A283678 Number of possible draws of 2n pairs of consecutive cards from a set of 4n + 1 cards, so that the card that initially occupies the central position is not selected.

1, 2, 54, 4500, 771750, 225042300, 99843767100, 62673358948200, 52880646612543750, 57733914846094987500, 79199384385873103852500, 133357363417740148141455000, 270426506783940730406180497500, 650063718230626755784087734375000, 1827886309419060919156885553671875000

Offset: 0

Author

Ignacio Larrosa Cañestro, Mar 14 2017

Keywords

• nonn
• easy

Comments

The probability that the middle card is not selected in a random draw of 2n consecutive card pairs between 4n + 1 cards is a(n)/(4n)!!.

Essentially a bisection of A001757. - Giovanni Resta, Mar 14 2017

Examples

			For n = 1, you have 5 cards (A, B, C, D, E) and you can make 2 draws of pairs of consecutive cards (AB, DE) and (DE, AB) without select C.
		

Links

• Ignacio Larrosa Cañestro, El as de corazones.

Crossrefs

Cf. A001147, A001757, A092563.

Programs

Maple

ogf := sqrt(x) * BesselI(0, sqrt(x)/4) * BesselK(0,sqrt(x)/4) / 2;
simplify(subs(x=1/x, asympt(ogf, x, 20))); # Mark van Hoeij, Oct 24 2017

Mathematica

Table[Binomial[2n, n] Product[2n + 1 - 2i, {i, 1, n}]^2, {n, 0, 15}] (* Indranil Ghosh, Mar 22 2017 *)

PARI

a(n)=binomial(2*n, n)*prod(i=1,n,2*n+1-2*i)^2 \\ Charles R Greathouse IV, Mar 14 2017

Python

from sympy import binomial, factorial2
print([binomial(2*n, n) * factorial2(2*n - 1)**2 for n in range(15)]) # Indranil Ghosh, Mar 22 2017

Formula

a(n) = binomial(2n, n)*((2n-1)!!)^2 = A092563(n)*A001147(n).

n*a(n) -2*(2*n-1)^3*a(n-1)=0. - R. J. Mathar, Jul 15 2017

a(n) ~ 2^(4*n+1)*(n/e)^(2*n)/sqrt(n*Pi). - Stefano Spezia, Mar 28 2025