A210108 -id:A210108 - cp's OEIS frontend

A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

1, 1, 5, 61, 1385, 50521, 2702765, 199360981, 19391512145, 2404879675441, 370371188237525, 69348874393137901, 15514534163557086905, 4087072509293123892361, 1252259641403629865468285, 441543893249023104553682821, 177519391579539289436664789665

Offset: 0

N. J. A. Sloane

Inverse Gudermannian gd^(-1)(x) = log(sec(x) + tan(x)) = log(tan(Pi/4 + x/2)) = arctanh(sin(x)) = 2 * arctanh(tan(x/2)) = 2 * arctanh(csc(x) - cot(x)). - Michael Somos, Mar 19 2011
a(n) is the number of downup permutations of [2n]. Example: a(2)=5 counts 4231, 4132, 3241, 3142, 2143. - David Callan, Nov 21 2011
a(n) is the number of increasing full binary trees on vertices {0,1,2,...,2n} for which the leftmost leaf is labeled 2n. - David Callan, Nov 21 2011
a(n) is the number of unordered increasing trees of size 2n+1 with only even degrees allowed and degree-weight generating function given by cosh(t). - Markus Kuba, Sep 13 2014
a(n) is the number of standard Young tableaux of skew shape (n+1,n,n-1,...,3,2)/(n-1,n-2,...2,1). - Ran Pan, Apr 10 2015
Since cos(z) has a root at z = Pi/2 and no other root in C with a smaller |z|, the radius of convergence of the e.g.f. (intended complex-valued) is Pi/2 = A019669 (see also A028296). - Stanislav Sykora, Oct 07 2016
All terms are odd. - Alois P. Heinz, Jul 22 2018
The sequence starting with a(1) is periodic modulo any odd prime p. The minimal period is (p-1)/2 if p == 1 mod 4 and p-1 if p == 3 mod 4 [Knuth & Buckholtz, 1967, Theorem 2]. - Allen Stenger, Aug 03 2020
Conjecture: taking the sequence [a(n) : n >= 1] modulo an integer k gives a purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 21 begins [1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, 20, 16, 2, 1, 5, 19, ...] with an apparent period of 6 = phi(21)/2. - Peter Bala, May 08 2023

			G.f. = 1 + x + 5*x^2 + 61*x^3 + 1385*x^4 + 50521*x^5 + 2702765*x^6 + 199360981*x^7 + ...
sec(x) = 1 + 1/2*x^2 + 5/24*x^4 + 61/720*x^6 + ...
From _Gary W. Adamson_, Jul 18 2011: (Start)
The first few rows of matrix M are:
   1,  1,  0,  0,  0, ...
   4,  4,  4,  0,  0, ...
   9,  9,  9,  9,  0, ...
  16, 16, 16, 16, 16, ... (End)

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Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 932.
J. M. Borwein and D. M. Bailey, Mathematics by Experiment, Peters, Boston, 2004; p. 49
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Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 420.
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L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
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N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
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J. V. Uspensky and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, NY, 1939, p. 269.

Seiichi Manyama, Table of n, a(n) for n = 0..242 (terms 0..99 from N. J. A. Sloane)
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Bishal Deb and Alan D. Sokal, Classical continued fractions for some multivariate polynomials generalizing the Genocchi and median Genocchi numbers, arXiv:2212.07232 [math.CO], 2022.
Bishal Deb and Alan D. Sokal, Continued fractions for cycle-alternating permutations, arXiv:2304.06545 [math.CO], 2023.
K. Dilcher and C. Vignat, Euler and the Strong Law of Small Numbers, Amer. Math. Mnthly, 123 (May 2016), 486-490.
Filippo Disanto and Emanuele Munarini, Local height in weighted Dyck models of random walks and the variability of the number of coalescent histories for caterpillar-shaped gene trees and species trees, SN Applied Sciences (2019), 1:578.
D. Dumont and J. Zeng, Polynomes d'Euler et les fractions continues de Stieltjes-Rogers, Ramanujan J. 2 (1998) 3, 387-410.
A. L. Edmonds and S, Klee, The combinatorics of hyperbolized manifolds, arXiv preprint arXiv:1210.7396 [math.CO], 2012. - From _N. J. A. Sloane_, Jan 02 2013
C. J. Fewster and D. Siemssen, Enumerating Permutations by their Run Structure, arXiv preprint arXiv:1403.1723 [math.CO], 2014.
P. Flajolet and R. Sedgewick, Analytic Combinatorics, 2009; see page 144.
D. Foata and M.-P. Schutzenberger, Nombres d'Euler et permutations alternantes, in J. N. Srivastava et al., eds., A Survey of Combinatorial Theory (North Holland Publishing Company, Amsterdam, 1973), pp. 173-187.
Dominique Foata and Guo-Niu Han, Seidel Triangle Sequences and Bi-Entringer Numbers, November 20, 2013.
Ghislain R. Franssens, On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and Stirling number triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.4.1.
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23. (Annotated scanned copy)
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Michael E. Hoffman, Derivative Polynomials, Euler Polynomials, and Associated Integer Sequences, The Electronic Journal of Combinatorics, vol.6, no.1, #R21, (1999).
Donald E. Knuth and Thomas J. Buckholtz, Computation of tangent, Euler and Bernoulli numbers, Math. Comp. 21 1967 663-688.
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Sam Wagstaff, Prime divisors of the Bernoulli and Euler numbers.
Eric Weisstein's World of Mathematics, Euler Number, Secant Number, Alternating Permutation.
Wolfram Research, Generating functions for E_n.
Index entries for "core" sequences.
Index entries for sequences related to boustrophedon transform.

