0,3

Each determinant is the numerator of the fraction x(n)/y(n) = [1!, 2!, ..., n!] (the simple continued fraction). The value x(n) is obtained by computing the determinant det(n X n) from the last column. The value y(n) is obtained by computing this determinant after removal of the first row and the first column (see example below).

Also denominator of fraction equal to the continued fraction [ 0; 1!, 2!, ... , n! ]. - Seiichi Manyama, Jun 05 2018

J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Collection ellipses (2004), p.115.

Seiichi Manyama, Table of n, a(n) for n = 0..43

a(0) = 1, a(1) = 1, a(n) = n! * a(n-1) + a(n-2). - Daniel Suteu, Dec 20 2016

a(n) ~ c * BarnesG(n+2), where c = 1.5943186620010986362991550255196986158205795892595646967623357407966... - Vaclav Kotesovec, Jun 05 2018

For n = 1, det[1] = 1.

For n = 2, det([[1,-1],[1,2]]) = 3, and the continued fraction expansion is 3/2 = [1!,2!].

For n = 3, det([[1,-1, 0],[1,2,-1],[0,1,6]]) = 19, and the continued fraction expansion is 19/det([[2,-1],[1,6]]) = 19/13 = [1!,2!,3!].

For n = 4, det([[1,-1,0,0],[1,2,-1,0],[0,1,6,-1],[0,0,1,24]]) = 459, and the continued fraction expansion is 459/det([[2,-1,0],[1,6,-1],[0,1,24]]) = 459/314 = [1!,2!,3!,4!].

for n from 15 by -1 to 1 do:x0:=n!:for p from n by -1 to 2 do : x0:= (p-1)! + 1/x0 :od:print(x0):od :

Cf. A001040, A036245, A036246, A084845.

Sequence in context: A228149 A358161 A079281 * A079306 A386794 A051381

Adjacent sequences: A176229 A176230 A176231 * A176233 A176234 A176235

Michel Lagneau, Apr 12 2010

a(0)=1 prepended by Alois P. Heinz, Dec 20 2016

approved