A135829 a(n) = F(n)*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.

0, 1, 1, 3, 10, 53, 434, 5695, 120029, 4086681, 224887484, 20019072757, 2882971364492, 671752346999393, 253253517790135653, 154485317604329747723, 152477261728991251138254, 243506341466516632397539361, 629220538826740707106492847078

Offset: 0

Author

Gary W. Adamson, Nov 29 2007

Keywords

• nonn
• easy

Comments

Essentially the same as A071895. [R. J. Mathar, Oct 28 2008]

From Michel Lagneau, Apr 12 2010: (Start)

Determinant of n+1 X n+1 matrix: ((F(0),-1,0,...,0),(1,F(1),-1,0,...,0),(0,1,F(2),-1,0,...,0),...,(0,0,...,1,F(n)). Each determinant is the numerator of the fraction x(n)/y(n) equal to the continued fraction expansion of the diagonal elements [F(0), F(1), ..., F(n)] of the n+1 X n+1 matrix. The value x(n) is obtained by computing the determinant det(n+1 X n+1) 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).

The sequence A001040 give the values of each determinant with numerator of continued fraction given by the expansion of the diagonal elements [n,n-1,...,3,2,1]. The same is true for the sequence A084845 with the expansion of the diagonal elements [n,n,...,n], and the sequence A036246 for the elements[ 0, 1, 4, ..., n^2 ].

Examples:

for n = 0, det[0] = 0; for n = 1, det(([[0,-1],[1,1]]) = 1;

for n = 2, det([[0,-1, 0],[1,1,-1],[0,1,1]]))=1;

for n = 3, det([[0,-1, 0,0],[1,1,-1,0],[0,1,1,-1],[0,0,1,2]])) = 3, and the continued fraction expansion is 3/det(([[1,-1, 0],[1,1,-1],[0,1,2]])) = 5/3 = 0 + 1 + 1/(1 + 1/2) => [0,1,1,2]. (End)

a(n) is the denominator of the continued fraction [F(1), F(2), ..., F(n)] for n > 0. - Seung Ju Lee, Aug 23 2020

Examples

			a(5) = 53 = F(5)*a(4) + a(3) = 5*10 + 3.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..98

Crossrefs

Cf. A000045.

Cf. A176232, A176233, A001040, A084845, A036246, A036245, A330638.

Programs

Maple

a:= proc(n) option remember; `if`(n<2, n,
      combinat[fibonacci](n)*a(n-1)+a(n-2))
    end:
seq(a(n), n=0..20);  # Alois P. Heinz, Jan 24 2021

Mathematica

RecurrenceTable[{a[0]==0,a[1]==1,a[n]==Fibonacci[n]*a[n-1]+a[n-2]}, a,{n,0,20}] (* Harvey P. Dale, Apr 26 2012 *)

Formula

a(n) = (-a(n-1)*a(n-4)*a(n-2) - a(n-1)*a(n-3)^2 + a(n-1)^2*a(n-3) + a(n-2)^2*a(n-3) + a(n-1)*a(n-2)^2)/(a(n-2)*a(n-3)). - Robert Israel, Dec 04 2016

a(n) ~ c * ((1 + sqrt(5))/2)^(n*(n+1)/2) / 5^(n/2), where c = 2.25240516839867905756631574518868900987391688308922490621152619277084562178... - Vaclav Kotesovec, Dec 29 2019

Extensions

More terms from Michel Lagneau, Apr 12 2010

Offset changed by N. J. A. Sloane, Apr 21 2010

Replaced n with n+1 where needed. - Seung Ju Lee, Aug 30 2020

Incorrect program removed by Alois P. Heinz, Jan 24 2021