Cf. A000111, A000182, A011248, A019669, A034947, A060075, A013525, A000816, A002436, A000464, A002105, A002439, A079144, A158690.
Essentially same as A028296 and A122045.
First column of triangle A060074.
Two main diagonals of triangle A060058 (as iterated sums of squares).
Absolute values of row sums of A160485. - Johannes W. Meijer, Jul 06 2009
Left edge of triangle A210108, see also A125053, A076552. Cf. A255881.
Bisection (even part) of A317139.
The sequences [(-k^2)^n*Euler(2*n, 1/k), n = 0, 1, ...] are: A000007 (k=1), A000364 (k=2), |A210657| (k=3), A000281 (k=4), A272158 (k=5), A002438 (k=6), A273031 (k=7).

series(sec(x),x,40): SERIESTOSERIESMULT(%): subs(x=sqrt(y),%): seriestolist(%);
# end of program
A000364_list := proc(n) local S,k,j; S[0] := 1;
for k from 1 to n do S[k] := k*S[k-1] od;
for k from  1 to n do
    for j from k to n do
        S[j] := (j-k)*S[j-1]+(j-k+1)*S[j] od od;
seq(S[j], j=1..n)  end:
A000364_list(16);  # Peter Luschny, Apr 02 2012
A000364 := proc(n)
    abs(euler(2*n)) ;
end proc: # R. J. Mathar, Mar 14 2013

Take[ Range[0, 32]! * CoefficientList[ Series[ Sec[x], {x, 0, 32}], x], {1, 32, 2}] (* Robert G. Wilson v, Apr 23 2006 *)
Table[Abs[EulerE[2n]], {n, 0, 30}] (* Ray Chandler, Mar 20 2007 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n}, m! SeriesCoefficient[ Sec[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[ n_] := If[ n < 0, 0, With[{m = 2 n + 1}, m! SeriesCoefficient[ InverseGudermannian[ x], {x, 0, m}]]]; (* Michael Somos, Nov 22 2013 *)
a[n_] := Sum[Sum[Binomial[k, m] (-1)^(n+k)/(2^(m-1)) Sum[Binomial[m, j]* (2j-m)^(2n), {j, 0, m/2}] (-1)^(k-m), {m, 0, k}], {k, 1, 2n}]; Table[ a[n], {n, 0, 16}] (* Jean-François Alcover, Jun 26 2019, after Vladimir Kruchinin *)
a[0] := 1; a[n_] := a[n] = -Sum[a[n - k]/(2 k)!, {k, 1, n}]; Map[(-1)^# (2 #)! a[#] &, Range[0, 16]] (* Oliver Seipel, May 18 2024 *)

a(n):=sum(sum(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*sum(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

a[n]:=if n=0 then 1 else sum(sum((i-k)^(2*n)*binomial(2*k, i)*(-1)^(i+k+n), i, 0, k-1)/ (2^(k-1)), k, 1, 2*n); makelist(a[n], n, 0, 16); /* Vladimir Kruchinin, Oct 05 2012 */

{a(n)=local(CF=1+x*O(x^n));if(n<0,return(0), for(k=1,n,CF=1/(1-(n-k+1)^2*x*CF));return(Vec(CF)[n+1]))} \\ Paul D. Hanna Oct 07 2005

{a(n) = if( n<0, 0, (2*n)! * polcoeff( 1 / cos(x + O(x^(2*n + 1))), 2*n))}; /* Michael Somos, Jun 18 2002 */

{a(n) = my(A); if( n<0, 0, n = 2*n+1 ; A = x * O(x^n); n! * polcoeff( log(1 / cos(x + A) + tan(x + A)), n))}; /* Michael Somos, Aug 15 2007 */

{a(n)=polcoeff(sum(m=0, n, (2*m)!/2^m * x^m/prod(k=1, m, 1+k^2*x+x*O(x^n))), n)} \\ Paul D. Hanna, Sep 20 2012

list(n)=my(v=Vec(1/cos(x+O(x^(2*n+1)))));vector(n,i,v[2*i-1]*(2*i-2)!) \\ Charles R Greathouse IV, Oct 16 2012

a(n)=subst(bernpol(2*n+1),'x,1/4)*4^(2*n+1)*(-1)^(n+1)/(2*n+1) \\ Charles R Greathouse IV, Dec 10 2014

a(n)=abs(eulerfrac(2*n)) \\ Charles R Greathouse IV, Mar 23 2022

\\ Based on an algorithm of Peter Bala, cf. link in A110501.
upto(n) = my(v1, v2, v3); v1 = vector(n+1, i, 0); v1[1] = 1; v2 = vector(n, i, i^2); v3 = v1; for(i=2, n+1, for(j=2, i-1, v1[j] += v2[i-j+1]*v1[j-1]); v1[i] = v1[i-1]; v3[i] = v1[i]); v3 \\ Mikhail Kurkov, Aug 30 2025

from functools import lru_cache
from math import comb
@lru_cache(maxsize=None)
def A000364(n): return 1 if n == 0 else (1 if n % 2 else -1)*sum((-1 if i % 2 else 1)*A000364(i)*comb(2*n,2*i) for i in range(n)) # Chai Wah Wu, Jan 14 2022

# after Mikhail Kurkov, based on an algorithm of Peter Bala, cf. link in A110501.
def euler_list(len: int) -> list[int]:
    if len == 0: return []
    v1 = [1] + [0] * (len - 1)
    v2 = [i**2 for i in range(1, len + 1)]
    result = [0] * len
    result[0] = 1
    for i in range(1, len):
        for j in range(1, i):
            v1[j] += v2[i - j] * v1[j - 1]
        v1[i] = v1[i - 1]
        result[i] = v1[i]
    return result
print(euler_list(1000))  # Peter Luschny, Aug 30 2025

# Algorithm of L. Seidel (1877)
# n -> [a(0), a(1), ..., a(n-1)] for n > 0.
def A000364_list(len) :
    R = []; A = {-1:0, 0:1}; k = 0; e = 1
    for i in (0..2*len-1) :
        Am = 0; A[k + e] = 0; e = -e
        for j in (0..i) : Am += A[k]; A[k] = Am; k += e
        if e < 0 : R.append(A[-i//2])
    return R
A000364_list(17) # Peter Luschny, Mar 31 2012

