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

0,2

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

Links

Table of n, a(n) for n=0..14.

Ignacio Larrosa, El as de corazones

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A001757.

Sequence in context: A123686 A122418 A069788 * A306266 A117681 A221603

Adjacent sequences: A283675 A283676 A283677 * A283679 A283680 A283681

Keyword

nonn,easy

Author

Ignacio Larrosa Cañestro, Mar 14 2017

Status

approved