A000364 -id:A000364 - cp's OEIS frontend

A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

1, 0, 1, 1, 0, 1, 0, 4, 0, 1, 5, 0, 10, 0, 1, 0, 40, 0, 20, 0, 1, 61, 0, 175, 0, 35, 0, 1, 0, 768, 0, 560, 0, 56, 0, 1, 1385, 0, 4996, 0, 1470, 0, 84, 0, 1, 0, 24320, 0, 22720, 0, 3360, 0, 120, 0, 1

Offset: 0

Paul Barry, Nov 05 2008

Production array is [cosh(x),x] beheaded. Inverse is A147308. Row sums are A000111(n+1).
Unsigned version of A147308. - N. J. A. Sloane, Nov 07 2008
From Peter Bala, Jan 26 2011: (Start)
Define a polynomial sequence {Z(n,x)} n >= 0 by means of the recursion
(1)... Z(n+1,x) = 1/2*x*{Z(n,x-1)+Z(n,x+1)}
with starting condition Z(0,x) = 1. We call Z(n,x) the zigzag polynomial of degree n. This table lists the coefficients of these polynomials (for n >= 1) in ascending powers of x, row indices shifted by 1. The first few polynomials are
... Z(1,x) = x
... Z(2,x) = x^2
... Z(3,x) = x + x^3
... Z(4,x) = 4*x^2 + x^4
... Z(5,x) = 5*x + 10*x^3 + x^5.
The value Z(n,1) equals the zigzag number A000111(n). The polynomials Z(n,x) occur in formulas for the enumeration of permutations by alternating descents A145876 and in the enumeration of forests of non-plane unary binary labeled trees A147315.
{Z(n,x)}n>=0 is a polynomial sequence of binomial type and so is analogous to the sequence of monomials x^n. Denoting Z(n,x) by x^[n] to emphasize this analogy, we have, for example, the following analog of Bernoulli's formula for the sum of integer powers:
(2)... 1^[m]+...+(n-1)^[m] = (1/(m+1))*Sum_{k=0..m} (-1)^floor(k/2)*binomial(m+1,k)*B_k*n^[m+1-k],
where {B_k} k >= 0 = [1, -1/2, 1/6, 0, -1/30, ...] is the sequence of Bernoulli numbers.
For similarly defined polynomial sequences to Z(n,x) see A185415, A185417 and A185419. See also A185424.
(End)
[gd(x)^(-1)]^m = Sum_{n>=m} Tg(n,m)*(m!/n!)*x^n, where gd(x) is Gudermannian function, Tg(n+1,m+1)=T(n,m). - Vladimir Kruchinin, Dec 18 2011
The Bell transform of abs(E(n)), E(n) the Euler numbers. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
   1;
   0,  1;
   1,  0,   1;
   0,  4,   0,  1;
   5,  0,  10,  0,  1;
   0, 40,   0, 20,  0, 1;
  61,  0, 175,  0, 35, 0, 1;

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000111 (row sums), A000364, A012123, A012259, A145876, A147315, A185415, A185417, A185419, A185421, A185424. - Peter Bala, Jan 26 2011

Z := proc(n, x) option remember;
description 'zigzag polynomials Z(n, x)'
if n = 0 return 1 else return 1/2*x*(Z(n-1, x-1)+Z(n-1, x+1)) end proc:
with(PolynomialTools):
for n from 1 to 10 CoefficientList(Z(n, x), x); end do; # Peter Bala, Jan 26 2011

t[n_, k_] := SeriesCoefficient[ 2^k*ArcTan[(E^x - 1)/(E^x + 1)]^k*n!/k!, {x, 0, n}]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten // Abs (* Jean-François Alcover, Jan 23 2015 *)

T(n, k)=local(X); if(k<1 || k>n, 0, X=x+x*O(x^n); n!*polcoeff(polcoeff((tan(X)+1/cos(X))^y, n), k)) \\ Paul D. Hanna, Feb 06 2011

R = PolynomialRing(QQ, 'x')
@CachedFunction
def zzp(n, x) :
    return 1 if n == 0 else x*(zzp(n-1, x-1)+zzp(n-1, x+1))/2
def A147309_row(n) :
    x = R.gen()
    L = list(R(zzp(n, x)))
    del L[0]
    return L
for n in (1..10) : print(A147309_row(n)) # Peter Luschny, Jul 22 2012

# uses[bell_matrix from A264428]
# Alternative: Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: abs(euler_number(n)), 10) # Peter Luschny, Jan 18 2016

