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Number of permutations of the set 1,2,..., 2n such that at least one pair of adjacent numbers in the permutation differs by n.
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%I #21 Nov 04 2023 10:07:48

%S 1,10,292,16152,1443616,189709600,34420171584,8241995095936,

%T 2517637537094656,955377719901439488,440888939541736115200,

%U 243144648530111594371072,157920570527279020394569728,119308432982412667510831095808,103738687936577909824307104989184

%N Number of permutations of the set 1,2,..., 2n such that at least one pair of adjacent numbers in the permutation differs by n.

%C IMO '89 problem 6 asks to show that a(n) > A002674(n)/2.

%H G. C. Greubel, <a href="/A159960/b159960.txt">Table of n, a(n) for n = 1..220</a>

%H IMO, <a href="http://www.imo-official.org/problems.aspx">International Mathematics Olympia</a>, see 1989, problem 6.

%F a(n) = Sum_{k=1..n} (-1)^(k-1)*binomial(2*n-k, k)*binomial(n, k)*2^k*(2*n-2*k)!.

%F Recurrence: (6*n - 17)*a(n) = 2*(n-1)*(36*n^2 - 156*n + 151)*a(n-1) - 4*(n-1)*(72*n^4 - 636*n^3 + 2062*n^2 - 2909*n + 1511)*a(n-2) + 4*(n-2)*(n-1)*(96*n^5 - 1280*n^4 + 6704*n^3 - 17208*n^2 + 21596*n - 10569)*a(n-3) + 8*(n-3)*(n-2)*(n-1)*(2*n - 7)*(6*n - 11)*a(n-4). - _Vaclav Kotesovec_, Mar 15 2014

%F a(n) ~ (1-BesselJ(0,2)) * sqrt(Pi) * 4^n * n^(2*n+1/2) / exp(2*n). - _Vaclav Kotesovec_, Mar 15 2014

%p f := proc (n) add((-1)^(k-1)*binomial(2*n-k, k)*binomial(n, k)*2^k*factorial(2*n-2*k), k = 1 .. n) end proc;

%t a[n_] := (2*n)!*(1-HypergeometricPFQ[{-n}, {1, -2*n}, -2])/2; Table[a[n], {n, 1, 15}] (* _Jean-François Alcover_, Jan 27 2014 *)

%o (PARI) a(n)=sum(k=1,n,(-1)^(k-1)*binomial(2*n-k,k)*binomial(n, k)<<k*(2*n-2*k)!)/2 \\ _Charles R Greathouse IV_, Jun 19 2013

%Y Cf. A002674.

%K easy,nice,nonn

%O 1,2

%A Ji Li (vieplivee(AT)hotmail.com), Apr 28 2009