{p=0..k} (-1)^p * (2*p+2*n-2*k-1)/(p+2*n-2*k-1) binomial(p+2*n-2*k-1,p) (A008292(2*n,k-p+1)+A008292(2*n,2*n+p-k) ) for n>0.
T(2*n+1,k) = 4^(n-k)* Sum_{p=0..k} (-1)^p * (p+n-k)/(p+2*n-2*k) binomial(p+2*n-2*k,p) (A008292(2*n+1,k-p+1)+A008292(2*n,2*n+p-k+1) ) for k<>n. (End)

Additional comments from Emeric Deutsch, May 08 2004
More terms from R. J. Mathar and Vladeta Jovovic, Jun 26 2007
Corrected by Emeric Deutsch, Jul 26 2009
Edited definition - N. J. A. Sloane, May 25 2023

A002430 Numerators in Taylor series for tan(x). Also from Taylor series for tanh(x).

Original entry on oeis.org

1, 1, 2, 17, 62, 1382, 21844, 929569, 6404582, 443861162, 18888466084, 113927491862, 58870668456604, 8374643517010684, 689005380505609448, 129848163681107301953, 1736640792209901647222, 418781231495293038913922

Offset: 1

N. J. A. Sloane

a(n) appears to be a multiple of A046990(n) (checked up to n=250). - Ralf Stephan, Mar 30 2004

The Taylor series for tan(x) appears to be identical to the sequence of quotients A160469(n)/A156769(n). - Johannes W. Meijer, May 24 2009

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... =
  x + (1/3)*x^3 + (2/15)*x^5 + (17/315)*x^7 + (62/2835)*x^9 + ... =
  Sum_{n >= 1} (2^(2n) - 1) * (2x)^(2n-1) * |bernoulli_2n| / (n*(2n-1)!).
tanh(x) = x - (1/3)*x^3 + (2/15)*x^5 - (17/315)*x^7 + (62/2835)*x^9 - (1382/155925)*x^11 + ...

G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 74.
H. A. Rothe, in C. F. Hindenburg, editor, Sammlung Combinatorisch-Analytischer Abhandlungen, Vol. 2, Chap. XI. Fleischer, Leipzig, 1800, p. 329.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 1..276 (first 100 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).
R. W. van der Waall, On a property of tan x, J. Number Theory 5 (1973) 242-244.
Eric Weisstein's World of Mathematics, Hyperbolic Tangent.
Eric Weisstein's World of Mathematics, Tangent.
Herbert S. Wilf, Generatingfunctionology, Academic Press, NY, 1990. See p. 50.

Cf. A036279 (denominators), A000182, A099612, A160469, A156769.

[Numerator( (-1)^(n-1)*4^n*(4^n-1)*Bernoulli(2*n)/Factorial(2*n) ): n in [1..20]]; // G. C. Greubel, Jul 03 2019

R := n -> (-1)^floor(n/2)*(4^n-2^n)*Zeta(1-n)/(n-1)!:
seq(numer(R(2*n)), n=1..20); # Peter Luschny, Aug 25 2015

a[n_]:= (-1)^Floor[n/2]*(4^n - 2^n)*Zeta[1-n]/(n-1)!; Table[Numerator@ a[2n], {n, 20}] (* Michael De Vlieger, Aug 25 2015 *)

a(n) = numerator( (-1)^(n-1)*4^n*(4^n-1)*bernfrac(2*n)/(2*n)! ); \\ G. C. Greubel, Jul 03 2019

[numerator( (-1)^(n-1)*4^n*(4^n-1)*bernoulli(2*n)/factorial(2*n)  ) for n in (1..20)] # G. C. Greubel, Jul 03 2019

a(n) is the numerator of (-1)^(n-1)*2^(2*n)*(2^(2*n) -1)* Bernoulli(2*n)/(2*n)!. - Johannes W. Meijer, May 24 2009

Let R(x) = (cos(x*Pi/2) + sin(x*Pi/2))*(4^x - 2^x)*Zeta(1-x)/(x-1)!. Then a(n) = numerator(R(2*n)) and A036279(n) = denominator(R(2*n)). - Peter Luschny, Aug 25 2015

More terms from Mark Hudson (mrmarkhudson(AT)hotmail.com), Jan 29 2003

A119879 Exponential Riordan array (sech(x),x).