From Peter Bala, Jan 26 2011: (Start)
GENERATING FUNCTION
The e.g.f., upon including a constant term of '1', is given by:
(1) F(x,t) = (tan(t) + sec(t))^x = Sum_{n>=0} Z(n,x)*t^n/n! = 1 + x*t + x^2*t^2/2! + (x+x^3)*t^3/3! + ....
Other forms include
(2) F(x,t) = exp(x*arcsinh(tan(t))) = exp(2*x*arctanh(tan(t/2))).
(3) F(x,t) = exp(x*(t + t^3/3! + 5*t^5/5! + 61*t^7/7! + ...)),
where the coefficients [1,1,5,61,...] are the secant or zig numbers A000364.
ROW GENERATING POLYNOMIALS
One easily checks from (1) that
d/dt(F(x,t)) = 1/2*x*(F(x-1,t) + F(x+1,t))
and so the row generating polynomials Z(n,x) satisfy the recurrence relation
(4) Z(n+1,x) = 1/2*x*{Z(n,x-1) + Z(n,x+1)}.
The e.g.f. for the odd-indexed row polynomials is
(5) sinh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n+1,x)*t^(2n+1)/(2n+1)!.
The e.g.f. for the even-indexed row polynomials is
(6) cosh(x*arcsinh(tan(t))) = Sum_{n>=0} Z(2n,x)*t^(2n)/(2n)!.
From sinh(2*x) = 2*sinh(x)*cosh(x) we obtain the identity
(7) Z(2n+1,2*x) = 2*Sum_{k=0..n} binomial(2n+1,2k)*Z(2k,x)*Z(2n-2k+1,x).
The zeros of Z(n,x) lie on the imaginary axis (use (4) and adapt the proof given in A185417 for the zeros of the polynomial S(n,x)).
BINOMIAL EXPANSION
The form of the e.g.f. shows that {Z(n,x)} n >= 0 is a sequence of polynomials of binomial type. In particular, we have the expansion
(8) Z(n,x+y) = Sum_{k=0..n} binomial(n,k)*Z(k,x)*Z(n-k,y).
The delta operator D* associated with this binomial type sequence is
(9) D* = D - D^3/3! + 5*D^5/5! - 61*D^7/7! + 1385*D^9/9! - ..., and satisfies
the relation
(10) tan(D*)+sec(D*) = exp(D).
The delta operator D* acts as a lowering operator on the zigzag polynomials:
(11) (D*)Z(n,x) = n*Z(n-1,x).
ANALOG OF THE LITTLE FERMAT THEOREM
For integer x and odd prime p
(12) Z(p,x) = (-1)^((p-1)/2)*x (mod p).
More generally, for k = 1,2,3,...
(13) Z(p+k-1,x) = (-1)^((p-1)/2)*Z(k,x) (mod p).
RELATIONS WITH OTHER SEQUENCES
Row sums [1,1,2,5,16,61,...] are the zigzag numbers A000111(n) for n >= 1.
Column 1 (with 0's omitted) is the sequence of Euler numbers A000364.
A145876(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(n+1,k-j)*Z(n,j).
A147315(n-1,k-1) = (1/k!)*Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A185421(n,k) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j).
A012123(n) = (-i)^n*Z(n,i) where i = sqrt(-1). A012259(n) = 2^n*Z(n,1/2).
(End)
T(n,m) = Sum(i=0..n-m, s(i)/(n-i)!*Sum(k=m..n-i, A147315(n-i,k)*Stirling1(k,m))), m>0, T(n,0) = s(n), s(n)=[1,0,1,0,5,0,61,0,1385,0,50521,...] (see A000364). - Vladimir Kruchinin, Mar 10 2011

A147315 L-matrix for Euler numbers A000111(n+1).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 11, 6, 1, 16, 45, 35, 10, 1, 61, 211, 210, 85, 15, 1, 272, 1113, 1351, 700, 175, 21, 1, 1385, 6551, 9366, 5901, 1890, 322, 28, 1, 7936, 42585, 70055, 51870, 20181, 4410, 546, 36, 1, 50521, 303271, 563970, 479345, 218925, 58107, 9240, 870, 45, 1

Offset: 0

Paul Barry, Nov 05 2008

This is the inverse of the coefficient array for the orthogonal polynomials p(n,x) defined by: p(n,x)=if(n=-1,0,if(n=0,1,(x-n)p(n-1,x)-C(n,2)p(n-2,x))).
The Hankel array H for A000111(n+1) satisfies H=L*D*U with U the transpose of L.
Row sums are A000772(n+1) with e.g.f. dif(exp(-1)exp(sec(x)+tan(x)),x).
From Peter Bala, Jan 31 2011: (Start)
The following comments refer to the table with an offset of 1: i.e., both the row and column indexing starts at 1.
An increasing tree is a labeled rooted tree with the property that the sequence of labels along any path starting from the root is increasing. A000111(n) for n>=1 enumerates the number of increasing unordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2 (plane unary-binary trees in the notation of [Bergeron et al.])
The entry T(n,k) of the present table gives the number of forests of k increasing unordered trees on the vertex set {1,2,...,n} in which all outdegrees are <=2. See below for some examples.
For ordered forests of such trees see A185421. For forests of increasing ordered trees on the vertex set {1,2,...,n}, rooted at 1, in which all outdegrees are <=2, see A185422.
The Stirling number of the second kind Stirling2(n,k) is the number of partitions of the set [n] into k blocks. Arranging the elements in each block in ascending numerical order provides an alternative combinatorial interpretation for Stirling2(n,k) as counting forests of k increasing unary trees on n nodes. Thus we may view the present array, which counts increasing unary-binary trees, as generalized Stirling numbers of the second kind associated with A000111 or with the zigzag polynomials Z(n,x) of A147309 - see especially formulas (2) and (3) below.
See A145876 for generalized Eulerian numbers associated with A000111. (End)
The Bell transform of A000111(n+1). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 18 2016

