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A205571

Expansion of e.g.f. 1/(1 - x*cosh(x)).

1, 1, 2, 9, 48, 305, 2400, 22057, 230272, 2708001, 35412480, 509177801, 7986468864, 135718942801, 2483729876992, 48699677975145, 1018542257111040, 22634000289407297, 532557637644976128, 13226748101381102473, 345792863300174479360, 9492229607399841038961

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OFFSET

0,3

COMMENTS

Radius of convergence of e.g.f. is |x| < r where r = 0.7650099545507... satisfies cosh(r) = 1/r. See A069814.

Number of ways to choose one element from each block of the labeled ordered partitions of an n-set into odd blocks (see Example). - Enrique Navarrete, Oct 03 2025

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..200

FORMULA

a(2*n-1) == 1 (mod 4), a(2*n+2) == 0 (mod 4), for n>=1.

a(n) ~ n!/(1+r*sqrt(1-r^2))*(1/r)^n, where r = A069814 = 0.7650099545507321... is the root of the equation r*cosh(r)=1. - Vaclav Kotesovec, Feb 13 2013

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/2)} binomial(n,2*k+1) * (2*k+1) * a(n-2*k-1). - Ilya Gutkovskiy, Mar 10 2022

a(n) = Sum_{k=0..n} k! * A185951(n,k). - Seiichi Manyama, Feb 17 2025

EXAMPLE

E.g.f.: A(x) = 1 + x + 2*x^2/2! + 9*x^3/3! + 48*x^4/4! + 305*x^5/5! +...

From Enrique Navarrete, Oct 03 2025: (Start)

a(5) = 305 since the number of ways to choose one element from each block of the labeled ordered partitions of a 5-set into odd blocks are:

Sample partition {1,2,3,4,5}: 1 partition, 5 ways.

Sample partition {1,2,3} {4} {5}: 60 such partitions, 180 ways.

Sample partition {1} {2} {3} {4} {5}: 120 such partitions, 120 ways. (End)

MATHEMATICA

CoefficientList[Series[1/(1-x*Cosh[x]), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Feb 13 2013 *)

PROG