A117513 Number of ways of arranging 2*n tokens in a row, with 2 copies of each token from 1 through n, such that between every pair of tokens labeled i (i = 1..n-1) there is exactly one taken labeled i+1.

1, 2, 12, 136, 2480, 66336, 2446528, 118984832, 7378078464, 568142287360, 53189920492544, 5949749335001088, 783686338494312448, 120058889459865165824, 21166245289132322242560, 4254864627502524070395904, 967406173145278971994898432, 247007221085479721384365129728

Offset: 1

Author

Nan Zang (nzang(AT)cs.ucsd.edu), Apr 28 2006

Keywords

• nonn

Comments

From Paul Barry, Oct 12 2009: (Start)

The aerated sequence is (2^(n/2 - 1) + 0^(n/2)/2)*((1 + (-1)^n)/2)*n!*[x^n](1 + x*tan(x/2)).

Multiples of the unsigned Genocchi numbers A110501: (1, 1, 3, 17, 155,...)*(1, 2, 4, 8, 16,...). (End)

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..258
• Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
• Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]

Crossrefs

Cf. A002105, A110501, A117514, A117515.

Programs

Maple

a := n -> (-2)^n*(1 - 2^(2*n))*bernoulli(2*n);
seq(a(n), n = 1..18); # Peter Luschny, Jul 26 2021

Mathematica

Array[(-2)^#*(1 - 2^(2 #))*BernoulliB[2 #] &, 18] (* Michael De Vlieger, Jul 26 2021 *)

Sage

# Algorithm of L. Seidel (1877)
# n -> [a(1), ..., a(n)] for n >= 1.
def A117513_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)
    return R
A117513_list(15) # Peter Luschny, Jun 29 2012

Formula

G.f.: 1/(1-2*x/(1-4*x/(1-8*x/(1-12*x/(1-18*x/(1-24*x/(1-32*x/(1-.../(1-2* floor((n+2)^2/4)*x/(1-... (continued fraction). - Paul Barry, Dec 03 2009

G.f.: T(0), where T(k) = 1 - x*(2*k+2)*(k+1)/( x*(2*k+2)*(k+1) - 1/( 1 - x*(2*k+2)*(k+2)/( x*(2*k+2)*(k+2) - 1/T(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Oct 24 2013

a(n) = (-2)^n*(1 - 2^(2*n))*Bernoulli(2*n). - Peter Luschny, Jul 26 2021

Extensions

More terms from Paul Barry, Oct 12 2009