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A001569 Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*(1-exp(x))^(1/2)).

%I A001569 M2161 N0861 #35 Dec 26 2021 21:04:49
%S A001569 1,-1,-1,2,37,329,1501,-31354,-1451967,-39284461,-737652869,560823394,
%T A001569 1103386777549,82520245792997,4398448305245905,168910341581721494,
%U A001569 998428794798272641,-720450682719825322809,-105099789680808769094057,-10594247095804692725600734
%N A001569 Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*(1-exp(x))^(1/2)).
%D A001569 S. M. Kerawala, Asymptotic solution of the "Probleme des menages", Bull. Calcutta Math. Soc., 39 (1947), 82-84.
%D A001569 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D A001569 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H A001569 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 A001569 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 A001569 a(n) = n!*Sum_{k=0..n} (-1)^k*Stirling2(n,k)/k!. - _Vladeta Jovovic_, Jul 17 2006
%F A001569 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 A001569 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 A001569 m = 20;
%t A001569 B[x_] = BesselI[0, x] + O[x]^(2 m) // Normal;
%t A001569 A[x_] = B[2(1 - E^x)^(1/2)] + O[x]^m;
%t A001569 CoefficientList[A[x], x]*Range[0, m-1]!^2 (* _Jean-François Alcover_, Oct 26 2019 *)
%o A001569 (PARI) a(n)=n!*sum(k=0,n,(-1)^k*stirling(n,k,2)/k!) \\ _Charles R Greathouse IV_, Apr 18 2016
%K A001569 sign,easy
%O A001569 0,4
%A A001569 _N. J. A. Sloane_
%E A001569 More terms from _Vladeta Jovovic_, Jul 17 2006