			Triangle begins
    1;
    1,    1;
    2,    3,    1;
    5,   11,    6,   1;
   16,   45,   35,  10,   1;
   61,  211,  210,  85,  15,  1;
  272, 1113, 1351, 700, 175, 21, 1;
  ...
The production array for L is the tridiagonal array
  1,  1;
  1,  2,  1;
  0,  3,  3,  1;
  0,  0,  6,  4,  1;
  0,  0,  0, 10,  5,  1;
  0,  0,  0,  0, 15,  6,  1;
  0,  0,  0,  0,  0, 21,  7,  1;
  0,  0,  0,  0,  0,  0, 28,  8,  1,;
  0,  0,  0,  0,  0,  0,  0, 36,  9,  1;
From _Peter Bala_, Jan 31 2011: (Start)
Examples of forests:
The diagrams below are drawn so that the leftmost child of a binary node has the maximum label.
T(4,1) = 5. The 5 forests consisting of a single non-plane increasing unary-binary tree on 4 nodes are
...4... ........ .......... ........... ...........
...|... ........ .......... ........... ...........
...3... .4...3.. .4........ ........4.. ........3..
...|... ..\./... ..\....... ......./... ......./...
...2... ...2.... ...3...2.. ..3...2.... ..4...2....
...|... ...|.... ....\./... ...\./..... ...\./.....
...1... ...1.... .....1.... ....1...... ....1......
T(4,2) = 11. The 11 forests consisting of two non-plane increasing unary-binary trees on 4 nodes are
......... ...3.....
.3...2... ...|.....
..\./.... ...2.....
...1...4. ...|.....
......... ...1...4.
.
......... ...4.....
.4...2... ...|.....
..\./.... ...2.....
...1...3. ...|.....
......... ...1...3.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...1...2. ...|.....
......... ...1...2.
.
......... ...4.....
.4...3... ...|.....
..\./.... ...3.....
...2...1. ...|.....
......... ...2...1.
.
......... ......... ..........
..2..4... ..3..4... ..4...3...
..|..|... ..|..|... ..|...|...
..1..3... ..1..2... ..1...2...
......... ......... .......... (End)

Alois P. Heinz, Rows n = 0..140, flattened
Paul Barry, A Note on Three Families of Orthogonal Polynomials defined by Circular Functions, and Their Moment Sequences, Journal of Integer Sequences, Vol. 15 (2012), #12.7.2.
F. Bergeron, Ph. Flajolet and B. Salvy, Varieties of increasing trees, Lecture Notes in Computer Science vol. 581, ed. J.-C. Raoult, Springer 1992, pp. 24-48.
Tom Copeland, Mathemagical Forests
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.

Cf. A145876, A147309, A185421, A185422, A185424.

A147315 := proc(n,k) n!*exp(x*(sec(t)+tan(t)-1)) - 1: coeftayl(%,t=0,n) ; coeftayl(%,x=0,k) ; end proc:
seq(seq(A147315(n,k),k=1..n),n=0..12) ; # R. J. Mathar, Mar 04 2011
# second Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
g:= proc(n) option remember; expand(`if`(n=0, 1, add(
      g(n-j)*x*binomial(n-1, j-1)*b(j, 0), j=1..n)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n+1))(g(n+1)):
seq(T(n), n=0..10);  # Alois P. Heinz, May 19 2021

t[n_, k_] := t[n, k] = t[n-1, k-1] + (k+1)*t[n-1, k] + 1/2*(k+1)*(k+2)*t[n-1, k+1]; t[n_, k_] /; (n < 0 || k < 0 || k > n) = 0; t[0, 0] = t[1, 0] = 1; Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 21 2011, after PARI prog. *)

Co(n,k):=sum(binomial(k,j)*(if oddp(n-k+j) then 0 else if (n-k+j)/2A147315(n,m):=1/m!*sum((if oddp(n-k) then 0 else 2^(1-k)*sum((-1)^(floor((n+k)/2)-i)*binomial(k,i)*(2*i-k)^n,i,0,floor(k/2)))*(sum(Co(i,m)*binomial(k-i+m-1,m-1),i,1,k)),k,m,n); /* Vladimir Kruchinin, Feb 17 2011 */

T(n,m):=(sum(binomial(k+m,m)*((-1)^(n-k-m)+1)*sum(binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)*stirling2(n+1,j+k+m+1), j,0,n-k-m), k,0,n-m))/(m+1)!; /* Vladimir Kruchinin, May 17 2011 */

{T(n,k)=if(k<0||k>n,0,if(n==0,1,T(n-1,k-1)+(k+1)*T(n-1,k)+(k+1)*(k+2)/2*T(n-1,k+1)))} /* offset=0 */

{T(n,k)=local(X=x+x*O(x^(n+2)));(n+1)!*polcoeff(polcoeff(exp(y*((1+sin(X))/cos(X)-1))-1,n+1,x),k+1,y)} /* offset=0 */

/* Generate from the production matrix P: */
{T(n,k)=local(P=matrix(n,n,r,c,if(r==c-1,1,if(r==c,c,if(r==c+1,c*(c+1)/2)))));if(k<0||k>n,0,if(n==k,1,(P^n)[1,k+1]))}

# uses[bell_matrix from A264428, A000111]
# Adds a column 1,0,0,0, ... at the left side of the triangle.
bell_matrix(lambda n: A000111(n+1), 10) # Peter Luschny, Jan 18 2016