E.g.f.: Sum_{n >= 0} a(n) * x^(2*n) / (2*n)! = sec(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n) * x^(2*n+1) / (2*n+1)! = gd^(-1)(x). - Michael Somos, Aug 15 2007
E.g.f.: Sum_{n >= 0} a(n)*x^(2*n+1)/(2*n+1)! = 2*arctanh(cosec(x)-cotan(x)). - Ralf Stephan, Dec 16 2004
Pi/4 - [Sum_{k=0..n-1} (-1)^k/(2*k+1)] ~ (1/2)*[Sum_{k>=0} (-1)^k*E(k)/(2*n)^(2k+1)] for positive even n. [Borwein, Borwein, and Dilcher]
Also, for positive odd n, log(2) - Sum_{k = 1..(n-1)/2} (-1)^(k-1)/k ~ (-1)^((n-1)/2) * Sum_{k >= 0} (-1)^k*E(k)/n^(2*k+1), where E(k) is the k-th Euler number, by Borwein, Borwein, and Dilcher, Lemma 2 with f(x) := 1/(x + 1/2), h := 1/2 and then replace x with (n-1)/2. - Peter Bala, Oct 29 2016
Let M_n be the n X n matrix M_n(i, j) = binomial(2*i, 2*(j-1)) = A086645(i, j-1); then for n>0, a(n) = det(M_n); example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385. - Philippe Deléham, Sep 04 2005
This sequence is also (-1)^n*EulerE(2*n) or abs(EulerE(2*n)). - Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 14 2006
a(n) = 2^n * E_n(1/2), where E_n(x) is an Euler polynomial.
a(k) = a(j) (mod 2^n) if and only if k == j (mod 2^n) (k and j  are even). [Stern; see also Wagstaff and Sun]
E_k(3^(k+1)+1)/4 = (3^k/2)*Sum_{j=0..2^n-1} (-1)^(j-1)*(2j+1)^k*[(3j+1)/2^n] (mod 2^n) where k is even and [x] is the greatest integer function. [Sun]
a(n) ~ 2^(2*n+2)*(2*n)!/Pi^(2*n+1) as n -> infinity. [corrected by Vaclav Kotesovec, Jul 10 2021]
a(n) = Sum_{k=0..n} A094665(n, k)*2^(n-k). - Philippe Deléham, Jun 10 2004
Recurrence: a(n) = -(-1)^n*Sum_{i=0..n-1} (-1)^i*a(i)*binomial(2*n, 2*i). - Ralf Stephan, Feb 24 2005
O.g.f.: 1/(1-x/(1-4*x/(1-9*x/(1-16*x/(...-n^2*x/(1-...)))))) (continued fraction due to T. J. Stieltjes). - Paul D. Hanna, Oct 07 2005
a(n) = (Integral_{t=0..Pi} log(tan(t/2)^2)^(2n)dt)/Pi^(2n+1). - Logan Kleinwaks (kleinwaks(AT)alumni.princeton.edu), Mar 15 2007
From Peter Bala, Mar 24 2009: (Start)
Basic hypergeometric generating function: 2*exp(-t)*Sum {n >= 0} Product_{k = 1..n} (1-exp(-(4*k-2)*t))*exp(-2*n*t)/Product_{k = 1..n+1} (1+exp(-(4*k-2)*t)) = 1 + t + 5*t^2/2! + 61*t^3/3! + .... For other sequences with generating functions of a similar type see A000464, A002105, A002439, A079144 and A158690.
a(n) = 2*(-1)^n*L(-2*n), where L(s) is the Dirichlet L-function L(s) = 1 - 1/3^s + 1/5^s - + .... (End)
Sum_{n>=0} a(n)*z^(2*n)/(4*n)!! = Beta(1/2-z/(2*Pi),1/2+z/(2*Pi))/Beta(1/2,1/2) with Beta(z,w) the Beta function. - Johannes W. Meijer, Jul 06 2009
a(n) = Sum_(Sum_(binomial(k,m)*(-1)^(n+k)/(2^(m-1))*Sum_(binomial(m,j)*(2*j-m)^(2*n),j,0,m/2)*(-1)^(k-m),m,0,k),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
If n is prime, then a(n)==1 (mod 2*n). - Vladimir Shevelev, Sep 04 2010
From Peter Bala, Jan 21 2011: (Start)
(1)... a(n) = (-1/4)^n*B(2*n,-1),
where {B(n,x)}n>=1 = [1, 1+x, 1+6*x+x^2, 1+23*x+23*x^2+x^3, ...] is the sequence of Eulerian polynomials of type B - see A060187. Equivalently,
(2)... a(n) = Sum_{k = 0..2*n} Sum_{j = 0..k} (-1)^(n-j) *binomial(2*n+1,k-j)*(j+1/2)^(2*n).
We also have
(3)... a(n) = 2*A(2*n,i)/(1+i)^(2*n+1),
where i = sqrt(-1) and where {A(n,x)}n>=1 = [x, x + x^2, x + 4*x^2 + x^3, ...] denotes the sequence of Eulerian polynomials - see A008292. Equivalently,
(4)... a(n) = i*Sum_{k = 1..2*n} (-1)^(n+k)*k!*Stirling2(2*n,k) *((1+i)/2)^(k-1)
= i*Sum_{k = 1..2*n} (-1)^(n+k)*((1+i)/2)^(k-1) Sum_{j = 0..k} (-1)^(k-j)*binomial(k,j)*j^(2*n).
Either this explicit formula for a(n) or (2) above may be used to obtain congruence results for a(n). For example, for prime p
(5a)... a(p) = 1 (mod p)
(5b)... a(2*p) = 5 (mod p)
and for odd prime p
(6a)... a((p+1)/2) = (-1)^((p-1)/2) (mod p)
(6b)... a((p-1)/2) = -1 + (-1)^((p-1)/2) (mod p).
(End)
a(n) = (-1)^n*2^(4*n+1)*(zeta(-2*n,1/4) - zeta(-2*n,3/4)). - Gerry Martens, May 27 2011
a(n) may be expressed as a sum of multinomials taken over all compositions of 2*n into even parts (Vella 2008): a(n) = Sum_{compositions 2*i_1 + ... + 2*i_k = 2*n} (-1)^(n+k)* multinomial(2*n, 2*i_1, ..., 2*i_k). For example, there are 4 compositions of the number 6 into even parts, namely 6, 4+2, 2+4 and 2+2+2, and hence a(3) = 6!/6! - 6!/(4!*2!) - 6!/(2!*4!) + 6!/(2!*2!*2!) = 61. A companion formula expressing a(n) as a sum of multinomials taken over the compositions of 2*n-1 into odd parts has been given by Malenfant 2011. - Peter Bala, Jul 07 2011
