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.

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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

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Author

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.

Crossrefs

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

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

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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.

Crossrefs

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

A005439 Genocchi medians (or Genocchi numbers of second kind).

Original entry on oeis.org

1, 1, 2, 8, 56, 608, 9440, 198272, 5410688, 186043904, 7867739648, 401293838336, 24290513745920, 1721379917619200, 141174819474169856, 13266093250285568000, 1415974941618255921152, 170361620874699124637696, 22948071824232932086513664, 3439933090471867097102680064
Offset: 0

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Author

Keywords

Comments

a(n) is the number of Boolean functions of n variables whose ROBDD (reduced ordered binary decision diagram) contains exactly n branch nodes, one for each variable. - Don Knuth, Jul 11 2007
The earliest known reference for these numbers is Seidel (1877, pages 185 and 186). - Don Knuth, Jul 13 2007
Hankel transform of 1,1,2,8,... is A168488. - Paul Barry, Nov 27 2009
According to Hetyei [2017], alternation acyclic tournaments "are counted by the median Genocchi numbers"; an alternation acyclic tournament "does not contain a cycle in which descents and ascents alternate." - Danny Rorabaugh, Apr 25 2017
The n-th Genocchi number of the second kind is also the number of collapsed permutations in (2n) letters. A permutation pi of size 2n is said to be collapsed if 1+floor(k/2) <= pi^{-1}(k) <= n + floor(k/2). There are 2 collapsed permutations of size 4, namely 1234 and 1324. - Arvind Ayyer, Oct 23 2020
For any positive integer n, a(n) is (-1)^n times the permanent of the 2n X 2n matrix M with M(j, k) = floor((2*j-k-1)/(2*n)). This former conjecture of Luschny, inspired by a conjecture of Zhi-Wei Sun in A036968, was proven by Fu, Lin and Sun (see link). - Peter Luschny, Sep 07 2021 [updated Sep 24 2021]

References

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

Crossrefs

Programs

  • GAP
    List([1..20],n->2*(-1)^n*Sum([0..n],k->Binomial(n,k)*(1-2^(n+k+1))*Bernoulli(n+k+1))); # Muniru A Asiru, Nov 29 2018
    
  • Magma
    [2*(-1)^n*(&+[Binomial(n, k)*(1-2^(n+k+1))*Bernoulli(n+k+1): k in [0..n]]): n in [1..20]]; // G. C. Greubel, Nov 28 2018
    
  • Maple
    seq(2*(-1)^n*add(binomial(n,k)*(1 - 2^(n+k+1))*bernoulli(n+k+1), k=0..n), n=0..20); # G. C. Greubel, Oct 18 2019
  • Mathematica
    a[n_]:= 2*(-1)^(n-2)*Sum[Binomial[n, k]*(1 -2^(n+k+1))*BernoulliB[n+k+1], {k, 0, n}]; Table[a[n], {n,16}] (* Jean-François Alcover, Jul 18 2011, after PARI prog. *)
  • PARI
    a(n)=2*(-1)^n*sum(k=0,n,binomial(n,k)*(1-2^(n+k+1))* bernfrac(n+k+1))
    
  • PARI
    a(n)=local(CF=1+x*O(x^(n+2)));if(n<0,return(0), for(k=1,n+1,CF=1/(1-((n-k+1)\2+1)^2*x*CF));return(Vec(CF)[n+2])) \\ Paul D. Hanna
    
  • Python
    from math import comb
    from sympy import bernoulli
    def A005439(n): return (-2 if n&1 else 2)*sum(comb(n,k)*(1-(1<Chai Wah Wu, Apr 14 2023
  • Sage
    # Algorithm of L. Seidel (1877)
    # n -> [a(1), ..., a(n)] for n >= 1.
    def A005439_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]
            if b : R.append(D[1])
            b = not b
        return R
    A005439_list(18) # Peter Luschny, Apr 01 2012
    
  • Sage
    [2*(-1)^n*sum(binomial(n,k)*(1-2^(n+k+1))*bernoulli(n+k+1) for k in (0..n)) for n in (1..20)] # G. C. Greubel, Oct 18 2019
    