From Peter Bala, Jan 31 2011: (Start)
The following formulas refer to the table with an offset of 1: i.e., both the row n and column k indexing start at 1.
GENERATING FUNCTION
E.g.f.:
(1)... exp(x*(sec(t)+tan(t)-1)) - 1 = Sum_{n>=1} R(n,x)*t^n/n!
= x*t + (x+x^2)*t^2/2! + (2*x+3*x^2+x^3)*t^3/3! + ....
TABLE ENTRIES
(2)... T(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*Z(n,j),
where Z(n,x) denotes the zigzag polynomials as described in A147309.
Compare (2) with the formula for the Stirling numbers of the second kind
(3)... Stirling2(n,k) = (1/k!)*Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^n.
Recurrence relation
(4)... T(n+1,k) = T(n,k-1) + k*T(n,k) + (1/2)*k(k+1)*T(n,k+1).
ROW POLYNOMIALS
The row polynomials R(n,x) begin
R(1,x) = x
R(2,x) = x+x^2
R(3,x) = 2*x+3*x^2+x^3
They satisfy the recurrence
(5)... R(n+1,x) = x*{R(n,x)+R'(n,x) + (1/2)*R''(n,x)},
where ' indicates differentiation with respect to x. This should be compared with the recurrence satisfied by the Bell polynomials Bell(n,x)
(6)... Bell(n+1,x) = x*(Bell(n,x) + Bell'(n,x)). (End)
From Vladimir Kruchinin, Feb 17 2011: (Start)
Sum_{m=1..n} T(n,m) = A000772(n).
Sum_{m=1..2n-1} T(2n-1,m)* Stirling1(m,1) = A000364(n).
Let Co(n,k) = Sum_{j=1..k} binomial(k,j)*(if (n-k+j) is odd then 0 else if (n-k+j)/2T(n,m) = m!* Sum_{k=m..n} (if n-k is odd then 0 else 2^(1-k)) * Sum_{i=0..floor(k/2)} (-1)^(floor((n+k)/2)-i) * binomial(k,i) * (2*i-k)^n)))) * Sum_{i=1..k} Co(i,m) * binomial(k-i+m-1,m-1), n>0.
(End)
T(n,m) = Sum_{k = 0..n-m} binomial(k+m,m)*((-1)^(n-k-m)+1)*Sum_{j=0..n-k-m} binomial(j+k+m,k+m)*(j+k+m+1)!*2^(-j-k-1)*(-1)^((n+k+m)/2+j+k+m)* Stirling2(n+1,j+k+m+1)/(m+1)!. - Vladimir Kruchinin, May 17 2011
The row polynomials R(n,x) are given by D^n(exp(x*t)) evaluated at t = 0, where D is the operator (1+t+t^2/2!)*d/dt. Cf. A008277 and A094198. See also A185422. - Peter Bala, Nov 25 2011

More terms from Michel Marcus, Mar 01 2014

A003701 Expansion of e.g.f. exp(x)/cos(x).

Original entry on oeis.org

1, 1, 2, 4, 12, 36, 152, 624, 3472, 18256, 126752, 814144, 6781632, 51475776, 500231552, 4381112064, 48656756992, 482962852096, 6034272215552, 66942218896384, 929327412759552, 11394877025289216, 174008703107274752, 2336793875186479104, 38928735228629389312

Offset: 0

R. H. Hardin

nonn

Binomial transform of A000364 (with interpolated zeros). Hankel transform is A055209. - Paul Barry, Jan 12 2009

			G.f. = 1 + x + 2*x^2 + 4*x^3 + 12*x^4 + 36*x^5 + 152*x^6 + 624*x^7 + 3472*x^8 + ...

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alois P. Heinz, Table of n, a(n) for n = 0..485 (first 101 terms from T. D. Noe)
T. Chow, Fair permutations and random k-sets, Problem 11523, Amer. Math. Monthly 117 (October 2010), 741; solution by Jim Simons, Amer. Math. Monthly 119 (November 2012), 801-803.
J. W. Layman, The Hankel Transform and Some of its Properties, J. Integer Sequences, 4 (2001), #01.1.5.

Cf. A009739, A062272, A062161.

Bisections are A000795 and A002084.

m:=50; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(Exp(x)/Cos(x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Oct 14 2018

G(x):= exp(x)*sec(x): f[0]:=G(x): for n from 1 to 54 do f[n]:= diff(f[n-1],x) od: x:=0: seq(f[n], n=0..22); # Zerinvary Lajos, Apr 05 2009
# second Maple program:
b:= proc(u, o) option remember;
      `if`(u+o=0, 1, add(b(o-1+j, u-j), j=1..u))
    end:
a:= n-> add(`if`(j::odd, 0, b(j, 0)*binomial(n, j)), j=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 12 2024

a[ n_] := If[ n < 0, 0, n! SeriesCoefficient[ Exp[ x ] / Cos[x], {x, 0, n}]] (* Michael Somos, Jun 06 2012 *)

x='x+O('x^66); Vec(serlaplace(exp(x)/cos(x))) \\ Joerg Arndt, May 07 2013

G.f.: 1/(1-x-x^2/(1-x-4x^2/(1-x-9x^2/(1-x-16x^2.... (continued fraction). - Paul Barry, Jan 12 2009
E.g.f.: exp(x)*sec(x). - Zerinvary Lajos, Apr 05 2009
E.g.f.: 1+x/H(0); H(k)=4k+1-x+x^2*(4k+1)/((2k+1)*(4k+3)-x^2+x*(2k+1)*(4k+3)/(2k+2-x+x*(2k+2)/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Nov 15 2011
G.f.: 1/G(0) where G(k)= 1 - 2*x*(k+1)/(1 + 1/(1 + 2*x*(k+1)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Nov 20 2012
G.f.: -1/x/Q(0), where Q(k)= 1 - 1/x - (k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Apr 26 2013
G.f.: (1-x)/Q(0), where Q(k)= (1-x)^2 - (1-x)^2*x^2*(k+1)^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 04 2013
a(n) ~ n! * ((-1)^n*exp(-Pi/2) + exp(Pi/2)) *(2/Pi)^(n+1). - Vaclav Kotesovec, Oct 08 2013
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)/( x*(2*k+1) - 1/(1 + x*(2*k+1)/( x*(2*k+1) + 1/(1 - x*(2*k+2)/( x*(2*k+2) - 1/(1 + x*(2*k+2)/( x*(2*k+2) + 1/Q(k+1) ))))))); (continued fraction). - Sergei N. Gladkovskii, Oct 22 2013
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)^2/( x^2*(k+1)^2 - (1-x)^2/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 21 2013

Extended and reformatted 03/97.

A160485 Triangle of the RBS1 polynomial coefficients.