Original entry on oeis.org

1, 0, 1, -1, 0, 1, 0, -3, 0, 1, 5, 0, -6, 0, 1, 0, 25, 0, -10, 0, 1, -61, 0, 75, 0, -15, 0, 1, 0, -427, 0, 175, 0, -21, 0, 1, 1385, 0, -1708, 0, 350, 0, -28, 0, 1, 0, 12465, 0, -5124, 0, 630, 0, -36, 0, 1, -50521, 0, 62325, 0, -12810, 0, 1050, 0, -45, 0, 1

Offset: 0

Paul Barry, May 26 2006

Row sums have e.g.f. exp(x)*sech(x) (signed version of A009006). Inverse of masked Pascal triangle A119467. Transforms the sequence with e.g.f. g(x) to the sequence with e.g.f. g(x)*sech(x).
Coefficients of the Swiss-Knife polynomials for the computation of Euler, tangent and Bernoulli number (triangle read by rows). Another version in A153641. - Philippe Deléham, Oct 26 2013
Relations to Green functions and raising/creation and lowering/annihilation/destruction operators are presented in Hodges and Sukumar and in Copeland's discussion of this sequence and 2020 pdf. - Tom Copeland, Jul 24 2020

			Triangle begins:
     1;
     0,    1;
    -1,    0,     1;
     0,   -3,     0,   1;
     5,    0,    -6,   0,   1;
     0,   25,     0, -10,   0,   1;
   -61,    0,    75,   0, -15,   0,   1;
     0, -427,     0, 175,   0, -21,   0,  1;
  1385,    0, -1708,   0, 350,   0, -28,  0,  1;

G. C. Greubel, Rows n=0..100 of triangle, flattened
Paul Barry, Riordan Arrays, Orthogonal Polynomials as Moments, and Hankel Transforms, J. Int. Seq. 14 (2011) # 11.2.2, example 28.
Tom Copeland, The Elliptic Lie Triad: Ricatti and KdV Equations, Infinigens, and Elliptic Genera, 2015.
Tom Copeland, Discussion of this sequence
Tom Copeland, SKipping over Dimensions, Juggling Zeros in the Matrix, 2020.
A. Hodges and C. V. Sukumar, Bernoulli, Euler, permutations and quantum algebras, Proc. R. Soc. A Oct. 2007 vol 463 no. 463 2086 2401-2414.
Peter Luschny, Additive decompositions of classical numbers via the Swiss-Knife polynomials [_Tom Copeland_, Oct 20 2015]
Miguel Méndez and Rafael Sánchez, On the combinatorics of Riordan arrays and Sheffer polynomials: monoids, operads and monops, arXiv:1707.00336 [math.CO], 2017, Section 4.3, Example 4.
Miguel A. Méndez and Rafael Sánchez Lamoneda, Monops, Monoids and Operads: The Combinatorics of Sheffer Polynomials, The Electronic Journal of Combinatorics 25(3) (2018), #P3.25.

Row sums are A155585. - Johannes W. Meijer, Apr 20 2011
Rows reversed: A081658.
Cf. A109449, A153641, A162660. - Philippe Deléham, Oct 26 2013
Cf. A000182, A046802, A119467, A133314.
Cf. A019538, A028246, A090582, A123125, A130850, A248727.

