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A176232
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!).
4
1, 1, 3, 19, 459, 55099, 39671739, 199945619659, 8061807424322619, 2925468678338137602379, 10615940739961495538937237819, 423754383328897950597328272711061579, 202979027621555455188781938315330372976764219
OFFSET
0,3
COMMENTS
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
REFERENCES
J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Collection ellipses (2004), p.115.
LINKS
FORMULA
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
EXAMPLE
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!].
MAPLE
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 :
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Apr 12 2010
EXTENSIONS
a(0)=1 prepended by Alois P. Heinz, Dec 20 2016
STATUS
approved