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Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*(1-exp(x))^(1/2)).
(Formerly M2161 N0861)
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%I M2161 N0861 #35 Dec 26 2021 21:04:49

%S 1,-1,-1,2,37,329,1501,-31354,-1451967,-39284461,-737652869,560823394,

%T 1103386777549,82520245792997,4398448305245905,168910341581721494,

%U 998428794798272641,-720450682719825322809,-105099789680808769094057,-10594247095804692725600734

%N Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*(1-exp(x))^(1/2)).

%D S. M. Kerawala, Asymptotic solution of the "Probleme des menages", Bull. Calcutta Math. Soc., 39 (1947), 82-84.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H S. M. Kerawala, <a href="/A001569/a001569.pdf">Asymptotic solution of the "Probleme des menages</a>, Bull. Calcutta Math. Soc., 39 (1947), 82-84. [Annotated scanned copy]

%F Let b(n) satisfy (n-2)*b(n) - n*(n-2)*b(n-1) - n*b(n-2) = 0; write b(n) = (n!/e^2)*(1 + Sum_{r>=1} a_r/n^r).

%F a(n) = n!*Sum_{k=0..n} (-1)^k*Stirling2(n,k)/k!. - _Vladeta Jovovic_, Jul 17 2006

%F E.g.f.: 1 + x*(1 - E(0))/(1-x) where E(k) = 1 + 1/(1-x*(k+1))/(k+1)/(1-x/(x-1/E(k+1) )); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Jan 19 2013

%F E.g.f.: 1 + x*(1 - S)/(1-x) where S = Sum_{k>=0} (1 + 1/(1-x-x*k)/(k+1)) * x^k / Product_{i=0..k-1} (1-x-x*i)*(i+1). - _Sergei N. Gladkovskii_, Jan 21 2013

%t m = 20;

%t B[x_] = BesselI[0, x] + O[x]^(2 m) // Normal;

%t A[x_] = B[2(1 - E^x)^(1/2)] + O[x]^m;

%t CoefficientList[A[x], x]*Range[0, m-1]!^2 (* _Jean-François Alcover_, Oct 26 2019 *)

%o (PARI) a(n)=n!*sum(k=0,n,(-1)^k*stirling(n,k,2)/k!) \\ _Charles R Greathouse IV_, Apr 18 2016

%K sign,easy

%O 0,4

%A _N. J. A. Sloane_

%E More terms from _Vladeta Jovovic_, Jul 17 2006