Original entry on oeis.org

1, 1, -2, 1, -8, 12, 1, -2, 60, -120, 1, -128, -168, 0, 1680, 1, 2638, 7320, -5040, -25200, -30240, 1, -98408, -300828, 52800, 1053360, 1330560, 665280, 1, 5307118, 17914260, 2522520, -56456400, -90810720, -60540480, -17297280

Offset: 1

Johannes W. Meijer, May 24 2009, Jul 06 2009, Sep 19 2012

In A160480 we defined the BS1 matrix by BS1[2*m-1,n=1] = 2*beta(2*m) and the recurrence relation BS1 [2*m-1,n] = (2*n-3)/(2*n-2)*(BS1[2*m-1,n-1]- BS1[2*m-3,n-1]/(2*n-3)^2), for positive and negative values of m and n= 1, 2, .. . As usual beta(m) = sum((-1)^k/(1+2*k)^m, k=0..infinity). It is well-known that BS1[1-2*m,n=1] = euler(2*m-2) for m = 1, 2, .., with euler(2*m-2) the Euler numbers A000364. These values together with the recurrence relation lead to BS1[ -1,n] = 1 for n = 1, 2, .. .
We discovered that the n-th term of the row coefficients BS1[1-2*m,n] for m = 1, 2, .., can be generated with the rather simple polynomials RBS1(1-2*m,n). Our discovery was enabled by the recurrence relation for the RBS1(1-2*m,n) polynomials which we derived from the recurrence relation for the BS[2*m-1,n] coefficients and the fact that RBS1(-1,n) = 1.
The RBS1 polynomials and the polynomials defined by sequence A083061 are related by a shift of +-1/2 and scaling by a power of 2 (see arXiv link). - Richard P. Brent, Jul 15 2014

			The first few rows of the triangle are:
[1]
[1, -2]
[1, -8, 12]
[1, -2, 60, -120]
[1, -128, -168, 0, 1680]
The first few RBS1(1-2*m,n) polynomials are:
RBS1(-1,n) = 1
RBS1(-3,n) = 1 - 2*n
RBS1(-5,n) = 1 - 8*n + 12*n^2
RBS1(-7,n) = 1 - 2*n + 60*n^2 - 120*n^3
From _Peter Bala_, Jan 22 2019: (Start)
Qbar(r,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n+k+1,k)*(2*k + 1)^(2*r):
Case r = 2: Qbar(2,n) = binomial(2*n+2,n+1)/2^(2*n+1) * ( 1 + 3^4*n/(n+2) + 5^4*n*(n-1)/((n+2)*(n+3)) + 7^4*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 12*n^2 + 8*n + 1, valid for n a nonnegative integer (when the series terminates). The identity is also valid for complex n with real part greater than 1 (provided the factor binomial(2*n,n) is replaced with the appropriate expression involving the gamma function).
Case r = 3: Qbar(3,n) = binomial(2*n+2,n+1)/(2^(2*n+1)) * ( 1 + 3^6*n/(n+2) + 5^6*n*(n-1)/((n+2)*(n+3)) + 7^6*n*(n-1)*(n-2)/((n+2)*(n+3)*(n+4)) + ... ) = 120*n^3 + 60*n^2 + 2*n + 1, valid for n a nonnegative integer. The identity is also valid for complex n with real part greater than 2.
Note, the case r = 0 is equivalent to the identity 1 = binomial(2*n,n)/2^(2*n-1) * ( 1 + (n-1)/(n+1) + (n-1)*(n-2)/((n+1)*(n+2)) + (n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)) + ... ), which is valid for complex n with real part greater than 0. This identity was found by Ramanujan. See Example 6, Chapter 10 in Berndt. (End)

B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, Chapter 10, p. 21.

P. Bala, A245244 and A160485 and some hypergeometric series evaluations of Ramanujan
R. P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.

A160480 is the Beta triangle.
A009389(2*n) equals the second left hand column divided by 2.
A001813 equals the first right hand column.
The absolute values of the row sums equal the Euler numbers A000364.
Cf. A083061, A036970, A245244.

nmax := 8; mmax := nmax: A(1, 1) := 1: RBS1(n, 2) := (2*n-1)^2*1-(2*n)*(2*n-1)*1: for m from 3 to mmax do for k from 0 to m-1 do A(m-1, k+1) := coeff(RBS1(n, m-1), n, k) od; RBS1(n+1, m-1) := 0: for k from 0 to m-1 do RBS1(n+1, m-1) := RBS1(n+1, m-1) + A(m-1, k+1)*(n+1)^k od: RBS1(n, m) := (2*n-1)^2*RBS1(n, m-1)-(2*n)*(2*n-1) * RBS1(n+1, m-1) od: for k from 0 to nmax-1 do A(nmax, k+1) := coeff(RBS1(n, nmax), n, k) od: seq(seq(A(n, m), m=1..n), n=1..nmax);

RBS1(1-2*m,n) = (2*n-1)^2*RBS1(3-2*m,n)-(2*n)*(2*n-1)*RBS1(3-2*m,n+1) for m = 2, 3, .., with RBS1(-1,n) =1 for n = 1, 2, .. .
From Peter Bala, Jan 22 2019: (Start)
The row polynomials RBS1 of the triangle are related to the polynomials Qbar(r,n), r = 0,1,2,..., introduced by Brent by Qbar(r,n) = RBS1(-2*r-1,-n).
Recurrence: Qbar(r+1,n) = (2*n + 1)^2*Qbar(r,n) - 2*n(2*n + 1)*Qbar(r,n-1) with Qbar(0,n) = 1 (Brent, equation 19, p.7).
Qbar(r,n) = binomial(2*n + 2,n + 1)/(2^(2*n + 1)) * Sum_{k = 0..n} binomial(n,k)/binomial(n + k + 1,k)*(2*k + 1)^(2*r); this follows easily from the above recurrence. Two examples are given below.
Qbar(r,n) = 1/4^n * Sum_{k = 0..n} binomial(2*n + 1,n - k)*(2*k + 1)^(2*r) For related polynomial sequences see A036970, A083061 and A245244. (End)

A005799 Generalized Euler numbers of type 2^n.

