A000182 -id:A000182 - cp's OEIS frontend

A003881 Decimal expansion of Pi/4.

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

Offset: 0

N. J. A. Sloane, Simon Plouffe

Also the ratio of the area of a circle to the circumscribed square. More generally, the ratio of the area of an ellipse to the circumscribed rectangle. Also the ratio of the volume of a cylinder to the circumscribed cube. - Omar E. Pol, Sep 25 2013
Also the surface area of a quarter-sphere of diameter 1. - Omar E. Pol, Oct 03 2013
Least positive solution to sin(x) = cos(x). - Franklin T. Adams-Watters, Jun 17 2014
Dirichlet L-series of the non-principal character modulo 4 (A101455) at 1. See e.g. Table 22 of arXiv:1008.2547. - R. J. Mathar, May 27 2016
This constant is also equal to the infinite sum of the arctangent functions with nested radicals consisting of square roots of two. Specifically, one of the Viete-like formulas for Pi is given by Pi/4 = Sum_{k = 2..oo} arctan(sqrt(2 - a_{k - 1})/a_k), where the nested radicals are defined by recurrence relations a_k = sqrt(2 + a_{k - 1}) and a_1 = sqrt(2) (see the article [Abrarov and Quine]). - Sanjar Abrarov, Jan 09 2017
Pi/4 is the area enclosed between circumcircle and incircle of a regular polygon of unit side. - Mohammed Yaseen, Nov 29 2023

			0.785398163397448309615660845819875721049292349843776455243736148...
N = 2, m = 6: Pi/4 = 4!*3^4 Sum_{k >= 0} (-1)^k/((2*k - 11)*(2*k - 5)*(2*k + 1)*(2*k + 7)*(2*k + 13)). - _Peter Bala_, Nov 15 2016

Jörg Arndt and Christoph Haenel, Pi: Algorithmen, Computer, Arithmetik, Springer 2000, p. 150.
Florian Cajori, A History of Mathematical Notations, Dover edition (2012), par. 437.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Sections 6.3 and 8.4, pp. 429 and 492.
Douglas R. Hofstadter, Gödel, Escher, Bach: An Eternal Golden Braid, Basic Books, p. 408.
J. Rivaud, Analyse, Séries, équations différentielles, Mathématiques supérieures et spéciales, Premier cycle universitaire, Vuibert, 1981, Exercice 3, p. 136.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 119.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sanjar M. Abrarov and Brendan M. Quine, A Viète-like formula for pi based on infinite sum of the arctangent functions with nested radicals, figshare, 4509014, (2017).
Peter Bala, Arctanh(z) and the Legendre polynomials
Jonathan M. Borwein, Peter B. Borwein, and Karl Dilcher, Pi, Euler numbers and asymptotic expansions, Amer. Math. Monthly, 96 (1989), 681-687.
Antonio Gracia Llorente, Shifting Constants Through Infinite Product Transformations, OSF Preprint, 2024.
Ronald K. Hoeflin, Titan Test.
Richard J. Mathar, Table of Dirichlet L-Series and Prime Zeta Modulo Functions for Small Moduli, arXiv:1008.2547 [math.NT], 2010-2015.
Literate Programs, Pi with Machin's formula (Haskell).
Michael Penn, A surprising appearance of pie!, YouTube video, 2020.
Michael Penn, Transforming normal identities into "crazy" ones, YouTube video, 2022.
Srinivasa Ramanujan, Question 353, J. Ind. Math. Soc.
Eric Weisstein's World of Mathematics, Prime Products.
Wikipedia, Leibniz formula for Pi.
Index entries for sequences from "Goedel, Escher, Bach".
Index entries for transcendental numbers.

Cf. A000796, A001586, A071904, A019669, A197723, A347152.
Cf. A000182, A000364, A024235, A278080, A278195.
Cf. A006752 (beta(2)=Catalan), A153071 (beta(3)), A175572 (beta(4)), A175571 (beta(5)), A175570 (beta(6)), A258814 (beta(7)), A258815 (beta(8)), A258816 (beta(9)).
Cf. A001622.

-- see link: Literate Programs
import Data.Char (digitToInt)
a003881_list len = map digitToInt $ show $ machin `div` (10 ^ 10) where
   machin = 4 * arccot 5 unity - arccot 239 unity
   unity = 10 ^ (len + 10)
   arccot x unity = arccot' x unity 0 (unity `div` x) 1 1 where
     arccot' x unity summa xpow n sign
    | term == 0 = summa
    | otherwise = arccot'
      x unity (summa + sign * term) (xpow `div` x ^ 2) (n + 2) (- sign)
    where term = xpow `div` n
-- Reinhard Zumkeller, Nov 20 2012

R:= RealField(100); Pi(R)/4; // G. C. Greubel, Mar 08 2018

Maple
```
evalf(Pi/4) ;
```

RealDigits[N[Pi/4,6! ]]  (* Vladimir Joseph Stephan Orlovsky, Dec 02 2009 *)
(* PROGRAM STARTS *)
(* Define the nested radicals a_k by recurrence *)
a[k_] := Nest[Sqrt[2 + #1] & , 0, k]
(* Example of Pi/4 approximation at K = 100 *)
Print["The actual value of Pi/4 is"]
N[Pi/4, 40]
Print["At K = 100 the approximated value of Pi/4 is"]
K := 100;  (* the truncating integer *)
N[Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 40] (* equation (8) *)
(* Error terms for Pi/4 approximations *)
Print["Error terms for Pi/4"]
k := 1; (* initial value of the index k *)
K := 10; (* initial value of the truncating integer K *)
sqn := {}; (* initiate the sequence *)
AppendTo[sqn, {"Truncating integer K ", " Error term in Pi/4"}];
While[K <= 30,
AppendTo[sqn, {K,
   N[Pi/4 - Sum[ArcTan[Sqrt[2 - a[k - 1]]/a[k]], {k, 2, K}], 1000] //
    N}]; K++]
Print[MatrixForm[sqn]]
(* Sanjar Abrarov, Jan 09 2017 *)

Pi/4 \\ Charles R Greathouse IV, Jul 07 2014

# Leibniz/Cohen/Villegas/Zagier/Arndt/Haenel
def FastLeibniz(n):
    b = 2^(2*n-1); c = b; s = 0
    for k in range(n-1,-1,-1):
        t = 2*k+1
        s = s + c/t if is_even(k) else s - c/t
        b *= (t*(k+1))/(2*(n-k)*(n+k))
        c += b
    return s/c
A003881 = RealField(3333)(FastLeibniz(1330))
print(A003881)  # Peter Luschny, Nov 20 2012

Equals Integral_{x=0..oo} sin(2x)/(2x) dx.
Equals lim_{n->oo} n*A001586(n-1)/A001586(n) (conjecture). - Mats Granvik, Feb 23 2011
Equals Integral_{x=0..1} 1/(1+x^2) dx. - Gary W. Adamson, Jun 22 2003
Equals Integral_{x=0..Pi/2} sin(x)^2 dx, or Integral_{x=0..Pi/2} cos(x)^2 dx. - Jean-François Alcover, Mar 26 2013
Equals (Sum_{x=0..oo} sin(x)*cos(x)/x) - 1/2. - Bruno Berselli, May 13 2013
Equals (-digamma(1/4) + digamma(3/4))/4. - Jean-François Alcover, May 31 2013
Equals Sum_{n>=0} (-1)^n/(2*n+1). - Geoffrey Critzer, Nov 03 2013
Equals Integral_{x=0..1} Product_{k>=1} (1-x^(8*k))^3 dx [cf. A258414]. - Vaclav Kotesovec, May 30 2015
Equals Product_{k in A071904} (if k mod 4 = 1 then (k-1)/(k+1)) else (if k mod 4 = 3 then (k+1)/(k-1)). - Dimitris Valianatos, Oct 05 2016
From Peter Bala, Nov 15 2016: (Start)
For N even: 2*(Pi/4 - Sum_{k = 1..N/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^(N/2)*(1/N - 1/N^3 + 5/N^5 - 61/N^7 + 1385/N^9 - ...), where the sequence of unsigned coefficients [1, 1, 5, 61, 1385, ...] is A000364. See Borwein et al., Theorem 1 (a).
For N odd: 2*(Pi/4 - Sum_{k = 1..(N-1)/2} (-1)^(k-1)/(2*k - 1)) ~ (-1)^((N-1)/2)*(1/N - 1/N^2 + 2/N^4 - 16/N^6 + 272/N^8 - ...), where the sequence of unsigned coefficients [1, 1, 2, 16, 272, ...] is A000182 with an extra initial term of 1.
For N = 0,1,2,... and m = 1,3,5,... there holds Pi/4 = (2*N)! * m^(2*N) * Sum_{k >= 0} ( (-1)^(N+k) * 1/Product_{j = -N..N} (2*k + 1 + 2*m*j) ); when N = 0 we get the Madhava-Gregory-Leibniz series for Pi/4.
For examples of asymptotic expansions for the tails of these series representations for Pi/4 see A024235 (case N = 1, m = 1), A278080 (case N = 2, m = 1) and A278195 (case N = 3, m = 1).
For N = 0,1,2,..., Pi/4 = 4^(N-1)*N!/(2*N)! * Sum_{k >= 0} 2^(k+1)*(k + N)!* (k + 2*N)!/(2*k + 2*N + 1)!, follows by applying Euler's series transformation to the above series representation for Pi/4 in the case m = 1. (End)
From Peter Bala, Nov 05 2019: (Start)
For k = 0,1,2,..., Pi/4 = k!*Sum_{n = -oo..oo} 1/((4*n+1)*(4*n+3)* ...*(4*n+2*k+1)), where Sum_{n = -oo..oo} f(n) is understood as lim_{j -> oo} Sum_{n = -j..j} f(n).
Equals Integral_{x = 0..oo} sin(x)^4/x^2 dx = Sum_{n >= 1} sin(n)^4/n^2, by the Abel-Plana formula.
Equals Integral_{x = 0..oo} sin(x)^3/x dx = Sum_{n >= 1} sin(n)^3/n, by the Abel-Plana formula. (End)
From Amiram Eldar, Aug 19 2020: (Start)
Equals arcsin(1/sqrt(2)).
Equals Product_{k>=1} (1 - 1/(2*k+1)^2).
Equals Integral_{x=0..oo} x/(x^4 + 1) dx.
Equals Integral_{x=0..oo} 1/(x^2 + 4) dx. (End)
With offset 1, equals 5 * Pi / 2. - Sean A. Irvine, Aug 19 2021
Equals (1/2)!^2 = Gamma(3/2)^2. - Gary W. Adamson, Aug 23 2021
Equals Integral_{x = 0..oo} exp(-x)*sin(x)/x dx (see Rivaud reference). - Bernard Schott, Jan 28 2022
From Amiram Eldar, Nov 06 2023: (Start)
Equals beta(1), where beta is the Dirichlet beta function.
Equals Product_{p prime >= 3} (1 - (-1)^((p-1)/2)/p)^(-1). (End)
Equals arctan( F(1)/F(4) ) + arctan( F(2)/F(3) ), where F(1), F(2), F(3), and F(4) are any four consecutive Fibonacci numbers. - Gary W. Adamson, Mar 03 2024
Pi/4 = Sum_{n >= 1} i/(n*P(n, i)*P(n-1, i)) = (1/2)*Sum_{n >= 1} (-1)^(n+1)*4^n/(n*A006139(n)*A006139(n-1)), where i = sqrt(-1) and P(n, x) denotes the n-th Legendre polynomial. The n-th summand of the series is O( 1/(3 + 2*sqrt(3))^n ). - Peter Bala, Mar 16 2024
Equals arctan( phi^(-3) ) + arctan(phi^(-1) ). - Gary W. Adamson, Mar 27 2024
Equals Sum_{n>=1} eta(n)/2^n, where eta(n) is the Dirichlet eta function. - Antonio Graciá Llorente, Oct 04 2024
Equals Product_{k>=2} ((k + 1)^(k*(2*k + 1))*(k - 1)^(k*(2*k - 1)))/k^(4*k^2). - Antonio Graciá Llorente, Apr 12 2025
Equals Integral_{x=sqrt(2)..oo} dx/(x*sqrt(x^2 - 1)). - Kritsada Moomuang, May 29 2025

