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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_{r>=1} a_r/n^r).
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 / Product_{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
KEYWORD
sign,easy
AUTHOR
EXTENSIONS
More terms from Vladeta Jovovic, Jul 17 2006
STATUS
approved

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Last modified May 13 01:02 EDT 2024. Contains 372497 sequences. (Running on oeis4.)