Original entry on oeis.org

1, 1, 2, 10, 104, 1816, 47312, 1714000, 82285184, 5052370816, 386051862272, 35917232669440, 3996998043812864, 524203898507631616, 80011968856686405632, 14061403972845412526080, 2818858067801804443910144

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also, a(n) equals the number of alternating permutations (p(1),...,p(2n)) of the multiset {1,1,2,2,...,n,n} satisfying p(1) <= p(2) > p(3) <= p(4) > p(5) <= ... <= p(2n). Hence, A275801(n) <= a(n) <= A275829(n). - Max Alekseyev, Aug 10 2016
This is the BinomialMean transform of A000364 (see A075271 for definition of transform). - John W. Layman, Dec 04 2002
This sequence appears to be middle column in Poupard's triangle A008301.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Peter Bala, Some S-fractions related to the expansions of sin(ax)/cos(bx) and cos(ax)/cos(bx)
Shishuo Fu, Zhicong Lin, and Zhi-Wei Sun, Proof of several conjectures relating permanents to Combinatorial sequences, arXiv:2109.11506v3 [math.CO], 2021-2023.
Shishuo Fu, Zhicong Lin, and Zhi-Wei Sun, Permanent identities, combinatorial sequences, and permutation statistics, Advances in Applied Mathematics, Volume 163, Part A, 102789 (2025).
Ira M. Gessel, Symmetric functions and P-recursiveness, J. Combin. Theory Ser. A 53 (1990), no. 2, 257-285.
H. Prodinger, On Touchard's continued fraction and extensions: combinatorics-free, self-contained proofs , arXiv:1102.5186 [math.CO], 2011.
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

Right edge of triangle A210108.

Cf. A000364, A008301, A275801, A275829, A154283, A000657.

T := proc(n, k) option remember;
if n < 0 or k < 0 then 0
elif n = 0 then euler(k, 1)
else T(n-1, k+1) - T(n-1, k) fi end:
a := n -> (-2)^n*T(n, n); seq(a(n), n=0..16); # Peter Luschny, Aug 23 2017

a[n_] := Sum[Binomial[n, i]Abs[EulerE[2i]], {i, 0, n}]/2^n

a(n) = (1/2^n) * Sum_{i=0..n} binomial(n, i) * A000364(i).
From Sergei N. Gladkovskii, Dec 27 2012, Oct 11 2013, Oct 27 2013, Jan 08 2014: (Start) Continued fractions:
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)*(2*k+1)/(1 - x*(k+1)*(2*k+1)/G(k+1)).
G.f.: Q(0)/(1-x), where Q(k) = 1 - x^2*(k+1)^2*(2*k+1)^2/(x^2*(k+1)^2*(2*k+1)^2 - (4*x*k^2 + 2*x*k + x - 1)*( 4*x*k^2 + 10*x*k + 7*x - 1)/Q(k+1)).
G.f.: R(0), where R(k) = 1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/(1 - x*(2*k+1)*(k+1)/(x*(2*k+1)*(k+1) - 1/R(k+1)))).
G.f.: 2/(x*Q(0)), where Q(k) = 2/x - 1 - (2*k+1)^2/(1 - (2*k+2)^2/Q(k+1)). (End)
a(n) ~ 2^(3*n+3) * n^(2*n+1/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, May 30 2015
a(n) = 2^n * Sum_{k=0..n} (-1)^k*binomial(n, k)*euler(n+k, 1). - Peter Luschny, Aug 23 2017
From Peter Bala, Dec 21 2019: (Start)
O.g.f. as a continued fraction: 1/(1 - x/(1 - x/(1 - 6*x/(1 - 6*x/(1 - 15*x/(1 - 15*x/(1 - ... - n*(2*n-1)*x/(1 - n*(2*n-1)*x/(1 - ...))))))))) - apply Bala, Proposition 3, with a = 0, b = 1 and replace x with x/2.
Conjectures:
E.g.f. as a continued fraction: 2/(2 - (1-exp(-4*t))/(2 - (1-exp(-8*t))/(2 - (1-exp(-12*t))/(2 - ... )))) = 1 + t + 2*t^2/2! + 10*t^3/3! + 104*t^4/4! + ....
Cf. A000657. [added April 18 2024: for a proof of this conjecture see Fu et al., Section 4.3.]
a(n) = (-2)^(n+1)*Sum_{k = 0..floor((n-1)/2)} binomial(n,2*k+1)*(2^(2*n-2*k) - 1)*Bernoulli(2*n-2*k)/(2*n-2*k) for n >= 1. (End)

Edited by Dean Hickerson, Dec 10 2002

A060627 1 + Sum_{n >= 1} Sum_{k = 0..n-1} (-1)^nT(n,k)y^(2k)x^(2n)/(2n)! = JacobiCN(x,y).

Original entry on oeis.org

1, 1, 4, 1, 44, 16, 1, 408, 912, 64, 1, 3688, 30768, 15808, 256, 1, 33212, 870640, 1538560, 259328, 1024, 1, 298932, 22945056, 106923008, 65008896, 4180992, 4096, 1, 2690416, 586629984, 6337665152, 9860488448, 2536974336, 67047424, 16384

Offset: 1

Vladeta Jovovic, Apr 13 2001

Essentially same triangle as triangle given by [1, 0, 9, 0, 25, 0, 49, 0, 81, 0, 121, ...] DELTA [0, 4, 0, 16, 0, 36, 0, 64, 0, 100, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jun 13 2004

For the recurrence of the row polynomials b_n(y^2) for cn(x|y^2) = Sum_{n >=0} b_n(y^2)*x^(2*n)/(2*n)! see the Fricke reference, where y=k. - Wolfdieter Lang, Jul 05 2016

			The first rows of triangle T(n, k), n >= 1, k = 0..n-1, are:
[1], [1, 4], [1, 44, 16], [1, 408, 912, 64], [1, 3688, 30768, 15808, 256], [1, 33212, 870640, 1538560, 259328, 1024], [1, 298932, 22945056, 106923008, 65008896, 4180992, 4096], [1, 2690416, 586629984, 6337665152, 9860488448, 2536974336, 67047424, 16384], ...