a(98) and a(99) corrected by Reinhard Zumkeller, Nov 20 2012

A001469 Genocchi numbers (of first kind); unsigned coefficients give expansion of x*tan(x/2).

Original entry on oeis.org

-1, 1, -3, 17, -155, 2073, -38227, 929569, -28820619, 1109652905, -51943281731, 2905151042481, -191329672483963, 14655626154768697, -1291885088448017715, 129848163681107301953, -14761446733784164001387, 1884515541728818675112649, -268463531464165471482681379

Offset: 1

N. J. A. Sloane

The Genocchi numbers satisfy Seidel's recurrence: for n>1, 0 = Sum_{j=0..[n/2]} C(n,2j)*a(n-j). - Ralf Stephan, Apr 17 2004
The (n+1)st Genocchi number is the number of Dumont permutations of the first kind on 2n letters. In a Dumont permutation of the first kind, each even integer must be followed by a smaller integer and each odd integer is either followed by a larger integer or is the last element. - Ralf Stephan, Apr 26 2004
According to Hetyei [2017], "alternation acyclic tournaments in which at least one ascent begins at each vertex, except for the largest one, are counted by the Genocchi numbers of the first kind." - Danny Rorabaugh, Apr 25 2017

L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
L. Euler, Institutionum Calculi Differentialis, volume 2 (1755), para. 181.
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. 73.
A. Genocchi, Intorno all'espressione generale de'numeri Bernulliani, Ann. Sci. Mat. Fis., 3 (1852), 395-405.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 528.
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).
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Problem 5.8.

Seiichi Manyama, Table of n, a(n) for n = 1..275 (first 100 terms from T. D. Noe)
F. Alayont and N. Krzywonos, Rook Polynomials in Three and Higher Dimensions, 2012.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
R. B. Brent, Generalizing Tuenter's Binomial Sums, J. Int. Seq. 18 (2015) # 15.3.2.
Alexander Burstein, Sergi Elizalde and Toufik Mansour, Restricted Dumont permutations, Dyck paths and noncrossing partitions, arXiv:math/0610234 [math.CO], 2006.
M. Domaratzki, A Generalization of the Genocchi Numbers with Applications to ...
M. Domaratzki, Combinatorial Interpretations of a Generalization of the Genocchi Numbers, Journal of Integer Sequences, Vol. 7 (2004), Article 04.3.6.
D. Dumont, Sur une conjecture de Gandhi concernant les nombres de Genocchi, (in French), Discrete Mathematics 1 (1972) 321-327.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
D. Dumont, Interprétations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318. (Annotated scanned copy)
Dominique Dumont and Arthur Randrianarivony, Sur une extension des nombres de Genocchi, European J. Combin. 16 (1995), 147-151.
Dominique Dumont and Arthur Randrianarivony, Dérangements et nombres de Genocchi, Discrete Math. 132 (1994), 37-49.
Richard Ehrenborg and Einar Steingrímsson, Yet another triangle for the Genocchi numbers, European J. Combin. 21 (2000), no. 5, 593-600. MR1771988 (2001h:05008).
J. M. Gandhi, Research Problems: A Conjectured Representation of Genocchi Numbers, Amer. Math. Monthly 77 (1970), no. 5, 505-506. MR1535914.
I. M. Gessel, Applications of the classical umbral calculus, arXiv:math/0108121 [math.CO], 2001.
Ira M. Gessel, On the Almkvist-Meurman Theorem for Bernoulli Polynomials, Integers (2023) Vol. 23, #A14.
Ira M. Gessel, A short proof of the Almkvist-Meurman theorem, arXiv:2310.15312 [math.NT], 2023.
René Gy, Bernoulli-Stirling Numbers, Integers (2020) Vol. 20, #A67.
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23.
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23. (Annotated scanned copy)
Gábor Hetyei, Alternation acyclic tournaments, arXiv:math/1704.07245 [math.CO], 2017.
G. Kreweras, Sur les permutations comptées par les nombres de Genocchi de 1-ière et 2-ième espèce, Europ. J. Comb., vol. 18, pp. 49-58, 1997.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
H. Liang and Wuyungaowa, Identities Involving Generalized Harmonic Numbers and Other Special Combinatorial Sequences, J. Int. Seq. 15 (2012) #12.9.6
Qui-Ming Luo, Fourier expansions and integral representations for Genocchi Polynomials, JIS 12 (2009) 09.1.4.
T. Mansour, Restricted 132-Dumont permutations, arXiv:math/0209379 [math.CO], 2002.
A. Randrianarivony and J. Zeng, Une famille de polynomes qui interpole plusieurs suites..., Adv. Appl. Math. 17 (1996), 1-26.
John Riordan and Paul R. Stein, Proof of a conjecture on Genocchi numbers, Discrete Math. 5 (1973), 381-388. MR0316372 (47 #4919).
N. J. A. Sloane, Rough notes on Genocchi numbers
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]
Hans J. H. Tuenter, Walking into an absolute sum, arXiv:math/0606080 [math.NT], 2006. Published version on Walking into an absolute sum, The Fibonacci Quarterly, 40(2):175-180, May 2002.
G. Viennot, Interprétations combinatoires des nombres d'Euler et de Genocchi, Séminaire de théorie des nombres, 1980/1981, Exp. No. 11, p. 41, Univ. Bordeaux I, Talence, 1982.
Eric Weisstein's World of Mathematics, Genocchi Number.
J. Worpitsky, Studien ueber die Bernoullischen und Eulerschen Zahlen, Journal für die reine undangewandte Mathematik (Crelle), 94 (1883), 203-232. See page 232. [Annotated scanned copy]
Index entries for sequences related to Bernoulli numbers.

Cf. A110501, A000182, A006846, A012509, A226158, A297703.

a(n) = -A065547(n, 1) and A065547(n+1, 2) for n >= 1.

[2*(1 - 4^n) * Bernoulli(2*n): n in [1..25]]; // Vincenzo Librandi, Oct 15 2018

A001469 := proc(n::integer) (2*n)!*coeftayl( 2*x/(exp(x)+1), x=0,2*n) end proc:
for n from 1 to 20 do print(A001469(n)) od : # R. J. Mathar, Jun 22 2006

a[n_] := 2*(1-4^n)*BernoulliB[2n]; Table[a[n], {n, 17}] (* Jean-François Alcover, Nov 24 2011 *)
a[n_] := 2*n*EulerE[2*n-1, 0]; Table[a[n], {n, 17}] (* Jean-François Alcover, Jul 02 2013 *)
Table[4 n PolyLog[1 - 2 n, -1], {n, 1, 19}] (* Peter Luschny, Aug 17 2021 *)

a(n)=if(n<1,0,n*=2; 2*(1-2^n)*bernfrac(n))