T := (n,k) -> binomial(n,k)*2^(n-k)*euler(n-k,1/2): # Peter Luschny, Jan 25 2009

T[n_, k_] := Binomial[n, k] 2^(n-k) EulerE[n-k, 1/2];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 20 2018, after Peter Luschny *)

{T(n,k) = binomial(n,k)*2^(n-k)*(2/(n-k+1))*(subst(bernpol(n-k+1, x), x, 1/2) - 2^(n-k+1)*subst(bernpol(n-k+1, x), x, 1/4))};
for(n=0,5, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 25 2019

@CachedFunction
def A119879_poly(n, x) :
    return 1 if n == 0  else add(A119879_poly(k, 0)*binomial(n, k)*(x^(n-k)-1+n%2) for k in range(n)[::2])
def A119879_row(n) :
    R = PolynomialRing(ZZ, 'x')
    return R(A119879_poly(n,x)).coeffs()  # Peter Luschny, Jul 16 2012
# Alternatively:

# uses[riordan_array from A256893]
riordan_array(sech(x), x, 9, exp=true) # Peter Luschny, Apr 19 2015

Number triangle whose k-th column has e.g.f. sech(x)*x^k/k!.
T(n,k) = C(n,k)*2^(n-k)*E_{n-k}(1/2) where C(n,k) is the binomial coefficient and E_{m}(x) are the Euler polynomials. - Peter Luschny, Jan 25 2009
The coefficients in ascending order of x^i of the polynomials p{0}(x) = 1 and p{n}(x) = Sum_{k=0..n-1; k even} binomial(n,k)*p{k}(0)*((n mod 2) - 1 + x^(n-k)). - Peter Luschny, Jul 16 2012
E.g.f.: exp(x*z)/cosh(x). - Peter Luschny, Aug 01 2012
Sum_{k=0..n} T(n,k)*x^k = A122045(n), A155585(n), A119880(n), A119881(n) for x = 0, 1, 2, 3 respectively. - Philippe Deléham, Oct 27 2013
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of A046802 (the h-polynomials of the stellahedra) are given by h_n(x) = A_n(x;1); the row polynomials of A248727 (the face polynomials of the stellahedra), by f_n(x) = A_n(1 + x;1); the Swiss-knife polynomials of this entry, A119879, by Sw_n(x) = A_n(-1;1 + x); and the row polynomials of the Worpitsky triangle (A130850), by w_n(x) = A(1 + x;0). Other specializations of A_n(x;y) give A090582 (the f-polynomials of the permutohedra, cf. also A019538) and A028246 (another version of the Worpitsky triangle). - Tom Copeland, Jan 24 2020
Triangle equals P*((I + P^2)/2)^(-1), where P denotes Pascal's triangle A007318. - Peter Bala, Mar 07 2024

A162005 The EG1 triangle.

Original entry on oeis.org

1, 2, 1, 16, 28, 1, 272, 1032, 270, 1, 7936, 52736, 36096, 2456, 1, 353792, 3646208, 4766048, 1035088, 22138, 1, 22368256, 330545664, 704357760, 319830400, 27426960, 199284, 1, 1903757312, 38188155904, 120536980224, 93989648000