CRC Standard Mathematical Tables and Formulae, 30th ed. 1996, p. 526.
I. P. Goulden and D. M. Jackson, Combinatorial Enumeration, Wiley, N.Y., 1983,(5.2.20).
H. S. Wall, Analytic Theory of Continued Fractions, Chelsea 1973, p. 374.

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972, p. 575, 16.22.2.
P. Bala, A triangle for calculating A060627
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/0501052 [math.CA], 2005.
R. Fricke, Die elliptischen Funktionen und ihre Anwendungen, Erster Teil, p. 399 with p. 397, Springer, Berlin, Heidelberg, 2012.
Eric W. Weisstein, Jacobi Elliptic Functions

Row sums: A000364.

Cf. A002754, A060628, A145271, A181612, A181613.

A060627 := proc(n,m) JacobiCN(z,k) ; coeftayl(%,z=0,2*n) ; (-1)^n*coeftayl(%,k=0,2*m)*(2*n)! ; end proc: # R. J. Mathar, Jan 30 2011

nmax = 8; se = Series[JacobiCN[x, y], {x, 0, 2*nmax} ]; t[n_, m_] := (-1)^n*Coefficient[se, x, 2*n] *(2*n)! // Coefficient[#, y, m]&; Table[t[n, m], {n, 1, nmax}, {m, 0, n-1}] // Flatten (* Jean-François Alcover, Mar 26 2013 *)

JacobiCN(x, y) = 1 - 1/2*x^2 + (1/24 + 1/6*y^2)*x^4 + ( - 1/720 - 11/180*y^2 - 1/45*y^4)*x^6 + (1/40320 + 17/1680*y^2 + 19/840*y^4 + 1/630*y^6)*x^8 + ( - 1/3628800 - 247/56700*y^6 - 461/453600*y^2 - 641/75600*y^4 - 1/14175*y^8)*x^10 + O(x^12).
From Peter Bala, Aug 23 2011: (Start)
The Taylor expansion of the Jacobian elliptic function cn(x,k) begins
  cn(x,k) = 1 - x^2/2! + (1+4*k^2)*x^4/4! - (1+44*k^2+16*k^4)*x^6/6! + ....
The coefficient polynomials in this expansion can be calculated using nested derivatives as follows (see [Dominici, Theorem 4.1 and Example 4.5]):
Let f(x) = sqrt(k^2-sin^2(x)). 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,k) of x^(2*n)/(2*n)! in the expansion of cn(x,k) is given by R(2*n,k) = D^(2*n)[f](0).
See A145271 for the coefficients in the expansion of D^n[f](x) in powers of f(x). See A181613 for the expansion of the reciprocal function 1/cn(x,k).
(End)
G.f. 1/(1 - x/(1 - (2*k)^2*x/(1 - 3^2*x/(1 - (4*k)^2*x/(1 - 5^2*x/(1 - ...)))))) = 1 + x + (1 + 4*k^2)*x^2 + (1 + 44*k^2 + 16*k^4)*x^3 + ... (see Wall, 94.19, p. 374). - Peter Bala, Apr 25 2017

A060063 Triangle of coefficients of certain polynomials used for G.f.s of columns of triangle A060058.

Original entry on oeis.org

1, 1, 1, 5, 26, 9, 61, 775, 1179, 225, 1385, 32516, 114318, 87156, 11025, 50521, 1894429, 11982834, 20371266, 9652725, 893025, 2702765, 148008446, 1472351967, 4417978068, 4546174779, 1502513550

Offset: 0

Wolfdieter Lang, Mar 16 2001

The row polynomials p(n,x) (rising powers of x) appear as numerators of the column g.f.s of triangle A060058.
First column (m=0) gives A000364 (Euler numbers). See A091742, A091743, A091744 for columns m=1..3.
The main diagonal gives A001818. The row sums give A052502. The alternating row sums give A091745.

			Triangle begins:
  {1};
  {1,1};
  {5,26,9};     <-- p(2,n)=5+26*x+9*x^2.
  {61,775,1179,225};
  ...

W. Lang, First 8 rows.

The row polynomials p(n, x) := Sum_{m=0..n} a(n, m)*x^m satisfy the differential equation: p(n, x) = x*((1-x)^2)*(d^2/dx^2)p(n-1, x) + (1+6*(n-1)*x+(5-6*n)*x^2)*(d/dx)p(n-1, x) + (3*n-2)*(1+(3*n-2)*x)*p(n-1, x), n >= 1, with input p(0, x)=1. - Wolfdieter Lang, Feb 13 2004

A092735 Decimal expansion of Pi^7.