{a(n)=polcoeff(sum(m=0, n, m!^2*(-x)^(m+1)/prod(k=1, m, 1-k^2*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 21 2011 */

from sympy import bernoulli
def A001469(n): return (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 A001469_list(n) :
    D = [0]*(n+2); D[1] = -1
    R = []; b = False
    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]
        else :
            for k in range(1, h+1, 1) :  D[k] -= D[k-1]
        b = not b
        if not b : R.append(D[h])
    return R
A001469_list(17) # Peter Luschny, Jun 29 2012

a(n) = 2*(1-4^n)*B_{2n} (B = Bernoulli numbers).
x*tan(x/2) = Sum_{n>=1} x^(2*n)*abs(a(n))/(2*n)! = x^2/2 + x^4/24 + x^6/240 + 17*x^8/40320 + 31*x^10/725760 + O(x^11).
E.g.f.: 2*x/(1 + exp(x)) = x + Sum_{n>=1} a(2*n)*x^(2*n)/(2*n)! = -x^2/2! + x^4/4! - 3 x^6/6! + 17 x^8/8! + ...
O.g.f.: Sum_{n>=0} n!^2*(-x)^(n+1) / Product_{k=1..n} (1-k^2*x). - Paul D. Hanna, Jul 21 2011
a(n) = Sum_{k=0..2n-1} 2^k*B(k)*binomial(2*n,k) where B(k) is the k-th Bernoulli number. - Benoit Cloitre, May 31 2003
abs(a(n)) = Sum_{k=0..2n} (-1)^(n-k+1)*Stirling2(2n, k)*A059371(k). - Vladeta Jovovic, Feb 07 2004
G.f.: -x/(1+x/(1+2x/(1+4x/(1+6x/(1+9x/(1+12x/(1+16x/(1+20x/(1+25x/(1+...(continued fraction). - Philippe Deléham, Nov 22 2011
E.g.f.: E(x) = 2*x/(exp(x)+1) = x*(1-(x^3+2*x^2)/(2*G(0)-x^3-2*x^2)); G(k) = 8*k^3 + (12+4*x)*k^2 + (4+6*x+2*x^2)*k + x^3 + 2*x^2 + 2*x - 2*(x^2)*(k+1)*(2*k+1)*(x+2*k)*(x+2*k+4)/G(k+1); (continued fraction, Euler's kind, 1-step). - Sergei N. Gladkovskii, Jan 18 2012
a(n) = (-1)^n*(2*n)!*Pi^(-2*n)*4*(1-4^(-n))*Li{2*n}(1). - Peter Luschny, Jun 29 2012
Asymptotic: abs(a(n)) ~ 8*Pi*(2^(2*n)-1)*(n/(Pi*exp(1)))^(2*n+1/2)*exp(1/2+(1/24)/n-(1/2880)/n^3+(1/40320)/n^5+...). - Peter Luschny, Jul 24 2013
G.f.: x/(T(0)-x) -1, where T(k) = 2*x*k^2 + 4*x*k + 2*x - 1 - x*(-1+x+2*x*k+x*k^2)*(k+2)^2/T(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2013
G.f.: -1 + x/(T(0)+x), where T(k) = 1 + (k+1)*(k+2)*x/(1+x*(k+2)^2/T(k+1)); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2013
a(n) = 4*n*PolyLog(1 - 2*n, -1). - Peter Luschny, Aug 17 2021

A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.

Original entry on oeis.org

1, 1, 0, -2, 0, 16, 0, -272, 0, 7936, 0, -353792, 0, 22368256, 0, -1903757312, 0, 209865342976, 0, -29088885112832, 0, 4951498053124096, 0, -1015423886506852352, 0, 246921480190207983616, 0, -70251601603943959887872, 0, 23119184187809597841473536, 0

Offset: 0

Paul D. Hanna, Jan 24 2009

Previous name was: a(n) = Sum_{k=0..n-1} (-1)^(k)*C(n-1,k)*a(n-1-k)*a(k) for n>0 with a(0)=1.
Factorials have a similar recurrence: f(n) = Sum_{k=0..n-1} C(n-1,k)*f(n-1-k)*f(k), n > 0.
Related to A102573: letting T(q,r) be the coefficient of n^(r+1) in the polynomial 2^(q-n)/n times Sum_{k=0..n} binomial(n,k)*k^q, then A155585(x) = Sum_{k=0..x-1} T(x,k)*(-1)^k. See Mathematica code below. - John M. Campbell, Nov 16 2011
For the difference table and the relation to the Seidel triangle see A239005. - Paul Curtz, Mar 06 2014
From Tom Copeland, Sep 29 2015: (Start)
Let z(t) = 2/(e^(2t)+1) = 1 + tanh(-t) = e.g.f.(-t) for this sequence = 1 - t + 2*t^3/3! - 16*t^5/5! + ... .
dlog(z(t))/dt = -z(-t), so the raising operators that generate Appell polynomials associated with this sequence, A081733, and its reciprocal, A119468, contain z(-d/dx) = e.g.f.(d/dx) as the differential operator component.
dz(t)/dt = z*(z-2), so the assorted relations to a Ricatti equation, the Eulerian numbers A008292, and the Bernoulli numbers in the Rzadkowski link hold.
From Michael Somos's formula below (drawing on the Edwards link), y(t,1)=1 and x(t,1) = (1-e^(2t))/(1+e^(2t)), giving z(t) = 1 + x(t,1). Compare this to the formulas in my list in A008292 (Sep 14 2014) with a=1 and b=-1,
A) A(t,1,-1) = A(t) = -x(t,1) = (e^(2t)-1)/(1+e^(2t)) = tanh(t) = t + -2*t^3/3! + 16*t^5/5! + -272*t^7/7! + ... = e.g.f.(t) - 1 (see A000182 and A000111)
B) Ainv(t) = log((1+t)/(1-t))/2 = tanh^(-1)(t) = t + t^3/3 + t^5/5 + ..., the compositional inverse of A(t)
C) dA/dt = (1-A^2), relating A(t) to a Weierstrass elliptic function
D) ((1-t^2)d/dt)^n t evaluated at t=0, a generator for the sequence A(t)
F) FGL(x,y)= (x+y)/(1+xy) = A(Ainv(x) + Ainv(y)), a related formal group law corresponding to the Lorentz FGL (Lorentz transformation--addition of parallel velocities in special relativity) and the Atiyah-Singer signature and the elliptic curve (1-t^2)*s = t^3 in Tate coordinates according to the Lenart and Zainoulline link and the Buchstaber and Bunkova link (pp. 35-37) in A008292.
A133437 maps the reciprocal odd natural numbers through the refined faces of associahedra to a(n).
A145271 links the differential relations to the geometry of flow maps, vector fields, and thereby formal group laws. See Mathworld for links of tanh to other geometries and statistics.
Since the a(n) are related to normalized values of the Bernoulli numbers and the Riemann zeta and Dirichlet eta functions, there are links to Witten's work on volumes of manifolds in two-dimensional quantum gauge theories and the Kervaire-Milnor formula for homotopy groups of hyperspheres (see my link below).
See A101343, A111593 and A059419 for this and the related generator (1 + t^2) d/dt and associated polynomials. (End)
With the exception of the first term (1), entries are the alternating sums of the rows of the Eulerian triangle, A008292. - Gregory Gerard Wojnar, Sep 29 2018

			E.g.f.: 1 + x - 2*x^3/3! + 16*x^5/5! - 272*x^7/7! + 7936*x^9/9! -+ ... = exp(x)/cosh(x).
O.g.f.: 1 + x - 2*x^3 + 16*x^5 - 272*x^7 + 7936*x^9 - 353792*x^11 +- ...
O.g.f.: 1 + x/(1+2*x) + 2!*x^2/((1+2*x)*(1+4*x)) + 3!*x^3/((1+2*x)*(1+4*x)*(1+6*x)) + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..200
P. 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
Tom Copeland, The Kervaire-Milnor formula
H. M. Edwards, A normal form for elliptic curves, Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 3, 393-422. see section 12, pp. 410-411.
Wenjie Fang, Hsien-Kuei Hwang, and Mihyun Kang, Phase transitions from exp(n^(1/2)) to exp(n^(2/3)) in the asymptotics of banded plane partitions, arXiv:2004.08901 [math.CO], 2020.
G. Rzadkowski, Bernoulli numbers and solitons-revisited, Journal of Nonlinear Math. Physics, 1711, pp. 121-126.

Cf. A001057, A009006, A102573.
Equals row sums of A119879. - Johannes W. Meijer, Apr 20 2011
Cf. A081733, A119468, A008292, A000182, A000111, A101343, A111593, A059419, A133437, A145271.
(-1)^n*a(n) are the alternating row sums of A123125. - Wolfdieter Lang, Jul 12 2017

A155585 := n -> 2^n*euler(n, 1): # Peter Luschny, Jan 26 2009
a := proc(n) option remember; `if`(n::even, 0^n, -(-1)^n - add((-1)^k*binomial(n,k) *a(n-k), k = 1..n-1)) end: # Peter Luschny, Jun 01 2016
# Or via the recurrence of the Fubini polynomials:
F := proc(n) option remember; if n = 0 then return 1 fi;
expand(add(binomial(n, k)*F(n-k)*x, k = 1..n)) end:
a := n -> (-2)^n*subs(x = -1/2, F(n)):
seq(a(n), n = 0..30); # Peter Luschny, May 21 2021

a[m_] := Sum[(-2)^(m - k) k! StirlingS2[m, k], {k, 0, m}] (* Peter Luschny, Apr 29 2009 *)
poly[q_] :=  2^(q-n)/n*FunctionExpand[Sum[Binomial[n, k]*k^q, {k, 0, n}]]; T[q_, r_] :=  First[Take[CoefficientList[poly[q], n], {r+1, r+1}]]; Table[Sum[T[x, k]*(-1)^k, {k, 0, x-1}], {x, 1, 16}] (* John M. Campbell, Nov 16 2011 *)
f[n_] := (-1)^n 2^(n+1) PolyLog[-n, -1]; f[0] = -f[0]; Array[f, 27, 0] (* Robert G. Wilson v, Jun 28 2012 *)

a(n)=if(n==0,1,sum(k=0,n-1,(-1)^(k)*binomial(n-1,k)*a(n-1-k)*a(k)))

a(n)=local(X=x+x*O(x^n));n!*polcoeff(exp(X)/cosh(X),n)

a(n)=polcoeff(sum(m=0,n,m!*x^m/prod(k=1,m,1+2*k*x+x*O(x^n))),n) \\ Paul D. Hanna, Jul 20 2011

{a(n) = local(A); if( n<0, 0, A = x * O(x^n); n! * polcoeff( 1 + sinh(x + A) / cosh(x + A), n))} /* Michael Somos, Jan 16 2012 */

a(n)=local(A=1+x);for(i=1,n,A=sum(k=0,n,intformal(subst(A,x,-x)+x*O(x^n))^k/k!));n!*polcoeff(A,n)
for(n=0,30,print1(a(n),", ")) \\ Paul D. Hanna, Nov 25 2013

from sympy import bernoulli
def A155585(n): return (((2<<(m:=n+1))-2)*bernoulli(m)<>1) if n&1 else (0 if n else 1) # Chai Wah Wu, Apr 14 2023

def A155585(n) :
    if n == 0 : return 1
    return add(add((-1)^(j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))
[A155585(n) for n in (0..26)] # Peter Luschny, Jul 23 2012

def A155585_list(n): # Akiyama-Tanigawa algorithm
    A = [0]*(n+1); R = []
    for m in range(n+1) :
        d = divmod(m+3, 4)
        A[m] = 0 if d[1] == 0 else (-1)^d[0]/2^(m//2)
        for j in range(m, 0, -1) :
            A[j - 1] = j * (A[j - 1] - A[j])
        R.append(A[0])
    return R
A155585_list(30) # Peter Luschny, Mar 09 2014