a(n) = the upper left term in M^n, where M is an infinite square production matrix; M[i,j] = A000290(i) = i^2, i >= 1 and 1 <= j <= i+1, and M[i,j] = 0, i >= 1 and j >= i+2 (see examples). - Gary W. Adamson, Jul 18 2011
E.g.f. A'(x) satisfies the differential equation A'(x)=cos(A(x)). - Vladimir Kruchinin, Nov 03 2011
From Peter Bala, Nov 28 2011: (Start)
a(n) = D^(2*n)(cosh(x)) evaluated at x = 0, where D is the operator cosh(x)*d/dx. a(n) = D^(2*n-1)(f(x)) evaluated at x = 0, where f(x) = 1+x+x^2/2! and D is the operator f(x)*d/dx.
Other generating functions: cosh(Integral_{t = 0..x} 1/cos(t)) dt = 1 + x^2/2! + 5*x^4/4! + 61*x^6/6! + 1385*x^8/8! + .... Cf. A012131.
A(x) := arcsinh(tan(x)) = log( sec(x) + tan(x) ) = x + x^3/3! + 5*x^5/5! + 61*x^7/7! + 1385*x^9/9! + .... A(x) satisfies A'(x) = cosh(A(x)).
B(x) := Series reversion( log(sec(x) + tan(x)) ) = x - x^3/3! + 5*x^5/5! - 61*x^7/7! + 1385*x^9/9! - ... = arctan(sinh(x)). B(x) satisfies B'(x) = cos(B(x)). (End)
HANKEL transform is A097476. PSUM transform is A173226. - Michael Somos, May 12 2012
a(n+1) - a(n) = A006212(2*n). - Michael Somos, May 12 2012
a(0) = 1 and, for n > 0, a(n) = (-1)^n*((4*n+1)/(2*n+1) - Sum_{k = 1..n} (4^(2*k)/2*k)*binomial(2*n,2*k-1)*A000367(k)/A002445(k)); see the Bucur et al. link. - L. Edson Jeffery, Sep 17 2012
O.g.f.: Sum_{n>=0} (2*n)!/2^n * x^n / Product_{k=1..n} (1 + k^2*x). - Paul D. Hanna, Sep 20 2012
From Sergei N. Gladkovskii, Oct 31 2011 to Oct 11 2013: (Start)
Continued fractions:
E.g.f.: (sec(x)) = 1+x^2/T(0), T(k) = 2(k+1)(2k+1) - x^2 + x^2*(2k+1)(2k+2)/T(k+1).
E.g.f.: 2/Q(0) where Q(k) = 1 + 1/(1 - x^2/(x^2 - 2*(k+1)*(2*k+1)/Q(k+1))).
G.f.: 1/Q(0) where Q(k) = 1 + x*k*(3*k-1) - x*(k+1)*(2*k+1)*(x*k^2+1)/Q(k+1).
E.g.f.: (2 + x^2 + 2*U(0))/(2 + (2 - x^2)*U(0)) where U(k)=  4*k + 4 + 1/( 1 + x^2/(2 - x^2 + (2*k+3)*(2*k+4)/U(k+1))).
E.g.f.: 1/cos(x) = 8*(x^2+1)/(4*x^2 + 8 - x^4*U(0)) where U(k) = 1 + 4*(k+1)*(k+2)/(2*k+3 - x^2*(2*k+3)/(x^2 - 8*(k+1)*(k+2)*(k+3)/U(k+1))).
G.f.: 1/U(0) where U(k) = 1 + x - x*(2*k+1)*(2*k+2)/(1 - x*(2*k+1)*(2*k+2)/U(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 + x - x*(2*k+2)*(2*k+3)/(1 - x*(2*k+2)*(2*k+3)/G(k+1)).
Let F(x) = sec(x^(1/2)) = Sum_{n>=0} a(n)*x^n/(2*n)!, then F(x)=2/(Q(0) + 1) where Q(k)= 1 - x/(2*k+1)/(2*k+2)/(1 - 1/(1 + 1/Q(k+1))).
G.f.: Q(0), where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)).
E.g.f.: 1/cos(x) = 1 + x^2/(2-x^2)*Q(0), where Q(k) = 1 - 2*x^2*(k+1)*(2*k+1)/( 2*x^2*(k+1)*(2*k+1)+ (12-x^2 + 14*k + 4*k^2)*(2-x^2 + 6*k + 4*k^2)/Q(k+1)). (End)
a(n) = Sum_{k=1..2*n} (Sum_{i=0..k-1} (i-k)^(2*n)*binomial(2*k,i)*(-1)^(i+k+n)) / 2^(k-1) for n>0, a(0)=1. - Vladimir Kruchinin, Oct 05 2012
It appears that a(n) = 3*A076552(n -1) + 2*(-1)^n for n >= 1. Conjectural congruences: a(2*n) == 5 (mod 60) for n >= 1 and a(2*n+1) == 1 (mod 60) for n >= 0. - Peter Bala, Jul 26 2013
From Peter Bala, Mar 09 2015: (Start)
O.g.f.: Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 - sqrt(-x)*(2*k + 1)) = Sum_{n >= 0} 1/2^n * Sum_{k = 0..n} (-1)^k*binomial(n,k)/(1 + x*(2*k + 1)^2).
O.g.f. is 1 + x*d/dx(log(F(x))), where F(x) = 1 + x + 3*x^2 + 23*x^3 + 371*x^4 + ... is the o.g.f. for A255881. (End)
Sum_(n >= 1, A034947(n)/n^(2d+1)) = a(d)*Pi^(2d+1)/(2^(2d+2)-2)(2d)! for d >= 0; see Allouche and Sondow, 2015. - Jonathan Sondow, Mar 21 2015
Asymptotic expansion: 4*(4*n/(Pi*e))^(2*n+1/2)*exp(1/2+1/(24*n)-1/(2880*n^3) +1/(40320*n^5)-...). (See the Luschny link.) - Peter Luschny, Jul 14 2015
a(n) = 2*(-1)^n*Im(Li_{-2n}(i)), where Li_n(x) is polylogarithm, i=sqrt(-1). - Vladimir Reshetnikov, Oct 22 2015
Limit_{n->infinity} ((2*n)!/a(n))^(1/(2*n)) = Pi/2. - Stanislav Sykora, Oct 07 2016
O.g.f.: 1/(1 + x - 2*x/(1 - 2*x/(1 + x - 12*x/(1 - 12*x/(1 + x - 30*x/(1 - 30*x/(1 + x - ... - (2*n - 1)*(2*n)*x/(1 - (2*n - 1)*(2*n)*x/(1 + x - ... ))))))))). - Peter Bala, Nov 09 2017
For n>0, a(n) = (-PolyGamma(2*n, 1/4) / 2^(2*n - 1) - (2^(2*n + 2) - 2) * Gamma(2*n + 1) * zeta(2*n + 1)) / Pi^(2*n + 1). - Vaclav Kotesovec, May 04 2020
a(n) ~ 2^(4*n + 3) * n^(2*n + 1/2) / (Pi^(2*n + 1/2) * exp(2*n)) * exp(Sum_{k>=1} bernoulli(k+1) / (k*(k+1)*2^k*n^k)). - Vaclav Kotesovec, Mar 05 2021
Peter Bala's conjectured congruences, a(2n) == 5 (mod 60) for n >= 1 and a(2n + 1) == 1 (mod 60), hold due to the results of Stern (mod 4) and Knuth & Buckholtz (mod 3 and 5). - Charles R Greathouse IV, Mar 23 2022