Offset: 1

Johannes W. Meijer, Jun 27 2009, Jul 02 2009, Aug 31 2009

We define the EG1 matrix by EG1[2m-1,1] = 2*eta(2m-1) and the recurrence relation EG1[2m-1,n] = EG1[2m-1,n-1] - EG1[2m-3,n-1]/(n-1)^2 for m = -2, -1, 0, 1, 2, .. and n = 2, 3, .., with eta(m) = (1-2^(1-m))*zeta(m) with eta(m) the Dirichlet eta function and zeta(m) the Riemann zeta function. For the EG2[2m,n] coefficients see A008955.
The n-th term of the row coefficients EG1[1-2*m,n] for m = 1, 2, .., can be generated with REG1(1-2*m,n) = (-1)^(m+1)*2^(1-m)*ECGP(1-2*m, n)*(1/n)*4^(-n)*(2*n)!/((n-1)!)^2 . For information about the ECGP polynomials see A094665 and the examples below.
We define the o.g.f.s. of the REG1(1-2*m,n) by GFREG1(z,1-2*m) = sum(REG1(1-2*m,n)* z^(n-1), n=1..infinity) for m = 1, 2, .., with GFREG1(z,1-2*m) = (-1)^(m+1)* RG(z,1-2*m)/ (2^(2*m-1)*(1-z)^((2*m+1)/2)). The RG(z,1-2m) polynomials led to the EG1 triangle.
We used the coefficients of the A156919 and A094665 triangles to determine those of the EG1 triangle, see the Maple program. The A156919 triangle gives information about the sums SF(p) = sum(n^(p-1)*4^(-n)*z^(n-1)*(2*n)!/((n-1)!)^2, n=1..infinity) for p= 0, 1, 2, .. .
Contribution from Johannes W. Meijer, Nov 23 2009: (Start)
The EG1 matrix is related to the ED2 array A167560 because sum(EG1(2*m-1,n)*z^(2*m-1), m=1..infinity) = ((2*n-1)!/(4^(n-1)*(n-1)!^2))*int(sinh(y*(2*z))/cosh(y)^(2*n),y=0..infinity).
(End)
Appears to equal triangle A322230 with rows read in reverse order. Triangle A322230 describes the e.g.f. S(x,k) = Integral C(x,k)*D(x,k)^2 dx, such that C(x,k)^2 - S(x,k)^2 = 1, and D(x,k)^2 - k^2*S(x,k)^2 = 1. - Paul D. Hanna, Dec 22 2018
Appears to equal triangle A325220, which has e.g.f. S(x,k) = -i * sn( i * Integral C(x,k) dx, k) such that C(x,k) = cn( i * Integral C(x,k) dx, k), where sn(x,k) and cn(x,k) are Jacobi Elliptic functions. - Paul D. Hanna, Apr 13 2019

			The first few rows of the EG1 triangle are :
[1]
[2, 1]
[16, 28, 1]
[272, 1032, 270, 1]
The first few RG(z,1-2*m) polynomials are:
RG(z,-1) = 1
RG(z,-3) = 2+z
RG(z,-5) = 16+28*z+z^2
RG(z,-7) = 272+1032*z+270*z^2+z^3
The first few GFREG1(z,1-2*m) are:
GFREG1(z,-1) = (1)*(1)/(2*(1-z)^(3/2))
GFREG1(z,-3) = (-1)*(2+z)/(2^3*(1-z)^(5/2))
GFREG1(z,-5) = (1)*(16+28*z+z^2)/( 2^5*(1-z)^(7/2))
GFREG1(z,-7) = (-1)*(272+1032*z+270*z^2+z^3)/(2^7*(1-z)^(9/2))
The first few REG1(1-2*m,n) are:
REG1(-1,n) = (1/1)*(1)*(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-3,n) = (-1/2)*(n) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-5,n) = (1/4) *(n+3*n^2) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
REG1(-7,n) = (-1/8)*(4*n+15*n^2+15*n^3) *(1/n)*4^(-n)*(2*n)!/(n-1)!^2
The first few ECGP(1-2*m,n) polynomials are:
ECGP(-1,n) = 1
ECGP(-3,n) = n
ECGP(-5,n) = n+3*n^2
ECGP(-7,n) = 4*n+15*n^2+15*n^3

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, Chapter 23, pp. 811-812.

A079484 equals the row sums.
A000182 (ZAG numbers), A162006 and A162007 equal the first three left hand columns.
A000012, A004004 (2x), A162008, A162009 and A162010 equal the first five right hand columns.
Related to A094665, A083061 and A156919 (DEF triangle).
Cf. A161198 [(1-x)^((-1-2*n)/2)], A008955 (EG2[2m, n])
Cf. A167560 (ED2 array).
Cf. A322230 (reversed rows), A325220.