E.g.f.: exp(x)*sech(x) = exp(x)/cosh(x). (See A009006.) - Paul Barry, Mar 15 2006
Sequence of absolute values is A009006 (e.g.f. 1+tan(x)).
O.g.f.: Sum_{n>=0} n! * x^n / Product_{k=1..n} (1 + 2*k*x). - Paul D. Hanna, Jul 20 2011
a(n) = 2^n*E_{n}(1) where E_{n}(x) are the Euler polynomials. - Peter Luschny, Jan 26 2009
a(n) = EL_{n}(-1) where EL_{n}(x) are the Eulerian polynomials. - Peter Luschny, Aug 03 2010
a(n+1) = (4^n-2^n)*B_n(1)/n, where B_{n}(x) are the Bernoulli polynomials (B_n(1) = B_n for n <> 1). - Peter Luschny, Apr 22 2009
G.f.: 1/(1-x+x^2/(1-x+4*x^2/(1-x+9*x^2/(1-x+16*x^2/(1-...))))) (continued fraction). - Paul Barry, Mar 30 2010
G.f.: -log(x/(exp(x)-1))/x = Sum_{n>=0} a(n)*x^n/(2^(n+1)*(2^(n+1)-1)*n!). - Vladimir Kruchinin, Nov 05 2011
E.g.f.: exp(x)/cosh(x) = 2/(1+exp(-2*x)) = 2/(G(0) + 1); G(k) = 1 - 2*x/(2*k + 1 - x*(2*k+1)/(x - (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 10 2011
E.g.f. is x(t,1) + y(t,1) where x(t,a) and y(t,a) satisfy y(t,a)^2 = (a^2 - x(t,a)^2) / (1 - a^2 * x(t,a)^2) and dx(t,a) / dt = y(t,a) * (1 - a * x(t,a)^2) and are the elliptic functions of Edwards. - Michael Somos, Jan 16 2012
E.g.f.: 1/(1 - x/(1+x/(1 - x/(3+x/(1 - x/(5+x/(1 - x/(7+x/(1 - x/(9+x/(1 +...))))))))))), a continued fraction. - Paul D. Hanna, Feb 11 2012
E.g.f. satisfies: A(x) = Sum_{n>=0} Integral( A(-x) dx )^n / n!. - Paul D. Hanna, Nov 25 2013
a(n) = -2^(n+1)*Li_{-n}(-1). - Peter Luschny, Jun 28 2012
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 to Dec 16 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 + (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: exp(x)/cosh(x) = 1 + x/S(0) where S(k) = (2*k+1) + x^2/S(k+1).
E.g.f.: 1 + x/(U(0)+x) where U(k) = 4*k+1 - x/(1 + x/(4*k+3 - x/(1 + x/U(k+1)))).
E.g.f.: 1 + tanh(x) = 4*x/(G(0)+2*x) where G(k) = 1 - (k+1)/(1 - 2*x/(2*x + (k+1)^2/G(k+1)));
G.f.: 1 + x/G(0) where G(k) = 1 + 2*x^2*(2*k+1)^2 - x^4*(2*k+1)*(2*k+2)^2*(2*k+3)/G(k+1) (due to Stieltjes).
E.g.f.: 1 + x/(G(0) + x) where G(k) = 1 - 2*x/(1 + (k+1)/G(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*(k+1)/( 1 - x*(k+1)/Q(k+1)).
G.f.: 2 - 1/Q(0) where Q(k) = 1 + x*k^2 + x/(1 - x*(k+1)^2/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - 2*x + x*(k+1)/(1-x*(k+1)/Q(k+1)).
G.f.: 1/Q(0) where Q(k) = 1 - x*(k+1)/(1 + x*(k+1)/Q(k+1)).
E.g.f.: 1 + x*Q(0) where Q(k) = 1 - x^2/( x^2 + (2*k+1)*(2*k+3)/Q(k+1)).
G.f.: 2 - T(0)/(1+x) where T(k) = 1 - x^2*(k+1)^2/(x^2*(k+1)^2 + (1+x)^2/T(k+1)).
E.g.f.: 1/(x - Q(0)) where Q(k) = 4*k^2 - 1 + 2*x + x^2*(2*k-1)*(2*k+3)/Q(k+1). (End)
G.f.: 1 / (1 - b(1)*x / (1 - b(2)*x / (1 - b(3)*x / ... ))) where b = A001057. - Michael Somos, Jan 03 2013
From Paul Curtz, Mar 06 2014: (Start)
a(2n) = A000007(n).
a(2n+1) = (-1)^n*A000182(n+1).
a(n) is the binomial transform of A122045(n).
a(n) is the row sum of A081658. For fractional Euler numbers see A238800.
a(n) + A122045(n) = 2, 1, -1, -2, 5, 16, ... = -A163982(n).
a(n) - A122045(n) = -A163747(n).
a(n) is the Akiyama-Tanigawa transform applied to 1, 0, -1/2, -1/2, -1/4, 0, ... = A046978(n+3)/A016116(n). (End)
a(n) = 2^(2*n+1)*(zeta(-n,1/2) - zeta(-n, 1)), where zeta(a, z) is the generalized Riemann zeta function. - Peter Luschny, Mar 11 2015
a(n) = 2^(n + 1)*(2^(n + 1) - 1)*Bernoulli(n + 1, 1)/(n + 1). (From Bill Gosper, Oct 28 2015) - N. J. A. Sloane, Oct 28 2015 [See the above comment from Peter Luschny, Apr 22 2009.]
a(n) = -(n mod 2)*((-1)^n + Sum_{k=1..n-1} (-1)^k*C(n,k)*a(n-k)) for n >= 1. - Peter Luschny, Jun 01 2016
a(n) = (-2)^n*F_{n}(-1/2), where F_{n}(x) is the Fubini polynomial. - Peter Luschny, May 21 2021

New name from Peter Luschny, Mar 12 2015

A110501 Unsigned Genocchi numbers (of first kind) of even index.

Original entry on oeis.org

1, 1, 3, 17, 155, 2073, 38227, 929569, 28820619, 1109652905, 51943281731, 2905151042481, 191329672483963, 14655626154768697, 1291885088448017715, 129848163681107301953, 14761446733784164001387, 1884515541728818675112649, 268463531464165471482681379

Offset: 1

Michael Somos, Jul 23 2005

The Genocchi numbers satisfy Seidel's recurrence: for n > 1, 0 = Sum_{j=0..floor(n/2)} (-1)^j*binomial(n, 2*j)*a(n-j). - Ralf Stephan, Apr 17 2004
The (n+1)-st Genocchi number is the number of Dumont permutations of the first kind on 2n letters. In a Dumont permutation of the first kind, each even integer must be followed by a smaller integer and each odd integer is either followed by a larger integer or is the last element. - Ralf Stephan, Apr 26 2004
The (n+1)-st Genocchi number is also the number of ways to place n rooks (attacking along planes; also called super rooks of power 2 by Golomb and Posner) on the three-dimensional Genocchi boards of size n. The Genocchi board of size n consists of cells of the form (i, j, k) where min{i, j} <= k and 1 <= k <= n. A rook placement on this board can also be realized as a pair of permutations of n the smallest number in the i-th position of the two permutations is not larger than i. - Feryal Alayont, Nov 03 2012
The (n+1)-st Genocchi number is also the number of Dumont permutations of the second kind, third kind, and fourth kind on 2n letters. In a Dumont permutation of the second kind, all odd positions are weak excedances and all even positions are deficiencies. In a Dumont permutation of the third kind, all descents are from an even value to an even value. In a Dumont permutation of the fourth kind, all deficiencies are even values at even positions. - Alexander Burstein, Jun 21 2019
The (n+1)-st Genocchi number is also the number of semistandard Young tableaux of skew shape (n+1,n,...,1)/(n-1,n-2,...,1) such that the entries in row i are at most i for i=1,...,n+1. - Alejandro H. Morales, Jul 26 2020
The (n+1)-st Genocchi number is also the number of positive terms of the Okounkov-Olshanski formula for the number of standard tableaux of skew shape (n+1,n,n-1,...,1)/(n-1,n-2,...,1), given by the (2n+1)-st Euler number A000111. - Alejandro H. Morales, Jul 26 2020
The (n+1)-st Genocchi number is also the number of collapsed permutations in (2n-1) letters. A permutation pi of size 2n-1 is said to be collapsed if ceil(k/2) <= pi^{-1}(k) <= n + floor(k/2). There are 3 collapsed permutations of size 3, namely 123, 132 and 213. - Arvind Ayyer, Oct 23 2020

			E.g.f.: x*tan(x/2) = x^2/2! + x^4/4! + 3*x^6/6! + 17*x^8/8! + 155*x^10/10! + ...
O.g.f.: A(x) = x + x^2 + 3*x^3 + 17*x^4 + 155*x^5 + 2073*x^6 + ...
where A(x) = x + x^2/(1+x) + 2!^2*x^3/((1+x)*(1+4*x)) + 3!^2*x^4/((1+x)*(1+4*x)*(1+9*x)) + 4!^2*x^5/((1+x)*(1+4*x)*(1+9*x)*(1+16*x)) + ... . - _Paul D. Hanna_, Jul 21 2011
From _Gary W. Adamson_, Jul 19 2011: (Start)
The first few rows of production matrix M are:
  1, 2,  0,  0,  0, 0, ...
  1, 3,  3,  0,  0, 0, ...
  1, 4,  6,  4,  0, 0, ...
  1, 5, 10, 10,  5, 0, ...
  1, 6, 15, 20, 15, 6, ... (End)

L. Carlitz, A conjecture concerning Genocchi numbers. Norske Vid. Selsk. Skr. (Trondheim) 1971, no. 9, 4 pp. MR0297697 (45 #6749)
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 49.
Leonhard Euler, Institutionum Calculi Differentialis, volume 2 (1755), para. 181.
A. Genocchi, Intorno all'espressione generale de'numeri Bernulliani, Ann. Sci. Mat. Fis., 3 (1852), 395-405.
R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2 (1999) p. 74; see Problem 5.8.

