

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
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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(n1) + a(n2).  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:= (p1)! + 1/x0 :od:print(x0):od :


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