nmax:=7; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1) * (x+1)*T1(i-1, x+1)-2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1=0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1)*A156919(n-1, m-1) end do end do: for n from 0 to nmax do SF(n) := sum(A156919(n, k1)*z^k1, k1=0..n)/(2^(n+1)*(1-z)^((2*n+3)/2)) od: GFREG1(z, -1) := A156919(0, 0)*A094665 (0, 0) / (2*(1-z)^(3/2)): for m from 2 to nmax do GFREG1(z, 1-2*m) := simplify((-1)^(m+1)*2^(1-m)* sum(A094665(m-1, k2)*SF(k2), k2=1..m-1)) od: for m from 1 to mmax do g(m) := sort((numer ((-1)^(m+1)* GFREG1(z, 1-2*m))), ascending) od: for n from 1 to nmax do for m from 1 to n do a(n, m) := abs(coeff(g(n), z, m-1)) od: od: seq(seq(a(n, m), m=1..n), n=1..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

A different form of the recurrence relation is EG1[1-2*m,n] = (EG1[3-2*m,n]-EG1[3-2*m,n+1])* (n^2) for m = 2, 3, .., with EG1[ -1,n] = (1/n)*4^(-n)*((2*n)!/(n-1)!^2).

A155100 Triangle read by rows: coefficients in polynomials P_n(u) arising from the expansion of D^(n-1) (tan x) in increasing powers of tan x for n>=1 and 1 for n=0.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 2, 2, 0, 8, 0, 6, 0, 16, 0, 40, 0, 24, 16, 0, 136, 0, 240, 0, 120, 0, 272, 0, 1232, 0, 1680, 0, 720, 272, 0, 3968, 0, 12096, 0, 13440, 0, 5040, 0, 7936, 0, 56320, 0, 129024, 0, 120960, 0, 40320, 7936, 0, 176896, 0, 814080, 0, 1491840

Offset: 0

N. J. A. Sloane, Nov 05 2009

The definition is d^(n-1) tan x / dx^n = P_n(tan x) for n>=1 and 1 for n=0.
Interpolates between factorials and tangent numbers.
From Peter Bala, Mar 02 2011: (Start)
Companion triangles are A104035 and A185896.
A combinatorial interpretation for the polynomial P_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 and put x_0 = -(n+1) and x_(n+1) = (-1)^n*(n+1). Then x_0,x_1,...,x_n,x_(n+1) is a snake of type S(n) when x_0 < x_1 > x_2 < ... x_(n+1). For example, -5 4 -3 -1 -2 5 is a snake of type S(4).
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 -5 4 -3 -1 -2 5 has sc = 3.
The polynomial P_(n+1)(t) is the generating function for the sign change statistic on snakes of type S(n): P_(n+1)(t) = sum {snakes in S(n)} t^sc.
See the example section below for the cases n=1 and n=2.
(End)
Equals A107729 when the first column is removed. - Georg Fischer, Jul 26 2023

			The polynomials P_{-1}(u) through P_6(u) with exponents in decreasing order:
      1
      u
      u^2 +    1
    2*u^3 +    2*u
    6*u^4 +    8*u^2 +    2
   24*u^5 +   40*u^3 +   16*u
  120*u^6 +  240*u^4 +  136*u^2 +  16
  720*u^7 + 1680*u^5 + 1232*u^3 + 272*u
  ...
Triangle begins:
  1
  0, 1
  1, 0, 1
  0, 2, 0, 2
  2, 0, 8, 0, 6
  0, 16, 0, 40, 0, 24
  16, 0, 136, 0, 240, 0, 120
  0, 272, 0, 1232, 0, 1680, 0, 720
  272, 0, 3968, 0, 12096, 0, 13440, 0, 5040
  0, 7936, 0, 56320, 0, 129024, 0, 120960, 0, 40320
  7936, 0, 176896, 0, 814080, 0, 1491840, 0, 1209600, 0, 362880
  0, 353792, 0, 3610112, 0, 12207360, 0, 18627840, 0, 13305600, 0, 3628800
  ...
From _Peter Bala_, Feb 07 2011: (Start)
Examples of sign change statistic sc on snakes of type S(n):
    Snakes     # sign changes sc  t^sc
  ===========  =================  ====
n=1:
  -2  1 -2 ........... 2 ........ t^2
  -2 -1 -2 ........... 0 ........ 1
                  yields P_2(t) = 1 + t^2;
n=2:
  -3  1 -2  3 ........ 3 ........ t^3
  -3  2  1  3 ........ 1 ........ t
  -3  2 -1  3 ........ 3 ........ t^3
  -3 -1 -2  3 ........ 1 ........ t
                  yields P_3(t) = 2*t + 2*t^3. (End)

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics, Addison-Wesley, Reading, MA, 2nd ed. 1998, p. 287.

