%I #28 Jun 05 2018 11:04:31
%S 1,1,3,19,459,55099,39671739,199945619659,8061807424322619,
%T 2925468678338137602379,10615940739961495538937237819,
%U 423754383328897950597328272711061579,202979027621555455188781938315330372976764219
%N Determinant of the n X n matrix with rows (1!,-1,0,...,0), (1, 2!,-1,0,...,0), (0,1,3!,-1,0,...,0), ..., (0,0,...,1,n!).
%C 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).
%C Also denominator of fraction equal to the continued fraction [ 0; 1!, 2!, ... , n! ]. - _Seiichi Manyama_, Jun 05 2018
%D J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Collection ellipses (2004), p.115.
%H Seiichi Manyama, <a href="/A176232/b176232.txt">Table of n, a(n) for n = 0..43</a>
%F a(0) = 1, a(1) = 1, a(n) = n! * a(n-1) + a(n-2). - _Daniel Suteu_, Dec 20 2016
%F a(n) ~ c * BarnesG(n+2), where c = 1.5943186620010986362991550255196986158205795892595646967623357407966... - _Vaclav Kotesovec_, Jun 05 2018
%e For n = 1, det[1] = 1.
%e For n = 2, det(([[1,-1],[1,2]]) = 3, and the continued fraction expansion is 3/2 = [1!,2!].
%e 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!].
%e 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!].
%p 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 :
%Y Cf. A001040, A036245, A036246, A084845.
%K nonn
%O 0,3
%A _Michel Lagneau_, Apr 12 2010
%E a(0)=1 prepended by _Alois P. Heinz_, Dec 20 2016