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A176232
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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!).
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4
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1, 1, 3, 19, 459, 55099, 39671739, 199945619659, 8061807424322619, 2925468678338137602379, 10615940739961495538937237819, 423754383328897950597328272711061579, 202979027621555455188781938315330372976764219
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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J. M. De Koninck, A. Mercier, 1001 problèmes en théorie classique des nombres. Collection ellipses (2004), p.115.
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LINKS
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FORMULA
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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
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EXAMPLE
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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!].
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MAPLE
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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 :
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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