G. C. Greubel, Rows n = 0..101 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.
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 [math.CO], 2021.

For other versions of this triangle see A008293, A101343.
A104035 is a companion triangle.
Highest order coefficients give factorials A000142. Constant terms give tangent numbers A000182. Other coefficients: A002301.
Setting u=1 in P_n gives A000831, u=2 gives A156073, u=3 gives A156075, u=4 gives A156076, u=1/2 gives A156102.
Setting u=sqrt(2) in P_n gives A156108 and A156122; setting u=sqrt(3) gives A156103 and A000436.
Cf. A008292, A019538, A107729, A185896.

P:=proc(n) option remember;
if n=-1 then RETURN(1); elif n=0 then RETURN(u); else RETURN(expand((u^2+1)*diff(P(n-1),u))); fi;
end;
for n from -1 to 12 do t1:=series(P(n),u,20); lprint(seriestolist(t1)); od:
# Alternatively:
with(PolynomialTools): seq(print(CoefficientList(`if`(i=0,1,D@@(i-1))(tan),tan)), i=0..7); # Peter Luschny, May 19 2015

p[n_, u_] := D[Tan[x], {x, n}] /. Tan[x] -> u /. Sec[x] -> Sqrt[1 + u^2] // Expand; p[-1, u_] = 1; Flatten[ Table[ CoefficientList[ p[n, u], u], {n, -1, 9}]] (* Jean-François Alcover, Jun 28 2012 *)
T[ n_, k_] := Which[n<0, Boole[n==-1 && k==0], n==0, Boole[k==1], True, (k-1)*T[n-1, k-1] + (k+1)*T[n-1, k+1]]; (* Michael Somos, Jul 09 2024 *)

{T(n, k) = if(n<0, n==-1 && k==0, n==0, k==1, (k-1)*T(n-1, k-1) + (k+1)*T(n-1, k+1))}; /* Michael Somos, Jul 09 2024 */

If the polynomials are denoted by P_n(u), we have the recurrence P_{-1}=1, P_0 = u, P_n = (u^2+1)*dP_{n-1}/du.
G.f.: Sum_{n >= 0} P_n(u) t^n/n! = (sin t + u*cos t)/(cos t - u sin t). [Hoffman]
From Peter Bala, Feb 07 2011: (Start)
RELATION WITH BERNOULLI NUMBERS A000367 AND A002445
Put T(n,t) = P_n(i*t), where i = sqrt(-1). We have the definite integral evaluation, valid when both m and n are >=1 and m+n >= 4:
int( T(m,t)*T(n,t)/(1-t^2), t = -1..1) = (-1)^((m-n)/2)*2^(m+n-1)*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.
RELATION WITH OTHER ROW POLYNOMIALS
The following three identities hold for n >= 1:
P_(n+1)(t) = (1+t^2)*R(n-1,t) where R(n,t) is the n-th row polynomial of A185896.
P_(n+1)(t) = (-2*i)^n*(t-i)*R(n,-1/2+1/2*i*t), where i = sqrt(-1) and R(n,x) is an ordered Bell polynomial, that is, the n-th row polynomial of A019538.
P_(n+1)(t) = (t-i)*(t+i)^n*A(n,(t-i)/(t+i)), 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 - see A008292. (End)
T(n,k) = cos((n+k)*Pi/2) * Sum_{p=0..n-1} A008292(n-1,p+1) Sum_{j=0..k}(-1)^(p+j+1) * binomial(p+1,k-j) *binomial(n-p-1,j) for n>1. - Ammar Khatab, Aug 15 2024

Name clarified by Peter Luschny, May 25 2015

A036279 Denominators in the Taylor series for tan(x).