Typo in name corrected by Anders Claesson, Dec 01 2015

A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.

Original entry on oeis.org

1, 1, 4, 34, 496, 11056, 349504, 14873104, 819786496, 56814228736, 4835447317504, 495812444583424, 60283564499562496, 8575634961418940416, 1411083019275488149504, 265929039218907754399744, 56906245479134057176170496, 13722623393637762299131396096, 3704005473270641755597685653504

Offset: 1

N. J. A. Sloane

Comments from R. L. Graham, Apr 25 2006 and Jun 08 2006: "This sequence also gives the number of ways of arranging 2n tokens in a row, with 2 copies of each token from 1 through n, such that the first token is a 1 and between every pair of tokens labeled i (i=1..n-1) there is exactly one token labeled i+1.
"For example, for n=3, there are 4 possibilities: 123123, 121323, 132312 and 132132 and indeed a(3) = 4. This is the work of my Ph. D. student Nan Zang. See also A117513, A117514, A117515.
"Develin and Sullivant give another occurrence of this sequence and show that their numbers have the same generating function, although they were unable to find a 1-1-mapping between their problem and Poupard's."
The sequence 1,0,1,0,4,0,34,0,496,0,11056, ... counts increasing complete binary trees with e.g.f. sec^2(x/sqrt 2). - Wenjin Woan, Oct 03 2007
a(n) = number of increasing full binary trees on vertex set [2n-1] with the left-largest property: the largest descendant of each non-leaf vertex occurs in its left subtree (Poupard). The first Mathematica recurrence below counts these trees by number 2k-1 of vertices in the left subtree of the root: the root is necessarily labeled 1 and n necessarily occurs in the left subtree and so there are Binomial[2n-3,2k-2] ways to choose the remaining labels for the left subtree. - David Callan, Nov 29 2007
Number of bilabeled unordered increasing trees with 2n labels. - Markus Kuba, Nov 18 2014
Conjecture: taking the sequence modulo an integer k gives an eventually purely periodic sequence with period dividing phi(k). For example, the sequence taken modulo 10 begins [1, 1, 4, 4, 6, 6, 4, 4, 6, 6, 4, 4, 6, 6, ...] with an apparent period [4, 4, 6, 6] of length 4 = phi(10) beginning at a(3). - Peter Bala, May 08 2023
Let c(1), c(2), c(3), ... be a geometric progression and s = (2*c(1)/c(2))^(1/2). Then c(1)*s*tan(x/s) = Sum_{n>0} a(n) * c(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Jan 15 2025

			G.f. = x + x^2 + 4*x^3 + 34*x^4 + 496*x^5 + 11056*x^6 + 349504*x^7 + ...

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).

T. D. Noe, Table of n, a(n) for n = 1..100
G. E. Andrews, J. Jimenez-Urroz, and K. Ono, q-series identities and values of certain L-functions, Duke Math J. Volume 108, Number 3 (2001), 395-419.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
Peter Bala, A continued fraction expansion related to A002105
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
M. P. Develin and S. P. Sullivant, Markov Bases of Binary Graph Models, Annals of Combinatorics 7 (2003) 441-466.
Dominique Foata and Guo-Niu Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432
Dominique Foata and Guo-Niu Han, Tree Calculus for Bivariable Difference Equations, 2012. - From _N. J. A. Sloane_, Feb 02 2013
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory Ser. A 115 (2008), no. 2, 293-303.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph. D. Thesis, Waterford Institute of Technology, Oct. 2011.
Markus Kuba and Alois Panholzer, Combinatorial families of multilabelled increasing trees and hook-length formulas, arXiv:1411.4587 [math.CO], 2014.
B. A. Kupershmidt, On skew-symmetric and general deformations of Lax pseudodifferential operators, J. Math. Sci. 151 (4) (2008) 3139, eq. (1.13).
Zhicong Lin, Shi-Mei Ma, David G. L. Wang, and Liuquan Wang, Positivity and divisibility of alternating descent polynomials, arXiv:2011.02685 [math.CO], 2020.
E. Norton, Symplectic Reflection Algebras in Positive Characteristic as Ore Extensions, arXiv preprint arXiv:1302.5411 [math.RA], 2013.
Ronald Orozco López, Solution of the Differential Equation y^(k)= e^(a*y), Special Values of Bell Polynomials and (k,a)-Autonomous Coefficients, Universidad de los Andes (Colombia 2021).
C. Poupard, Deux proprietes des arbres binaires ordonnes stricts, European J. Combin., 10 (1989), 369-374.
M. Safaryan, Problem 11725, Amer. Math. Monthly, 120 (2013), 661.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 66-67.
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 64-67. [Annotated scanned copy]
A. Vieru, Agoh's conjecture: its proof, its generalizations, its analogues, arXiv preprint arXiv:1107.2938 [math.NT], 2011.
Index entries for sequences related to Bernoulli numbers.

Row sums of A008301.
Left edge of triangle A210108.
Cf. A000111, A000217, A000364, A000464, A002439, A008281, A079144, A158690.