Alan Sokal, Table of n, a(n) for n = 1..250 (terms up to a(100) from Alois P. Heinz)
F. Alayont and N. Krzywonos, Rook Polynomials in Three and Higher Dimensions, to appear in Involve
F. Alayont, R. Moger-Reischer and R. Swift, Rook Number Interpretations of Generalized Central Factorial and Genocchi Numbers, preprint, 2012.
R. C. Archibald, Review of Terrill-Terrill paper, Math. Comp., 1 (1945), pp. 385-386.
A. Ayyer, D. Hathcock and P. Tetali, Toppleable Permutations, Excedances and Acyclic Orientations, arXiv:2010.11236 [math.CO], 2020.
Peter Bala, A triangle for calculating the Genocchi numbers
Ange Bigeni, A bijection between the irreducible k-shapes and the surjective pistols of height k-1, arXiv preprint arXiv:1402.1383 [math.CO] (2014). Also Discrete Math., 338 (2015), 1432-1448.
Ange Bigeni, Enumerating the symplectic Dellac configurations, arXiv:1705.03804 [math.CO], 2017.
Ange Bigeni, The universal sl2 weight system and the Kreweras triangle, arXiv:1712.05475 [math.CO], 2017.
Ange Bigeni, Combinatorial interpretations of the Kreweras triangle in terms of subset tuples, arXiv:1712.01929 [math.CO], 2017.
Ange Bigeni, A generalization of the Kreweras triangle through the universal sl_2 weight system, Journal of Combinatorial Theory, Series A (2019) Vol. 161, 309-326.
A. Burstein, M. Josuat-Vergès and W. Stromquist, New Dumont permutations, Pure Math. Appl. (Pu.M.A.) 21 (2010), no. 2, 177-206.
E. Clark and R. Ehrenborg, The excedance algebra, Discr. Math., 313 (2013), 1429-1435.
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. See pp. 4, 12.
Bishal Deb, Continued fractions using a Laguerre digraph interpretation of the Foata-Zeilberger bijection and its variants, arXiv:2304.14487 [math.CO], 2023. See pp. 4, 11.
D. Dumont, Sur une conjecture de Gandhi concernant les nombres de Genocchi, Discrete Mathematics 1 (1972) 321-327.
D. Dumont, Interpretations combinatoires des nombres de Genocchi, Duke Math. J., 41 (1974), 305-318.
Dominique Dumont and Dominique Foata, Une propriété de symétrie des nombres de Genocchi Bull. Soc. Math. France 104 (1976), no. 4, 433-451. MR0434830 (55 #7794)
D. Dumont and G. Viennot, A combinatorial interpretation of the Seidel generation of Genocchi numbers, Preprint, Annotated scanned copy.
Dominique Dumont and Gérard Viennot, A combinatorial interpretation of the Seidel generation of Genocchi numbers, Combinatorial mathematics, optimal designs and their applications (Proc. Sympos. Combin. Math. and Optimal Design, Colorado State Univ., Fort Collins, Colo., 1978). Ann. Discrete Math. 6 (1980), 77-87. MR0593524 (82j:10024).
A. L. Edmonds and S, Klee, The combinatorics of hyperbolized manifolds, arXiv:1210.7396 [math.CO], 2012.
Richard Ehrenborg and Einar Steingrímsson, Yet another triangle for the Genocchi numbers, European J. Combin. 21 (2000), no. 5, 593-600. MR1771988 (2001h:05008)
Sen-Peng Eu, Tung-Shan Fu, Hsin-Hao Lai, and Yuan-Hsun Lo, Gamma-positivity for a Refinement of Median Genocchi Numbers, arXiv:2103.09130 [math.CO], 2021.
Vincent Froese and Malte Renken, Terrain-like Graphs and the Median Genocchi Numbers, arXiv:2210.16281 [math.CO], 2022.
J. M. Gandhi, Research Problems: A Conjectured Representation of Genocchi Numbers, Amer. Math. Monthly 77 (1970), no. 5, 505-506. MR1535914
Ira M. Gessel, On the Almkvist-Meurman Theorem for Bernoulli Polynomials, Integers (2023) Vol. 23, #A14.
S. W. Golomb and E. C. Posner, Rook Domains, Latin Squares, Affine Planes, and Error-Distributing Codes, Transactions of the Information Theory Group of the IEEE, Vol. 10, No. 3 (1964), 196-208.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Guo-Niu Han and Jing-Yi Liu, Combinatorial proofs of some properties of tangent and Genocchi numbers, European Journal of Combinatorics, Vol. 71 (2018), pp. 99-110; arXiv preprint, arXiv:1707.08882 [math.CO], 2017-2018.
Florent Hivert and Olivier Mallet, Combinatorics of k-shapes and Genocchi numbers, in FPSAC 2011, Reykjavik, Iceland DMTCS proc. AO, 2011, 493-504.
Alexander Lazar and Michelle L. Wachs, The Homogenized Linial Arrangement and Genocchi Numbers, arXiv:1910.07651 [math.CO], 2019.
Zhicong Lin and Sherry H.F. Yan, Cycles on a multiset with only even-odd drops, arXiv:2108.03790 [math.CO], 2021. See also Disc. Math. (2022) Vol. 345, No. 2, 112683.
A. H. Morales and D. G. Zhu, On the Okounkov--Olshanski formula for standard tableaux of skew shape, arXiv:2007.05006 [math.CO], 2020.
Qiongqiong Pan and Jiang Zeng, Cycles of even-odd drop permutations and continued fractions of Genocchi numbers, arXiv:2108.03200 [math.CO], 2021.
John Riordan and Paul R. Stein, Proof of a conjecture on Genocchi numbers, Discrete Math. 5 (1973), 381-388. MR0316372 (47 #4919) - From _N. J. A. Sloane_, Jun 12 2012
Alan Sokal, Table of n, a(n) for n = 1..10000 [315 MB file]
H. M. Terrill and E. M. Terrill, Tables of numbers related to the tangent coefficients, J. Franklin Inst., 239 (1945), 66-67.

Cf. A000182, A001469, A002105, A005439, A036968, A211183, A297703.

[Abs(2*(4^n-1)*Bernoulli(2*n)): n in [1..20]]; // Vincenzo Librandi, Jul 28 2017

A110501 := proc(n)
    2*(-1)^n*(1-4^n)*bernoulli(2*n) ;
end proc:
seq(A110501(n),n=0..10) ; # R. J. Mathar, Aug 02 2013

a[n_] := 2*(4^n - 1) * BernoulliB[2n] // Abs; Table[a[n], {n, 19}] (* Jean-François Alcover, May 23 2013 *)

{a(n) = if( n<1, 0, 2 * (-1)^n * (1 - 4^n) * bernfrac( 2*n))};

{a(n) = if( n<1, 0, (2*n)! * polcoeff( x * tan(x/2 + x * O(x^(2*n))), 2*n))};

{a(n)=polcoeff(sum(m=0,n,m!^2*x^(m+1)/prod(k=1,m, 1+k^2*x+x*O(x^n))),n)} /* Paul D. Hanna, Jul 21 2011 */

upto(n) = my(v1, v2, v3); v1 = vector(n, i, 0); v1[1] = 1; v2 = vector(n-1, i, ((i+1)^2)\4); v3 = v1; for(i=2, n, 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 28 2025

from sympy import bernoulli
def A110501(n): return ((2<<(m:=n<<1))-2)*abs(bernoulli(m)) # Chai Wah Wu, Apr 14 2023

# Algorithm of L. Seidel (1877)
# n -> [a(1), ..., a(n)] for n >= 1.
def A110501_list(n) :
    D = []; [D.append(0) for i in (0..n+2)]; D[1] = 1
    R = [] ; 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]
        else :
            for k in range(1,h+1,1) :  D[k] += D[k-1]
        b = not b
        if b : R.append(D[h])
    return R
A110501_list(19) # Peter Luschny, Apr 01 2012

[2*(-1)^n*(1-4^n)*bernoulli(2*n) for n in (1..20)] # G. C. Greubel, Nov 28 2018

(-1)^n * a(n) = A036968(2*n) = A001469(n).
a(n) = 2*(-1)^n*(1-4^n)*B_{2*n} (B = A027641/A027642 are Bernoulli numbers).
A002105(n) = 2^(n-1)/n * a(n). - Don Knuth, Jan 16 2007
A000111(2*n-1) = a(n)*2^(2*n-2)/n. - Alejandro H. Morales, Jul 26 2020
E.g.f.: x * tan(x/2) = Sum_{k > 0} a(k) * x^(2*k) / (2*k)!.
E.g.f.: x * tan(x/2) = x^2 / (2 - x^2 / (6 - x^2 / (... 4*k+2 - x^2 / (...)))). - Michael Somos, Mar 13 2014
O.g.f.: Sum_{n >= 0} n!^2 * x^(n+1) / Product_{k = 1..n} (1 + k^2*x). - Paul D. Hanna, Jul 21 2011
a(n) = Sum_{k = 0..2*n} (-1)^(n-k+1)*Stirling2(2*n, k)*A059371(k). - Vladeta Jovovic, Feb 07 2004
O.g.f.: A(x) = x/(1-x/(1-2*x/(1-4*x/(1-6*x/(1-9*x/(1-12*x/(... -[(n+1)/2]*[(n+2)/2]*x/(1- ...)))))))) (continued fraction). - Paul D. Hanna, Jan 16 2006
a(n) = Pi^(-2*n)*integral(log(t/(1-t))^(2*n)-log(1-1/t)^(2*n) dt,t=0,1). - Gerry Martens, May 25 2011
a(n) = the upper left term of M^(n-1); M is an infinite square production matrix with M[i,j] = C(i+1,j-1), i.e., Pascal's triangle without the first two rows and right border, see the examples and Maple program. - Gary W. Adamson, Jul 19 2011
G.f.: 1/U(0) where U(k) = 1 + 2*(k^2)*x - x*((k+1)^2)*(x*(k^2)+1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Sep 15 2012
a(n+1) = Sum_{k=0..n} A211183(n, k)*2^(n-k). - Philippe Deléham, Feb 03 2013
G.f.: 1 + x/(G(0)-x) where G(k) =  2*x*(k+1)^2 + 1 - x*(k+2)^2*(x*k^2+2*x*k+x+1)/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Feb 10 2013
G.f.: G(0) where G(k) = 1 + x*(2*k+1)^2/( 1 + x + 4*x*k + 4*x*k^2 - 4*x*(k+1)^2*(1 + x + 4*x*k + 4*x*k^2)/(4*x*(k+1)^2 + (1 + 4*x + 8*x*k + 4*x*k^2)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 11 2013
G.f.: R(0), where R(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/(1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/R(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 27 2013
E.g.f. (offset 1): sqrt(x)*tan(sqrt(x)/2) = Q(0)*x/2, where Q(k) = 1 - x/(x - 4*(2*k+1)*(2*k+3)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Jan 06 2014
Pi^2/6 = 2*Sum_{k=1..N} (-1)^(k-1)/k^2 + (-1)^N/N^2(1 - 1/N + 1/N^3 - 3/N^5 + 17/N^7 - 155/N^9 +- ...), where the terms in the parenthesis are (-1)^n*a(n)/N^(2n-1). - M. F. Hasler, Mar 11 2015
a(n) = 2*n*|euler(2*n-1, 0)|. - Peter Luschny, Jun 09 2016
a(n) = 4^(1-n) * (4^n-1) * Pi^(-2*n) * (2*n)! * zeta(2*n). - Daniel Suteu, Oct 14 2016
a(n) ~ 8*Pi*(2^(2*n)-1)*(n/(Pi*exp(1)))^(2*n+1/2)*exp(1/2+(1/24)/n-(1/2880)/n^3+(1/40320)/n^5+...). [Given in A001469 by Peter Luschny, Jul 24 2013, copied May 14 2022.]
a(n) = A000182(n) * n / 4^(n-1) (Han and Liu, 2018). - Amiram Eldar, May 17 2024

