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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.

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

Views

Author

Michael Somos, Jul 23 2005

Keywords

Comments

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

Examples

			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)

References

  • 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.

Links

Crossrefs

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

Programs

  • Magma
    [Abs(2*(4^n-1)*Bernoulli(2*n)): n in [1..20]]; // Vincenzo Librandi, Jul 28 2017
  • Maple
    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
  • Mathematica
    a[n_] := 2*(4^n - 1) * BernoulliB[2n] // Abs; Table[a[n], {n, 19}] (* Jean-François Alcover, May 23 2013 *)
  • PARI
    {a(n) = if( n<1, 0, 2 * (-1)^n * (1 - 4^n) * bernfrac( 2*n))};
  • PARI
    {a(n) = if( n<1, 0, (2*n)! * polcoeff( x * tan(x/2 + x * O(x^(2*n))), 2*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 */
  • PARI
    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
  • Python
    from sympy import bernoulli
def A110501(n): return ((2<<(m:=n<<1))-2)*abs(bernoulli(m)) # Chai Wah Wu, Apr 14 2023
  • Sage
    # 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
  • Sage
    [2*(-1)^n*(1-4^n)*bernoulli(2*n) for n in (1..20)] # G. C. Greubel, Nov 28 2018

Formula

(-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

Extensions

Edited by M. F. Hasler, Mar 22 2015

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