A283678 - OEIS

%I #27 May 03 2020 03:46:10

%S 1,2,54,4500,771750,225042300,99843767100,62673358948200,

%T 52880646612543750,57733914846094987500,79199384385873103852500,

%U 133357363417740148141455000,270426506783940730406180497500,650063718230626755784087734375000,1827886309419060919156885553671875000

%N 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.

%C 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)!!.

%C Essentially a bisection of A001757. - _Giovanni Resta_, Mar 14 2017

%H Ignacio Larrosa, <a href="https://groups.google.com/d/msg/es.ciencia.matematicas/Q2F1lCN9fI4/a30fnrpwBQAJ">El as de corazones</a>

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

%F n*a(n) -2*(2*n-1)^3*a(n-1)=0. - _R. J. Mathar_, Jul 15 2017

%e 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.

%p ogf := sqrt(x) * BesselI(0, sqrt(x)/4) * BesselK(0,sqrt(x)/4) / 2;

%p simplify(subs(x=1/x, asympt(ogf, x, 20))); # _Mark van Hoeij_, Oct 24 2017

%t Table[Binomial[2n, n] Product[2n + 1 - 2i, {i, 1, n}]^2, {n, 0, 15}] (* _Indranil Ghosh_, Mar 22 2017 *)

%o (PARI) a(n)=binomial(2*n, n)*prod(i=1,n,2*n+1-2*i)^2 \\ _Charles R Greathouse IV_, Mar 14 2017

%o (Python)

%o from sympy import binomial, factorial2

%o print([binomial(2*n, n) * factorial2(2*n - 1)**2 for n in range(15)]) # _Indranil Ghosh_, Mar 22 2017

%Y Cf. A001757.

%K nonn,easy

%O 0,2

%A _Ignacio Larrosa Cañestro_, Mar 14 2017