A002105:= func< n | (-1)^(n+1)*2^n*(4^n - 1)*Bernoulli(2*n)/n >;
[A002105(n): n in [1..30]]; // G. C. Greubel, Sep 20 2024

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: A002105 := n -> S(2*n-1, 2*n-1)/2^(n-1): seq(A002105(i),i=1..16); # Peter Luschny, Jul 08 2012
# The above function written as a formula: a(n) = A008281(2*n-1, 2*n-1)/2^(n-1).
# Alternatively, based on the triangular numbers A000217:
T := proc(n, k) option remember; if k = 0 then 1 else if k = n then T(n, k-1) else
A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) fi fi end:
a := n -> T(n, n): seq(a(n), n = 0..18);  # Peter Luschny, Sep 30 2023

u[1] = 1; u[n_]/;n>=2 := u[n] = Sum[Binomial[2n-3,2k-2]u[k]u[n-k],{k,n-1}]; Table[u[n],{n,8}] (* Poupard and also Develin and Sullivant, give a different recurrence that involves a symmetric sum: v[1] = 1; v[n_]/;n>=2 := v[n] = 1/2 Sum[Binomial[2n-2,2k-1]v[k]v[n-k],{k,n-1}] *) (*David Callan, Nov 29 2007 *)
a[n_] := (-1)^n 2^(n+1) PolyLog[1-2n, -1]; Array[a, 10] (* Vladimir Reshetnikov, Jan 23 2011 *)
Table[(-1)^(n+1)*2^n*(2^(2n)-1)*BernoulliB[2n]/n,{n,1,20}] (* Vaclav Kotesovec, Nov 03 2014 *)
eulerCF[f_, len_] := Module[{g}, g[len-1]=1; g[k_]:=g[k]=1-f[k]/(f[k]-1/g[k+1]); CoefficientList[g[0] + O[x]^len, x]]; A002105List[len_] := eulerCF[(1/2) x (#+1) (#+2)&, len]; A002105List[19] (* Peter Luschny, Aug 08 2015 after Sergei N. Gladkovskii *)
Table[PolyGamma[2n-1, 1/2] 2^(2-n)/Pi^(2n), {n, 1, 10}] (* Vladimir Reshetnikov, Oct 18 2015 *)
Table[EulerE[2n-1, 0] (-2)^n, {n, 1, 10}] (* Vladimir Reshetnikov, Oct 21 2015 *)

{a(n) = if( n<1, 0, ((-2)^n - (-8)^n) * bernfrac(2*n) / n)}; /* Michael Somos, Jun 22 2002 */

{a(n) = if( n<0, 0, (2*n)! * polcoeff( -2 * log( cos(x / quadgen(8) + O(x^(2*n + 1)))), 2*n))}; /* Michael Somos, Jul 17 2003 */

{a(n) = if( n<0, 0, -(-2)^(n+1) * sum(i=1, 2*n, 2^-i * sum(j=1, i, (-1)^j * binomial( i-1, j-1) * j^(2*n - 1))))}; /* Michael Somos, Sep 07 2013 */

{a(n)=local(CF=1+x*O(x^n));if(n<1,return(0), for(k=1,n,CF=1/(1-(n-k+1)*(n-k+2)/2*x*CF));return(Vec(CF)[n]))}  /* Paul D. Hanna */

{a(n)=local(X=x+x*O(x^n),Egf);Egf=sum(m=0,n,prod(k=1,m,tanh(k*X)));n!*polcoeff(Egf,n)} /* Paul D. Hanna, May 11 2010 */

from sympy import bernoulli
def A002105(n): return abs(((2-(2<<(m:=n<<1)))*bernoulli(m)<Chai Wah Wu, Apr 14 2023

# Algorithm of L. Seidel (1877)
# n -> [a(1), ..., a(n)] for n >= 1.
def A002105_list(n) :
    D = [0]*(n+2); D[1] = 1
    R = []; z = 1/2; b = True
    for i in(0..2*n-1) :
        h = i//2 + 1
        if b :
            for k in range(h-1, 0, -1) : D[k] += D[k+1]
            z *= 2
        else :
            for k in range(1, h+1, 1) :  D[k] += D[k-1]
        b = not b
        if b : R.append(D[h]*z/h)
    return R
A002105_list(16) # Peter Luschny, Jun 29 2012

def A002105(n): return (-1)^(n+1)*2^n*(4^n -1)*bernoulli(2*n)/n
[A002105(n) for n in range(1,31)] # G. C. Greubel, Sep 20 2024