Edited by M. F. Hasler, Mar 22 2015

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 3, 3, 1, 5, 8, 6, 4, 1, 16, 25, 20, 10, 5, 1, 61, 96, 75, 40, 15, 6, 1, 272, 427, 336, 175, 70, 21, 7, 1, 1385, 2176, 1708, 896, 350, 112, 28, 8, 1, 7936, 12465, 9792, 5124, 2016, 630, 168, 36, 9, 1, 50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1

Offset: 0

Philippe Deléham, Aug 27 2005

The boustrophedon transform {t} of a sequence {s} is given by t_n = Sum_{k=0..n} T(n,k)*s(k). Triangle may be called the boustrophedon triangle.
The 'signed version' of the triangle is the exponential Riordan array [sech(x) + tanh(x), x]. - Peter Luschny, Jan 24 2009
Up to signs, the matrix is self-inverse: T^(-1)(n,k) = (-1)^(n+k)*T(n,k). - R. J. Mathar, Mar 15 2013

			Triangle starts:
      1;
      1,     1;
      1,     2,     1;
      2,     3,     3,     1;
      5,     8,     6,     4,     1;
     16,    25,    20,    10,     5,    1;
     61,    96,    75,    40,    15,    6,    1;
    272,   427,   336,   175,    70,   21,    7,   1;
   1385,  2176,  1708,   896,   350,  112,   28,   8,  1;
   7936, 12465,  9792,  5124,  2016,  630,  168,  36,  9,  1;
  50521, 79360, 62325, 32640, 12810, 4032, 1050, 240, 45, 10, 1; ...

Reinhard Zumkeller, Rows n = 0..125 of table, flattened
Peter Luschny, The Swiss-Knife polynomials.
J. Millar, N. J. A. Sloane and N. E. Young, A new operation on sequences: the Boustrophedon transform, J. Combin. Theory, 17A 44-54 1996 (Abstract, pdf, ps).
Wikipedia, Boustrophedon transform
Index entries for sequences related to boustrophedon transform

Cf. A000111, A000667, A000795, A002084, A003719, A007318, A009747.
See also : A000182, A000964, A009739, A062161, A062272.
Cf. A153641, A162660.
Cf. A000667 (row sums), A247453.

a109449 n k = a109449_row n !! k
a109449_row n = zipWith (*)
                (a007318_row n) (reverse $ take (n + 1) a000111_list)
a109449_tabl = map a109449_row [0..]
-- Reinhard Zumkeller, Nov 02 2013

f:= func< n,x | Evaluate(BernoulliPolynomial(n+1), x) >;
A109449:= func< n,k | k eq n select 1 else 2^(2*n-2*k+1)*Binomial(n,k)*Abs(f(n-k,3/4) - f(n-k,1/4) + f(n-k,1) - f(n-k,1/2))/(n-k+1) >;
[A109449(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Jul 10 2025

From Peter Luschny, Jul 10 2009, edited Jun 06 2022: (Start)
A109449 := (n,k) -> binomial(n, k)*A000111(n-k):
seq(print(seq(A109449(n, k), k=0..n)), n=0..9);
B109449 := (n,k) -> 2^(n-k)*binomial(n, k)*abs(euler(n-k, 1/2)+euler(n-k, 1)) -`if`(n-k=0, 1, 0): seq(print(seq(B109449(n, k), k=0..n)), n=0..9);
R109449 := proc(n, k) option remember; if k = 0 then A000111(n) else R109449(n-1, k-1)*n/k fi end: seq(print(seq(R109449(n, k), k=0..n)), n=0..9);
E109449 := proc(n) add(binomial(n, k)*euler(k)*((x+1)^(n-k)+ x^(n-k)), k=0..n) -x^n end: seq(print(seq(abs(coeff(E109449(n), x, k)), k=0..n)), n=0..9);
sigma := n -> ifelse(n=0, 1, [1,1,0,-1,-1,-1,0,1][n mod 8 + 1]/2^iquo(n-1,2)-1):
L109449 := proc(n) add(add((-1)^v*binomial(k, v)*(x+v+1)^n*sigma(k), v=0..k), k=0..n) end: seq(print(seq(abs(coeff(L109449(n), x, k)), k=0..n)), n=0..9);
X109449 := n -> n!*coeff(series(exp(x*t)*(sech(t)+tanh(t)), t, 24), t, n): seq(print(seq(abs(coeff(X109449(n), x, k)), k=0..n)), n=0..9);
(End)

lim = 10; s = CoefficientList[Series[(1 + Sin[x])/Cos[x], {x, 0, lim}], x] Table[k!, {k, 0, lim}]; Table[Binomial[n, k] s[[n - k + 1]], {n, 0, lim}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 24 2015, after Jean-François Alcover at A000111 *)
T[n_, k_] := (n!/k!) SeriesCoefficient[(1 + Sin[x])/Cos[x], {x, 0, n - k}]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 27 2019 *)

A109449(n,k)=binomial(n,k)*if(n>k,2*abs(polylog(k-n,I)),1) \\ M. F. Hasler, Oct 05 2017

R = PolynomialRing(ZZ, 'x')
@CachedFunction
def skp(n, x) :
    if n == 0 : return 1
    return add(skp(k, 0)*binomial(n, k)*(x^(n-k)-(n+1)%2) for k in range(n)[::2])
def A109449_row(n):
    x = R.gen()
    return [abs(c) for c in list(skp(n,x)-skp(n,x-1)+x^n)]
for n in (0..10) : print(A109449_row(n)) # Peter Luschny, Jul 22 2012

Sum_{k>=0} T(n, k) = A000667(n).
Sum_{k>=0} T(2n, 2k) = A000795(n).
Sum_{k>=0} T(2n, 2k+1) = A009747(n).
Sum_{k>=0} T(2n+1, 2k) = A003719(n).
Sum_{k>=0} T(2n+1, 2k+1) = A002084(n).
Sum_{k>=0} T(n, 2k) = A062272(n).
Sum_{k>=0} T(n, 2k+1) = A062161(n).
Sum_{k>=0} (-1)^(k)*T(n, k) = A062162(n). - Johannes W. Meijer, Apr 20 2011
E.g.f.: exp(x*y)*(sec(x)+tan(x)). - Vladeta Jovovic, May 20 2007
T(n,k) = 2^(n-k)*C(n,k)*|E(n-k,1/2) + E(n-k,1)| - [n=k] where C(n,k) is the binomial coefficient, E(m,x) are the Euler polynomials and [] the Iverson bracket. - Peter Luschny, Jan 24 2009
From Reikku Kulon, Feb 26 2009: (Start)
A109449(n, 0) = A000111(n), approx. round(2^(n + 2) * n! / Pi^(n + 1)).
A109449(n, n - 1) = n.
A109449(n, n) = 1.
For n > 0, k > 0: A109449(n, k) = A109449(n - 1, k - 1) * n / k. (End)
From Peter Luschny, Jul 10 2009: (Start)
Let p_n(x) = Sum_{k=0..n} Sum_{v=0..k} (-1)^v C(k,v)*F(k)*(x+v+1)^n, where F(0)=1 and for k>0 F(k)=-1 + s_k 2^floor((k-1)/2), s_k is 0 if k mod 8 in {2,6}, 1 if k mod 8 in {0,1,7} and otherwise -1. T(n,k) are the absolute values of the coefficients of these polynomials.
Another way to express the polynomials p_n(x) is
p_n(x) = -x^n + Sum_{k=0..n} binomial(n,k)*Euler(k)((x+1)^(n-k) + x^(n-k)). (End)
From Peter Bala, Jan 26 2011: (Start)
An explicit formula for the n-th row polynomial is
x^n + i*Sum_{k=1..n}((1+i)/2)^(k-1)*Sum_{j=0..k} (-1)^j*binomial(k,j)*(x+i*j)^n, where i = sqrt(-1). This is the triangle of connection constants between the polynomial sequences {Z(n,x+1)} and {Z(n,x)}, where Z(n,x) denotes the zigzag polynomials described in A147309.
Denote the present array by M. The first column of the array (I-x*M)^-1 is a sequence of rational functions in x whose numerator polynomials are the row polynomials of A145876 - the generalized Eulerian numbers associated with the zigzag numbers. (End)
Let skp{n}(x) denote the Swiss-Knife polynomials A153641. Then
T(n,k) = [x^(n-k)] |skp{n}(x) - skp{n}(x-1) + x^n|. - Peter Luschny, Jul 22 2012
T(n,k) = A007318(n,k) * A000111(n - k), k = 0..n. - Reinhard Zumkeller, Nov 02 2013
T(n,k) = abs(A247453(n,k)). - Reinhard Zumkeller, Sep 17 2014

Edited, formula corrected, typo T(9,4)=2016 (before 2816) fixed by Peter Luschny, Jul 10 2009

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

A002425 Denominator of Pi^(2n)/(Gamma(2n)(1-2^(-2n))zeta(2n)).