Original entry on oeis.org

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 2900518163668125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875

Offset: 1

N. J. A. Sloane

The n-th coefficient of Taylor series for tan(x) appears to be identical to the quotient A160469(n)/A156769(n). - Johannes W. Meijer, May 24 2009

			tan(x) = x + 2*x^3/3! + 16*x^5/5! + 272*x^7/7! + ... = x + (1/3)*x^3 + (2/15)*x^5 + (17/315)*x^7 + (62/2835)*x^9 + ... =
  Sum_{n >= 1} (2^(2n) - 1) * (2x)^(2n-1) * |Bernoulli_2n| / (n*(2n-1)!).
The coefficients in the expansion of tan x are 0, 1, 0, 1/3, 0, 2/15, 0, 17/315, 0, 62/2835, 0, 1382/155925, 0, 21844/6081075, 0, 929569/638512875, 0, ... = A002430/A036279
tanh(x) = x - (1/3)*x^3 + (2/15)*x^5 - (17/315)*x^7 + (62/2835)*x^9 - (1382/155925)*x^11 + ...
The coefficients in the expansion of tanh x are 0, 1, 0, -1/3, 0, 2/15, 0, -17/315, 0, 62/2835, 0, -1382/155925, 0, 21844/6081075, 0, -929569/638512875, 0, 6404582/10854718875, 0, -443861162/1856156927625, ... = A002430/A036279

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).
G. W. Caunt, Infinitesimal Calculus, Oxford Univ. Press, 1914, p. 477.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 74.
H. A. Rothe, in C. F. Hindenburg, editor, Sammlung Combinatorisch-Analytischer Abhandlungen, Vol. 2, Chap. XI. Fleischer, Leipzig, 1800, p. 329.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 34, equation 34:6:1 at page 323.

Seiichi Manyama, Table of n, a(n) for n = 1..253 (first 100 terms from T. D. Noe)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 75 (4.3.67).
Eric Weisstein's World of Mathematics, Hyperbolic Tangent.
Eric Weisstein's World of Mathematics, Tangent.
Herbert S. Wilf, Generatingfunctionology, Academic Press, NY, 1990. See p. 50.

Cf. A002430, A000182, A099612, A156769, A160469.

R := n -> (-1)^floor(n/2)*(4^n-2^n)*Zeta(1-n)/(n-1)!:
seq(denom(R(2*n)), n=1..18); # Peter Luschny, Aug 25 2015

f[n_] := (-1)^Floor[n/2] (4^n - 2^n) Zeta[1 - n]/(n - 1)!; Table[Denominator@ f[2 n], {n, 17}] (* Michael De Vlieger, Aug 25 2015 *)

a(n) = denominator((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)*Bernoulli(2*n)/(2*n)!). - Johannes W. Meijer, May 24 2009

Let R(x) = (cos(x*Pi/2) + sin(x*Pi/2))*(4^x-2^x)*Zeta(1-x)/(x-1)!. Then a(n) = denominator(R(2*n)) and A002430(n) = numerator(R(2*n)). - Peter Luschny, Aug 25 2015

Incorrect comment by Stephen Crowley deleted by Johannes W. Meijer, Jan 19 2009

cp's OEIS Frontend

A000831 Expansion of e.g.f. (1 + tan(x))/(1 - tan(x)).

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A156919 Table of coefficients of polynomials related to the Dirichlet eta function.

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A060628 Triangle of coefficients in expansion of elliptic function sn(u) in powers of u and k.

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A080635 Number of permutations on n letters without double falls and without initial falls.

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A008303 Triangle read by rows: T(n,k) (n >= 1, 0 <= k <= ceiling(n/2)-1) = number of permutations of [n] with k peaks.

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A002430 Numerators in Taylor series for tan(x). Also from Taylor series for tanh(x).

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