E.g.f.: 2*log(sec(x / sqrt(2))) = Sum_{n>0} a(n) * x^(2*n) / (2*n)!. - Michael Somos, Jun 22 2002
A000182(n) = 2^(n-1) * a(n). - Michael Somos, Jun 22 2002
a(n) = 2^(n-1)/n * A110501(n). - Don Knuth, Jan 16 2007
a(n+1) = Sum_{k = 0..n} A094665(n, k). - Philippe Deléham, Jun 11 2004
O.g.f.: A(x) = x/(1-x/(1-3*x/(1-6*x/(1-10*x/(1-15*x/(... -n*(n+1)/2*x/(1 - ...))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
sqrt(2) tan( x/sqrt(2)) = Sum_(n>=0) (x^(2n+1)/(2n+1)!) a_n. - Dominique Foata and Guo-Niu Han, Oct 24 2008
Basic hypergeometric generating function: Sum_{n>=0} Product {k = 1..n} (1-exp(-2*k*t))/Product {k = 1..n} (1+exp(-2*k*t)) = 1 + t + 4*t^2/2! + 34*t^3/3! + 496*t^4/4! + ... [Andrews et al., Theorem 4]. For other sequences with generating functions of a similar type see A000364, A000464, A002439, A079144 and A158690. - Peter Bala, Mar 24 2009
E.g.f.: Sum_{n>=0} Product_{k=1..n} tanh(k*x) = Sum_{n>=0} a(n)*x^n/n!. - Paul D. Hanna, May 11 2010
a(n) = (-1)^(n+1)*sum(j!*stirling2(2*n+1,j)*2^(n+1-j)*(-1)^(j),j,1,2*n+1), n>=0. - Vladimir Kruchinin, Aug 23 2010
From Gary W. Adamson, Jul 14 2011: (Start)
a(n) = upper left term in M^n, a(n+1) = sum of top row terms in M^n; where M = the infinite square production matrix:
   1, 3, 0,  0,  0, 0, 0, ...
   1, 3, 6,  0,  0, 0, 0, ...
   1, 3, 6, 10,  0, 0, 0, ...
   1, 3, 6, 10, 15, 0, 0, ... (End)
E.g.f. A(x) satisfies differential equation A''(x)=exp(A(x)). - Vladimir Kruchinin, Nov 18 2011
E.g.f.: For E(x)=sqrt(2)* tan( x/sqrt(2))=x/G(0); G(k)= 4*k + 1 - x^2/(8*k + 6 - x^2/G(k+1)); (from continued fraction Lambert's, 2-step). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = (-1)^n*2^(n+1)*Li_{1-2*n}(-1). (See also the Mathematica prog. by Vladimir Reshetnikov.) - Peter Luschny, Jun 28 2012
G.f.: 1/G(0) where G(k) = 1 - x*( 4*k^2 + 4*k + 1 ) - x^2*(k+1)^2*( 4*k^2 + 8*k + 3)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Jan 14 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*(k+2)/2*x/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: (1/G(0))/sqrt(x) - 1/sqrt(x), where G(k) = 1 - sqrt(x)*(2*k+1)/(1 + sqrt(x)*(2*k+1)/(1 + sqrt(x)*(k+1)/(1 - sqrt(x)*(k+1)/G(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Jul 07 2013
log(2) - 1/1 + 1/2 - 1/3 + ... + (-1)^n / n = (-1)^n / 2 * (1/n - 1 / (2*n^2) + 1 / (2*n^2)^2 - 4 / (2*n^2)^3 + ... + (-1)^k * a(k) / (2*n^2)^k + ...) asymptotic expansion. - Michael Somos, Sep 07 2013
G.f.: T(0), where T(k) = 1-x*(k+1)*(k+2)/(x*(k+1)*(k+2)-2/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013
a(n) ~ 2^(3*n+2) * n^(2*n-1/2) / (exp(2*n) * Pi^(2*n-1/2)). - Vaclav Kotesovec, Nov 03 2014
From Peter Bala, Sep 10 2015: (Start)
The e.g.f. A(x) = sqrt(2)*tan(x/sqrt(2)) satisfies A''(x) = A(x)*A'(x), hence the recurrence a(0) = 0, a(1) = 1, else a(n) = Sum_{i = 0..n-2} binomial(n-2,i)*a(i)*a(n-1-i) for the aerated sequence [0,1,0,1,0,4,0,34,0,496,...].
Note that the same recurrence, but with the initial conditions a(0) = 1 and a(1) = 1, produces the sequence [1,1,1,2,5,16,61,272,...] = A000111. (End)
a(n) = polygamma(2*n-1, 1/2)*2^(2-n)/Pi^(2*n). - Vladimir Reshetnikov, Oct 18 2015
E.g.f.: sqrt(2)*tan(x/sqrt(2)) = Sum_{n>0} a(n) * x^(2*n-1) / (2*n-1)!. - Michael Somos, Mar 05 2017
From Peter Bala, May 05 2017: (Start)
Let B(x) = A(x)/x = 1 + x + 4*x^2 + 34*x^3 + ... denote the shifted o.g.f. Then B(x) = 1/(1 + 2*x - 3*x/(1 - x/(1 + 2*x - 10*x/(1 - 6*x/(1 + 2*x - 21*x/(1 - 15*x/(1 + 2*x - 36*x/(1 - 28*x/(1 + 2*x - ...))))))))), where the coefficient sequence [3, 1, 10, 6, 21, 15, 36, 28, ...] in the partial numerators of the continued fraction is obtained by swapping adjacent triangular numbers. Cf. A079144.
It follows (by means of an equivalence transformation) that the second binomial transform of B(x), with g.f. equal to 1/(1 - 2*x)*B(x/(1 - 2*x)), has the S-fraction representation 1/(1 - 3*x/(1 - x/(1 - 10*x/(1 - 6*x/(1 - 21*x/(1 - 15*x/(1 - 36*x/(1 - 28*x/(1 - ...))))))))). Compare with the S-fraction representation of the g.f. A(x) given above by Hanna, dated Oct 07 2005. (End)
The computation can be based on the triangular numbers, a(n) = T(n, n) where T(n, k) = A000217(n - k + 1) * T(n, k - 1) + T(n - 1, k) for 0 < k < n, and T(n, 0) = 1, T(n, n) = T(n, k-1) if k > 0. This is equivalent to Paul D. Hanna's continued fraction 2005. - Peter Luschny, Sep 30 2023

Additional comments from Michael Somos, Jun 25 2002

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

A008301 Poupard's triangle: triangle of numbers arising in enumeration of binary trees.

Original entry on oeis.org

1, 1, 2, 1, 4, 8, 10, 8, 4, 34, 68, 94, 104, 94, 68, 34, 496, 992, 1420, 1712, 1816, 1712, 1420, 992, 496, 11056, 22112, 32176, 40256, 45496, 47312, 45496, 40256, 32176, 22112, 11056, 349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000

Offset: 0

N. J. A. Sloane

The doubloon polynomials evaluated at q=1. [Note the error in (D1) of the Foata-Han article in the Ramanujan journal which should read d_{1,j}(q) = delta_{2,j}.] - R. J. Mathar, Jan 27 2011

T(n,k), 0 <= k <= 2n-2, is the number of increasing 0-2 trees on vertices [0,2n] in which the parent of 2n is k (Poupard). A little more generally, for fixed m in [k+1,2n], T(n,k) is the number of trees in which m is a leaf with parent k. (The case m=2n is Poupard's result.) T(n,k) is the number of increasing 0-2 trees on vertices [0,2n] in which the minimal path from the root ends at k+1 (Poupard). - David Callan, Aug 23 2011

			[1],
[1, 2, 1],
[4, 8, 10, 8, 4],
[34, 68, 94, 104, 94, 68, 34],
[496, 992, 1420, 1712, 1816, 1712, 1420, 992, 496],
[11056, 22112, 32176, 40256, 45496, 47312, 45496, 40256, 32176, 22112, 11056],
[349504, 699008, 1026400, 1309568, 1528384, 1666688, 1714000, 1666688, 1528384, 1309568, 1026400, 699008, 349504], ...