Original entry on oeis.org

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 968383680827, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961

Offset: 1

N. J. A. Sloane

Differs from the absolute values of A275994 the first time at index 60.
Consider the C(k)-summation process for divergent series: the series Sum((-1)^n*(n+1)^k) == 1 - 2^k + 3^k - 4^k + ..., summable C(1) to the value 1/2 for k = 0, is for each k >= 1 exactly summable C(k+1) to the sum s(k+1) = (2^(k+1)-1)*B(k+1)/ (k+1) and so a(n) = abs(numerator(s(2n))). - Benoit Cloitre, Apr 27 2002
Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - Ralf Stephan, Dec 21 2004
(-1)^n*a(n+1) is the numerator of Euler(2n+1,1). - N. J. A. Sloane, Nov 10 2009 (a misprint corrected by Vladimir Shevelev, Sep 18 2017)
a(n) is the absolute value of the constant term of the Euler polynomial E_{2n-1} times the even part of 2n. - Peter Luschny, Nov 26 2010
From Vladimir Shevelev, Aug 31 2017: (Start)
Let E_m(x) = x^m + Sum_{odd k=1..m} e_k(m)*x^(m-k) be the Euler polynomial, let 2*n-1 <= m. Show that the expression c(m,n) = |e_(2*n-1)(m)|/binomial(m,2*n-1) does not depend on m and c(m,n) = a(n)/A006519(2*n). Indeed, by the formula in the Shevelev link |e_(2*n-1)(m)| = binomial(m,2*n-1)*(4^n-1)*B_(2*n)/n. On the other hand, by Cloitre's formula, we have a(n) = (4^n-1)*|B_(2*n)|*2^A001511(n) /n. Taking into account that 2^A001511 = A006519(2*n) we obtain the claimed equality. Since sign(e_k(n)) = (-1)^((k+1)/2), we have the following application of the sequence: e_k(n) = (-1)^((k+1)/2))*a((k+1)/2)*binomial(n,k)/A006519(k+1). (End)

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. 73.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.
Konrad Knopp, Theory and application of infinite series, Divergent series, Dover, p. 479
L. Oettinger, Archiv. Math. Phys., 26 (1856), see esp. p. 5.
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).

N. J. A. Sloane, Table of n, a(n) for n = 1..300
H. Cohn, Some elementary aspects of modular functions in several variables, Bull. Am. Math. Soc., Sept. 1965, 681ff, esp. p. 688.
Ren Guan, K_0 groups of noncommutative R^2n, arXiv:2208.06253 [math.RA], 2022. See p. 22.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51. [Annotated scanned copy of pages 38-51 only, plus notes]
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Vladimir Shevelev, On a Luschny question, arXiv:1708.08096 [math.NT], 2017.

Numerator given by A037239.
Different from A089171, A275994.
Cf. A048896, A089170, A089171, A160469, A335956.

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

A002425 := n -> (-1)^n*euler(2*n-1,0)*2^padic[ordp](2*n,2); # Peter Luschny, Nov 26 2010
A002425_list := proc(n) 1/(1+1/exp(z)); series(%,z,2*n+4);
seq(numer((-1)^i*(2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end;
A002425_list(17); # Peter Luschny, Jul 12 2012

a[n_]:= (-1)^(n-1) * Numerator[EulerE[2n-1, 1]]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Sep 20 2011, after N. J. A. Sloane's comment *)
a[n_]:= If[n<1, 0, With[{m = 2n-1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* Michael Somos, Sep 14 2013 *)
Table[2*(4^n-1)*Zeta[1-2n] // Abs // Numerator, {n, 1, 20}] (* Jean-François Alcover, Oct 16 2013 *)

for(n=1,20,print1(abs(numerator(2*bernfrac(2*n)*(4^n-1)/(2*n))),","))

a(n)=if(n<1,0,(-1)^n/n*(1-4^n)*bernfrac(2*n)*2^valuation(2*n,2))

a(n)=(-1)^n*4*bitand(n,-n)*polylog(1-2*n,-1); \\ Peter Luschny, Nov 22 2012

def A002425_list(n):
    T = [0]*n; T[0] = 1; S = [0]*n; k2 = 0
    for k in (1..n-1): T[k] = k*T[k-1]
    for k in (1..n):
        if is_odd(k): S[k-1] = 4*k2; k2 += 1
        else: S[k-1] = S[k2-1]+2*k2-1
        for j in (k..n-1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
    return [T[j]>>S[j] for j in (0..n-1)]
A002425_list(20)  # Peter Luschny, Nov 17 2012

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

a(n) = (-1)^n/n*(1 - 4^n)*B(2*n)*2^A001511(n) where B(k) denotes the k-th Bernoulli number. - Benoit Cloitre, Dec 30 2003
This is different from the sequence of numerators of the expansion of cosec(x) - cot(x) - see A089171.
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))).
Equals A160469(n)/A048896(n-1).
Equals A089171(n)*A089170(n-1). (End)
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - Peter Luschny, Jul 12 2012
a(n) = numerator(abs(2*(4^n-1)*zeta(1-2*n))). - Jean-François Alcover, Oct 16 2013
For every positive integers n,k we have a(n) = (-1)^(n+k)*N(2*n-1,k) + 2*(-1)^(n-1)*A006519(2*n)*(1^(2*n-1)-2^(2*n-1)+..+(-1)^k*(k-1)^(2*n-1)), where N(n,k) is the numerator of Euler(n,k). So, the right hand side is an invariant of k. - Vladimir Shevelev, Sep 19 2017
a(n) = numerator(r(n)) where r(n) = (-1)^binomial(2*n, 2)*Sum_{k=1..2*n}(-1)^k*Stirling2(2*n, k)*2^(-k)*(k-1)!. - Peter Luschny, May 24 2020
a(n) = 2*(-1)^n*A335956(2*n)*zeta(1-2*n). - Peter Luschny, Aug 30 2020

The n=15 term was formerly incorrectly given as 86125672563301143.

Formula and cross-references edited by Johannes W. Meijer, May 21 2009

A009006 Expansion of e.g.f.: 1 + tan(x).

Original entry on oeis.org

1, 1, 0, 2, 0, 16, 0, 272, 0, 7936, 0, 353792, 0, 22368256, 0, 1903757312, 0, 209865342976, 0, 29088885112832, 0, 4951498053124096, 0, 1015423886506852352, 0, 246921480190207983616, 0, 70251601603943959887872, 0, 23119184187809597841473536, 0

Offset: 0

R. H. Hardin

R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 1, 1997; See Exercise 1.41(d).

G. C. Greubel, Table of n, a(n) for n = 0..451
Kwang-Wu Chen, An Interesting Lemma for Regular C-fractions, J. Integer Seqs., Vol. 6, 2003.

A000182(n) = a(2n-1).

m:=30; R:=PowerSeriesRing(Rationals(), m); b:=Coefficients(R!(1 + Tan(x))); [Factorial(n-1)*b[n]: n in [1..m]]; // G. C. Greubel, Jul 21 2018

u:=proc(n) if n=0 then 1 else -add(u(k)*binomial(n,k)/2*2^(n-k),k=0..n-1) fi end;seq(u(n),n=0..15); # Robert FERREOL, Dec 30 2006

a[m_] := Abs[Sum[(-2)^(m-k) k! StirlingS2[m,k], {k,0,m}]]; Table[a[i], {i,0,20}] (* Peter Luschny, Apr 29 2009 *)
A009006[n_] :=  Cos[Pi (n-1) / 2] (4^(n+1) - 2^(n+1)) * BernoulliB[n+1] / (n+1); a[0] := 1; Table[A009006[n], {n, 0, 30}] (* Peter Luschny, Jun 14 2021 *)

a(n)=if(n<1,n==0,n!*polcoeff(tan(x+x*O(x^n)),n))

def A009006(n) :
    if n == 0 : return 1
    return add(add((-1)^(n//2+j+1)*binomial(n+1,k-j)*j^n for j in (0..k)) for k in (1..n))
[A009006(n) for n in (0..26)] # Peter Luschny, Jul 23 2012

