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 A001569 Sum_{n>=0} a(n)*x^n/n!^2 = BesselI(0,2*(1-exp(x))^(1/2)). (Formerly M2161 N0861) 8
 1, -1, -1, 2, 37, 329, 1501, -31354, -1451967, -39284461, -737652869, 560823394, 1103386777549, 82520245792997, 4398448305245905, 168910341581721494, 998428794798272641, -720450682719825322809, -105099789680808769094057, -10594247095804692725600734 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,4 REFERENCES S. M. Kerawala, Asymptotic solution of the "Probleme des menages", Bull. Calcutta Math. Soc., 39 (1947), 82-84. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS S. M. Kerawala, Asymptotic solution of the "Probleme des menages, Bull. Calcutta Math. Soc., 39 (1947), 82-84. [Annotated scanned copy] FORMULA 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 a_r/n^r, r=1..inf). a(n) = n!*Sum_{k=0..n} (-1)^k*Stirling2(n,k)/k!. - Vladeta Jovovic, Jul 17 2006 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 E.g.f.: 1 + x*(1 - S)/(1-x) where S=sum(k>=0, ( 1+1/(1-x-x*k)/(k+1) )*x^k/prod(i=0..k-1, (1-x-x*i)*(i+1) ) ). - Sergei N. Gladkovskii, Jan 21 2013 MATHEMATICA m = 20; B[x_] = BesselI[0, x] + O[x]^(2 m) // Normal; A[x_] = B[2(1 - E^x)^(1/2)] + O[x]^m; CoefficientList[A[x], x]*Range[0, m-1]!^2 (* Jean-François Alcover, Oct 26 2019 *) PROG (PARI) a(n)=n!*sum(k=0, n, (-1)^k*stirling(n, k, 2)/k!) \\ Charles R Greathouse IV, Apr 18 2016 CROSSREFS Sequence in context: A078976 A200911 A243101 * A092853 A297796 A300542 Adjacent sequences:  A001566 A001567 A001568 * A001570 A001571 A001572 KEYWORD sign,easy AUTHOR EXTENSIONS More terms from Vladeta Jovovic, Jul 17 2006 STATUS approved

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Last modified April 16 07:59 EDT 2021. Contains 343030 sequences. (Running on oeis4.)