cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

References

  • 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.

Links

Crossrefs

Cf. A110501, A000182, A006846, A012509, A226158, A297703.
a(n) = -A065547(n, 1) and A065547(n+1, 2) for n >= 1.

Programs

  • Magma
    [2*(1 - 4^n) * Bernoulli(2*n): n in [1..25]]; // Vincenzo Librandi, Oct 15 2018
  • Maple
    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
  • Mathematica
    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 *)
  • PARI
    a(n)=if(n<1,0,n*=2; 2*(1-2^n)*bernfrac(n))
  • PARI
    {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 */
  • Python
    from sympy import bernoulli
def A001469(n): return (2-(2<<(m:=n<<1)))*bernoulli(m) # Chai Wah Wu, Apr 14 2023
  • Sage
    # 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

Formula

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

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.