Let b(n) be a(n) shifted one place to the left with b(2+4k) = -a(3+4k), k=0, 1, .. Then b(n) is the expansion of sech(x)^2. - Mario Catalani (mario.catalani(AT)unito.it), Feb 08 2003
g(x) = x + x^2 - 2*x^4 + 16*x^6 - 272*x^8 + ... satisfies g(x/(1+2x)) = -g(-x).
E.g.f.: 1 + tan(x).
E.g.f. exp(x)*sech(x) is 1,1,0,-2,0,16,0,-272,... (A155585). - Paul Barry, Mar 15 2006
From Robert FERREOL, Dec 30 2006: (Start)
a(n) = 2^n*abs(Euler(n,0)) where Euler(n,x) is the n-th Eulerian polynomial.
a(n) = abs(u(n)) where u(n) = -Sum_{k=0..n-1} u(k)*binomial(n, k)*2^(n-k-1) with u(0) = 1.  (End)
Sum_{k=0..n} A075263(n, k)*2^k = 1,-1,0,2,0,-16,0,272,0,-7936,0,... for n = 0, 1, 2, 3, 4, ..., respectively. - Philippe Deléham, Aug 20 2007
E.g.f. -log(cos(x)), for n > 0. - Vladimir Kruchinin, Aug 09 2010
a(n) = Sum_{k=1..n} Sum_{j=0..k} (-1)^(floor(n/2)+j+1)*binomial(n+1,k-j)*j^n for n > 0. - Peter Luschny, Jul 23 2012
From Sergei N. Gladkovskii, Oct 25 2012 - Dec 20 2013: (Start)
Continued fractions:
G.f.: 1 + x/T(0) where T(k) = 1 - (k+1)*(k+2)*x^2/T(k+1).
E.g.f.: 1 + tan(x) = 1+x/(U(0)-x) where U(k)= 4*k+1 + x/(1+x/(4*k+3 - x/(1- x/U(k+1)))).
E.g.f.: 1+tan(x) = 1 - 3*x/(U(0) + 3*x^2) where U(k) = 64*k^3 + 48*k^2 - 4*k*(2*x^2+1) - 2*x^2 - 3 - x^4*(4*k-1)*(4*k+7)/U(k+1).
E.g.f.: 1+x*G(0) where G(k) = 1 - x^2/(x^2 - (2*k+1)*(2*k+3)/G(k+1)).
G.f.: 1 + x/G(0) where G(k) = 1 - 2*x^2*(4*k^2+4*k+1)-4*x^4*(k+1)^2*(4*k^2+8*k+3) /G(k+1).
G.f.: 1 + x*Q(0) where Q(k) = 1 - x^2*(k+1)*(k+2)/(x^2*(k+1)*(k+2) - 1/Q(k+1)).
G.f.: Q(0) where Q(k) = 1 + x*(k+1)/(x*(k+1)+1/(1- x*(k+1)/(x*(k+1) - 1/Q(k+1)))).
E.g.f.: 2 - 1/Q(0) where Q(k) = 1 + x/(4*k+1 - x/(1 - x/(4*k+3 + x/Q(k+1)))). (End)
a(n) ~ 2*n!*(2/Pi)^(n+1) if n is odd. - Vaclav Kotesovec, Jun 01 2013
a(n) = i^(n+1) * 2^n * ((-1)^n-1) * (2^(n+1)-1) * Bernoulli(n+1)/(n+1), n > 0. - Benedict W. J. Irwin, May 27 2016
a(0) = a(1) = 1; a(n) = -2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k) * a(n-k-1). - Ilya Gutkovskiy, Jul 05 2020

Reformatted Mar 15 1997
Definition corrected by Joerg Arndt, Apr 29 2011
Terms a(26) onward added by G. C. Greubel, Jul 21 2018

A198631 Numerators of the rational sequence with e.g.f. 1/(1+exp(-x)).

Original entry on oeis.org

1, 1, 0, -1, 0, 1, 0, -17, 0, 31, 0, -691, 0, 5461, 0, -929569, 0, 3202291, 0, -221930581, 0, 4722116521, 0, -968383680827, 0, 14717667114151, 0, -2093660879252671, 0, 86125672563201181, 0, -129848163681107301953, 0, 868320396104950823611, 0

Offset: 0

Wolfdieter Lang, Oct 31 2011

Numerators of the row sums of the Euler triangle A060096/A060097.
The corresponding denominator sequence looks like 2*A006519(n+1) when n is odd.
Also numerator of the value at the origin of the n-th derivative of the standard logistic function. - Enrique Pérez Herrero, Feb 15 2016

			The rational sequence r(n) = a(n) / A006519(n+1) starts:
1, 1/2, 0, -1/4, 0, 1/2, 0, -17/8, 0, 31/2, 0, -691/4, 0, 5461/2, 0, -929569/16, 0, 3202291/2, 0, -221930581/4, 0, 4722116521/2, 0, -968383680827/8, 0, 14717667114151/2, 0, -2093660879252671/4, ...

Robert Israel, Table of n, a(n) for n = 0..550
Eric Weisstein's World of Mathematics, Sigmoid Function.
Wikipedia, Logistic Function.

Cf. A000182, A060096, A060097, A006519, A002425, A089171, A090681.

seq(denom(euler(i,x))*euler(i,1),i=0..33); # Peter Luschny, Jun 16 2012

Join[{1},Table[Numerator[EulerE[n,1]/(2^n-1)], {n, 34}]] (* Peter Luschny, Jul 14 2013 *)

def A198631_list(n):
    x = var('x')
    s = (1/(1+exp(-x))).series(x,n+2)
    return [(factorial(i)*s.coefficient(x,i)).numerator() for i in (0..n)]
A198631_list(34) # Peter Luschny, Jul 12 2012

# Alternatively:
def A198631_list(len):
    e, f, R, C = 2, 1, [], [1]+[0]*(len-1)
    for n in (1..len-1):
        for k in range(n, 0, -1):
            C[k] = -C[k-1] / (k+1)
        C[0] = -sum(C[k] for k in (1..n))
        R.append(numerator((e-1)*f*C[0]))
        f *= n; e <<= 1
    return R
print(A198631_list(36)) # Peter Luschny, Feb 21 2016

a(n) = numerator(sum(E(n,m),m=0..n)), n>=0, with the Euler triangle E(n,m)=A060096(n,m)/A060097(n,m).
E.g.f.: 2/(1+exp(-x)) (see a comment in  A060096).
r(n) := sum(E(n,m),m=0..n) = ((-1)^n)*sum(((-1)^m)*m!*S2(n,m)/2^m, m=0..n), n>=0, where S2 are the Stirling numbers of the second kind A048993. From the e.g.f. with y=exp(-x), dx=-y*dy, putting y=1 at the end. - Wolfdieter Lang, Nov 03 2011
a(n) = numerator(euler(n,1)/(2^n-1)) for n > 0. - Peter Luschny, Jul 14 2013
a(n) = numerator(2*(2^n-1)*B(n,1)/n) for n > 0, B(n,x) the Bernoulli polynomials. - Peter Luschny, May 24 2014
Numerators of the Taylor series coefficients 4*(2^(n+1)-1)*B(n+1)/(n+1) for n>0 of 1 + 2 * tanh(x/2) (cf. A000182 and A089171). - Tom Copeland, Oct 19 2016
a(n) = -2*zeta(-n)*A335956(n+1). - Peter Luschny, Jul 21 2020
Conjecture: r(n) = Sum_{k=0..n} A001147(k) * A039755(n, k) * (-1)^k / (k+1) where r(n) = a(n) / A006519(n+1) = (n!) * ([x^n] (2 / (1 + exp(-x)))), for n >= 0. - Werner Schulte, Feb 16 2024

New name, a simpler standalone definition by Peter Luschny, Jul 13 2012

Second comment corrected by Robert Israel, Feb 21 2016

A349264 Generalized Euler numbers, a(n) = n![x^n](sec(4x)(sin(4x) + 1)).

Original entry on oeis.org

1, 4, 16, 128, 1280, 16384, 249856, 4456448, 90767360, 2080374784, 52975108096, 1483911200768, 45344872202240, 1501108249821184, 53515555843342336, 2044143848640217088, 83285910482761809920, 3605459138582973251584, 165262072909347030040576, 7995891855149741436305408

Offset: 0

Peter Luschny, Nov 20 2021

nonn

			Exponential generating functions of generalized Euler numbers in context:
egf1 = sec(1*x)*(sin(x) + 1).
   [A000111, A000364, A000182]
egf2 = sec(2*x)*(sin(x) + cos(x)).
   [A001586, A000281, A000464]
egf3 = sec(3*x)*(sin(2*x) + cos(x)).
   [A007289, A000436, A000191]
egf4 = sec(4*x)*(sin(4*x) + 1).
   [A349264, A000490, A000318]
egf5 = sec(5*x)*(sin(x) + sin(3*x) + cos(2*x) + cos(4*x)).
   [A349265, A000187, A000320]
egf6 = sec(6*x)*(sin(x) + sin(5*x) + cos(x) + cos(5*x)).
   [A001587, A000192, A000411]
egf7 = sec(7*x)*(-sin(2*x) + sin(4*x) + sin(6*x) + cos(x) + cos(3*x) - cos(5*x)).
   [A349266, A064068, A064072]
egf8 = sec(8*x)*2*(sin(4*x) + cos(4*x)).
   [A349267, A064069, A064073]
egf9 = sec(9*x)*(4*sin(3*x) + 2)*cos(3*x)^2.
   [A349268, A064070, A064074]

Matthew House, Table of n, a(n) for n = 0..389
William Y. C. Chen, Neil J. Y. Fan, Jeffrey Y. T. Jia, The generating function for the Dirichlet series Lm(s), Mathematics of Computation, Vol. 81, No. 278, pp. 1005-1023, April 2012.
Ruth Lawrence and Don Zagier, Modular forms and quantum invariants of 3-manifolds, Asian J. Math. 3 (1999), no. 1, 93-107.
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967) 689-694.
D. Shanks, Corrigendum: Generalized Euler and class numbers, Math. Comp. 22, (1968) 699.
D. Shanks, Generalized Euler and class numbers, Math. Comp. 21 (1967), 689-694; 22 (1968), 699. [Annotated scanned copy]

Row 4 of A349271.
Cf. A000111, A001586, A007289, A349265, A001587, A349266, A349267, A349268.
Secant  bisections: A000364, A000281, A000436, A000490, A000187, A000192, A064068, A064069, A064070.
Tangent bisections: A000182, A000464, A000191, A000318, A000320, A000411, A064072, A064073, A064074.
Cf. A225147, A349429.

sec(4*x)*(sin(4*x) + 1): series(%, x, 20): seq(n!*coeff(%, x, n), n = 0..19);

m = 19; CoefficientList[Series[Sec[4*x] * (Sin[4*x] + 1), {x, 0, m}], x] * Range[0, m]! (* Amiram Eldar, Nov 20 2021 *)

seq(n)={my(x='x + O('x^(n+1))); Vec(serlaplace((sin(4*x) + 1)/cos(4*x)))} \\ Andrew Howroyd, Nov 20 2021

cp's OEIS Frontend

A003881 Decimal expansion of Pi/4.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

PARI

SageMath

Formula

Extensions

A001469 Genocchi numbers (of first kind); unsigned coefficients give expansion of x*tan(x/2).

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

Sage

Formula

A155585 a(n) = 2^n*E(n, 1) where E(n, x) are the Euler polynomials.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

PARI

PARI

Python

Sage

Sage

Formula

Extensions

A110501 Unsigned Genocchi numbers (of first kind) of even index.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

Python

Sage

Sage

Formula

Extensions

A109449 Triangle read by rows, T(n,k) = binomial(n,k)*A000111(n-k), 0 <= k <= n.

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

A002425 Denominator of Pi^(2n)/(Gamma(2n)(1-2^(-2n))zeta(2n)).