Reinhard Zumkeller, Rows n = 0..100 of triangle, flattened
Neil J. Y. Fan, Liao He, The Complete cd-Index of Boolean Lattices, Electron. J. Combin., 22 (2015), #P2.45.
D. Foata, G.-N. Han, The doubloon polynomial triangle, Ram. J. 23 (2010), 107-126.
Dominique Foata and Guo-Niu Han, Doubloons and new q-tangent numbers, Quart. J. Math. 62 (2) (2011) 417-432.
D. Foata and G.-N. Han, Tree Calculus for Bivariable Difference Equations, 2012. - From _N. J. A. Sloane_, Feb 02 2013
D. Foata and G.-N. Han, Tree Calculus for Bivariable Difference Equations, arXiv:1304.2484 [math.CO], 2013.
R. L. Graham and Nan Zang, Enumerating split-pair arrangements, J. Combin. Theory, Ser. A, 115 (2008), pp. 293-303.
C. Poupard, Deux propriétés des arbres binaires ordonnés stricts, European J. Combin., 10 (1989), 369-374.

Cf. A107652. Leading diagonal and row sums = A002105.

Cf. A210108 (left half).

a008301 n k = a008301_tabf !! n !! k
a008301_row n = a008301_tabf !! n
a008301_tabf = iterate f [1] where
   f zs = zs' ++ tail (reverse zs') where
     zs' = (sum zs) : h (0 : take (length zs `div` 2) zs) (sum zs) 0
     h []        = []
     h (x:xs) y' y = y'' : h xs y'' y' where y'' = 2*y' - 2*x - y
-- Reinhard Zumkeller, Mar 17 2012

doubloon := proc(n,j,q) option remember; if n = 1 then if j=2 then 1; else 0; end if; elif j >= 2*n+1 or ( n>=1 and j<=1 ) then 0 ; elif j=2 and n>=1 then add(q^(k-1)*procname(n-1,k,q),k=1..2*n-2) ; elif n>=2 and 3<=j and j<=2*n then 2*procname(n,j-1,q)-procname(n,j-2,q)-(1-q)*add( q^(n+i+1-j)*procname(n-1,i,q),i=1..j-3) - (1+q^(n-1))*procname(n-1,j-2,q)+(1-q)*add(q^(i-j+1)*procname(n-1,i,q),i=j-1..2*n-1) ; else error; end if; expand(%) ; end proc:
A008301 := proc(n,k) doubloon(n+1,k+2,1) ; end proc:
seq(seq(A008301(n,k),k=0..2*n),n=0..12) ; # R. J. Mathar, Jan 27 2011
# Second program based on the Poupard numbers g_n(k) (A236934).
T := proc(n,k) option remember; local j;
  if n = 1 then 1
elif k = 1 then 0
elif k = 2 then 2*add(T(n-1, j), j=1..2*n-3)
elif k > n then T(n, 2*n-k)
else 2*T(n, k-1)-T(n, k-2)-4*T(n-1, k-2)
  fi end:
A008301 := (n,k) -> T(n+1,k+1)/2^n;
seq(print(seq(A008301(n,k), k=1..2*n-1)), n=1..6); # Peter Luschny, May 12 2014

doubloon[1, 2, q_] = 1; doubloon[1, j_, q_] = 0; doubloon[n_, j_, q_] /; j >= 2n+1 || n >= 1 && j <= 1 = 0; doubloon[n_ /; n >= 1, 2, q_] := doubloon[n, 2, q] = Sum[ q^(k-1)*doubloon[n-1, k, q], {k, 1, 2n-2}]; doubloon[n_, j_, q_] /; n >= 2 <= j && j <= 2n := doubloon[n, j, q] = 2*doubloon[n, j-1, q] - doubloon[n, j-2, q] - (1-q)*Sum[ q^(n+i+1-j)*doubloon[n-1, i, q], {i, 1, j-3}] - (1 + q^(n-1))*doubloon[n-1, j-2, q] + (1-q)* Sum[ q^(i-j+1)*doubloon[n-1, i, q], {i, j-1, 2n-1}]; A008301[n_,k_] := doubloon[n+1, k+2, 1]; Flatten[ Table[ A008301[n, k], {n, 0, 6}, {k, 0, 2n}]] (* Jean-François Alcover, Jan 23 2012, after R. J. Mathar *)
T[n_, k_] := T[n, k] = Which[n==1, 1, k==1, 0, k==2, 2*Sum[T[n-1, j], {j, 1, 2*n-3}], k>n, T[n, 2*n-k], True, 2*T[n, k-1] - T[n, k-2] - 4*T[n-1, k - 2]]; A008301[n_, k_] := T[n+1, k+1]/2^n; Table[A008301[n, k], {n, 1, 6}, {k, 1, 2*n-1}] // Flatten (* Jean-François Alcover, Nov 28 2015, after Peter Luschny *)

Recurrence relations are given on p. 370 of the Poupard paper; however, in line -5 the summation index should be k and in line -4 the expression 2_h^{k-1} should be replaced by 2d_h^(k-1). - Emeric Deutsch, May 03 2004
If we write the triangle like this:
                 0,    1,   0
            0,   1,    2,   1,   0
       0,   4,   8,   10,   8,   4,   0
  0,  34,  68,  94,  104,  94,  68,  34,  0
then the first nonzero term is the sum of the previous row and the remaining terms in each row are obtained by the rule illustrated by 104 = 2*94 - 2*8 - 1*68. - N. J. A. Sloane, Jun 10 2005
Continuing Sloane's remark: If we also set the line "... 1 ..." on the top of the pyramid, then we obtain T(n,k) = A236934(n+1,k+1)/2^n for n >= 1 and 1 <= k <= 2n-1 (see the second Maple program). - Peter Luschny, May 12 2014

More terms from Emeric Deutsch, May 03 2004

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A000364 Euler (or secant or "Zig") numbers: e.g.f. (even powers only) sec(x) = 1/cos(x).

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A002105 Reduced tangent numbers: 2^n*(2^{2n} - 1)*|B_{2n}|/n, where B_n = Bernoulli numbers.

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A005799 Generalized Euler numbers of type 2^n.

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A008301 Poupard's triangle: triangle of numbers arising in enumeration of binary trees.

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A125054 Central terms of triangle A125053.

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A002105 Reduced tangent numbers: 2^n(2^{2n} - 1)|B_{2n}|/n, where B_n = Bernoulli numbers.