Formula

a(n) = T(n, 1) where T(1, x) = 1; T(n, x) = (x+1)*((x+1)*T(n-1, x+1)-x*T(n-1, x)); see A058942.
a(n) = A000366(n)*2^(n-1).
a(n) = 2 * (-1)^n * Sum_{k=0..n} binomial(n, k)*(1-2^(n+k+1))*B(n+k+1), with B(n) the Bernoulli numbers. - Ralf Stephan, Apr 17 2004
O.g.f.: 1 + x*A(x) = 1/(1-x/(1-x/(1-4*x/(1-4*x/(1-9*x/(1-9*x/(... -[(n+1)/2]^2*x/(1-...)))))))) (continued fraction). - Paul D. Hanna, Oct 07 2005
G.f.: (of 1,1,2,8,...) 1/(1-x-x^2/(1-5*x-16*x^2/(1-13*x-81*x^2/(1-25*x-256*x^2/(1-41*x-625*x^2/(1-... (continued fraction). - Paul Barry, Nov 27 2009
O.g.f.: Sum_{n>=0} n!*(n+1)! * x^(n+1) / Product_{k=1..n} (1 + k*(k+1)*x). - Paul D. Hanna, May 10 2012
From Sergei N. Gladkovskii, Dec 14 2011, Dec 27 2012, May 29 2013, Oct 09 2013, Oct 24 2013, Oct 27 2013: (Start)
Continued fractions:
G.f.: A(x) = 1/S(0), S(k) = 1 - x*(k+1)*(k+2)/(1 - x*(k+1)*(k+2)/S(k+1)).
G.f.: A(x) = -1/S(0), S(k) = 2*x*(k+1)^2 - 1 - x^2*(k+1)^2*(k+2)^2/S(k+1).
G.f.: A(x) = 1/G(0) where G(k) = 1 - x*(k+1)^2/(1 - x*(k+1)^2/G(k+1)).
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 1/(1 - 1/(4*x*(k+1)) + 1/G(k+1))).
G.f.: Q(0)/x - 1/x, where Q(k) = 1 - x*(k+1)^2/( x*(k+1)^2 - 1/(1 - x*(k+1)^2/( x*(k+1)^2 - 1/Q(k+1)))).
G.f.: T(0)/(1-2*x), where T(k) = 1 - x^2*((k + 2)*(k+1))^2/(x^2*((k + 2)*(k+1))^2 - (1 - 2*x*k^2 - 4*x*k - 2*x)*(1 - 2*x*k^2 - 8*x*k - 8*x)/T(k+1)).
G.f.: R(0), where R(k) = 1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/(1 - x*(k+1)*(k+2)/( x*(k+1)*(k+2) - 1/R(k+1) ))). (End)
a(n) ~ 2^(2*n+4) * n^(2*n+3/2) / (exp(2*n) * Pi^(2*n+1/2)). - Vaclav Kotesovec, Oct 28 2014
Rewriting the above: a(n) ~ 4*(2*n+1)! / Pi^(2*n+1). Compare to Genocchi numbers A110501(n) = g_n ~ 4*(2*n)! / Pi^(2*n). So these are indeed like "Genocchi medians" g_{n + 1/2}. - Alan Sokal, May 13 2022
Asymptotic expansion: a(n) ~ 4*(2*n+1)! * Pi^(-(2*n+1)) * (1 + (Pi^2/16)/n + (Pi^2 (Pi^2 - 16)/512)/n^2 + (Pi^2 (Pi^4 + 384)/24576)/n^3 + (Pi^2 (Pi^6 + 96*Pi^4 + 768*Pi^2 - 12288)/1572864)/n^4 + (Pi^2 (Pi^8 + 320*Pi^6 + 12800*Pi^4 + 491520)/125829120)/n^5 + ...) --- Proof uses binomial sum for Genocchi medians in terms of Genocchi or Bernoulli numbers, combined with leading term of convergent sum (with exponentially small corrections) for the latter. Can also check against the 10000 term a-file. - Alan Sokal, May 23 2022.
a(n) = n!^2 * [x^n*y^n] exp(x)*f(x-y), where f(x) is the derivative of the Genocchi number generating function 2*x/(exp(x)+1). - Ira M. Gessel, Jul 23 2024

Extensions

More terms and additional comments from David W. Wilson, Jan 11 2001
a(0)=1 prepended by Peter Luschny, Apr 14 2023
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