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 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. 0
 1, 2, 54, 4500, 771750, 225042300, 99843767100, 62673358948200, 52880646612543750, 57733914846094987500, 79199384385873103852500, 133357363417740148141455000, 270426506783940730406180497500, 650063718230626755784087734375000, 1827886309419060919156885553671875000 (list; graph; refs; listen; history; text; internal format)
 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 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 xrange(0, 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 CaĆ±estro, Mar 14 2017 STATUS approved

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Last modified October 22 19:53 EDT 2019. Contains 328319 sequences. (Running on oeis4.)