Original entry on oeis.org

3, 0, 2, 0, 2, 9, 3, 2, 2, 7, 7, 7, 6, 7, 9, 2, 0, 6, 7, 5, 1, 4, 2, 0, 6, 4, 9, 3, 0, 7, 2, 0, 4, 1, 8, 3, 1, 9, 1, 7, 4, 3, 2, 4, 7, 5, 2, 9, 5, 4, 0, 2, 2, 6, 2, 7, 5, 4, 2, 3, 4, 4, 9, 2, 3, 8, 3, 1, 3, 4, 6, 6, 7, 2, 9, 3, 6, 1, 1, 8, 8, 0, 9, 3, 8, 4, 5, 2, 6, 2, 3, 0, 9, 0, 0, 0, 9, 7, 3, 5, 5, 6, 8, 6, 3

Offset: 4

Mohammad K. Azarian, Apr 12 2004

Wentworth (1903) shows how to compute the tangent of 15 degrees (A019913) to five decimal places by the laborious process of adding up the first few terms of Pi/12 + Pi^3/5184 + 2Pi^5/3732480 + 17Pi^7/11287019520 + ... - Alonso del Arte, Mar 13 2015

			3020.293227776792067514206493...

George Albert Wentworth, New Plane and Spherical Trigonometry, Surveying, and Navigation. Boston: The Atheneum Press (1903): 240.

G. C. Greubel, Table of n, a(n) for n = 4..10000
Index entries for transcendental numbers

Cf. A000796, A002161, A019692, A091925, A092731, A000364, A002437.

R:= RealField(100); (Pi(R))^7; // G. C. Greubel, Mar 09 2018

RealDigits[Pi^7, 10, 100][[1]] (* Alonso del Arte, Mar 13 2015 *)

PARI
```
Pi^7 \\ G. C. Greubel, Mar 09 2018
```

From Peter Bala, Oct 30 2019: (Start)
Pi^7 = (6!/(2*33367)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/6)^7 + 1/(n + 5/6)^7 ), where 33367 = ((3^7 + 1)/4)*A000364(3) = A002437(3).
Pi^7 = (6!/(2*1191391)) * Sum_{n >= 0} (-1)^n*( 1/(n + 1/10)^7 - 1/(n + 3/10)^7 - 1/(n + 7/10)^7 + 1/(n + 9/10)^7 ), where 1191391 = ((5^7 - 1)/4)*A000364(3). (End)

A132049 Numerator of 2nA000111(n-1)/A000111(n): approximations of Pi, using Euler (up/down) numbers.

Original entry on oeis.org

2, 4, 3, 16, 25, 192, 427, 4352, 12465, 158720, 555731, 8491008, 817115, 626311168, 2990414715, 60920233984, 329655706465, 7555152347136, 45692713833379, 232711080902656, 7777794952988025, 217865914337460224

Offset: 1

Wolfdieter Lang Sep 14 2007

The denominators are given in A132050.

a(n)/n = 2, 2, 1, 4, 5, 32, 61, 544, ... are integers for n<=19. a(20)/20 = 58177770225664/5. - Paul Curtz, Mar 25 2013, Apr 04 2013

			Rationals r(n): [3, 16/5, 25/8, 192/61, 427/136, 4352/1385, 12465/3968, 158720/50521, ...].

J.-P. Delahaye, Pi - die Story (German translation), Birkhäuser, 1999 Basel, p. 31. French original: Le fascinant nombre Pi, Pour la Science, Paris, 1997.

Leonhard Euler, On the sums of series of reciprocals, (Presented to the St. Petersburg Academy on December 5, 1735), arXiv:math/0506415 [math.HO], 2005-2008.
Wolfdieter Lang, Rationals and some values
Wikipedia, Bernoulli number

Cf. triangle A008281 (main diagonal give zig-zag numbers A000111), A223925.

S := proc(n, k) option remember;
if k=0 then `if`(n=0,1,0) else S(n,k-1)+S(n-1,n-k) fi end:
R := n -> 2*n*S(n-1,n-1)/S(n,n);
A132049 := n -> numer(R(n)); A132050 := n -> denom(R(n));
seq(A132049(i),i=3..22); # Peter Luschny, Aug 04 2011

e[n_] := If[EvenQ[n], Abs[EulerE[n]], Abs[(2^(n+1)*(2^(n+1) - 1)*BernoulliB[n+1])/(n+1)]]; r[n_] := 2*n*(e[n-1]/e[n]); a[n_] := Numerator[r[n]]; Table[a[n], {n, 3, 22}] (* Jean-François Alcover, Mar 18 2013 *)

from itertools import islice, count, accumulate
from fractions import Fraction
def A132049_gen(): # generator of terms
    yield 2
    blist = (0,1)
    for n in count(2):
        yield Fraction(2*n*blist[-1],(blist:=tuple(accumulate(reversed(blist),initial=0)))[-1]).numerator
A132049_list = list(islice(A132049_gen(),40)) # Chai Wah Wu, Jun 09-11 2022

a(n)=numerator(r(n)) with the rationals r(n)=2*n*e(n-1)/e(n), where e(n)=A000111(n) ("zig-zag" or "up-down" numbers), i.e., e(2*k)=A000364(k) (Euler numbers, secant numbers, "zig"-numbers) and e(2*k+1)=A000182(k+1),k>=0, (tangent numbers, "zag"-numbers). Rationals in lowest terms.

Entries confirmed by N. J. A. Sloane, May 10 2012
More explicit definition from M. F. Hasler, Apr 03 2013
a(1) and a(2) prepended by Paul Curtz, Apr 04 2013

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A103327 Triangle read by rows: T(n,k) = binomial(2n+1, 2k+1).

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A147309 Riordan array [sec(x), log(sec(x) + tan(x))].

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A147315 L-matrix for Euler numbers A000111(n+1).

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A003701 Expansion of e.g.f. exp(x)/cos(x).

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A160485 Triangle of the RBS1 polynomial coefficients.

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A060627 1 + Sum_{n >= 1} Sum_{k = 0..n-1} (-1)^nT(n,k)y^(2k)x^(2n)/(2n)! = JacobiCN(x,y).

A132049 Numerator of 2nA000111(n-1)/A000111(n): approximations of Pi, using Euler (up/down) numbers.