login
a(n) = F(n)*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.
4

%I #44 Apr 01 2022 21:24:03

%S 0,1,1,3,10,53,434,5695,120029,4086681,224887484,20019072757,

%T 2882971364492,671752346999393,253253517790135653,

%U 154485317604329747723,152477261728991251138254,243506341466516632397539361,629220538826740707106492847078

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

%C Essentially the same as A071895. [_R. J. Mathar_, Oct 28 2008]

%C From _Michel Lagneau_, Apr 12 2010: (Start)

%C 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).

%C 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 ].

%C Examples:

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

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

%C 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)

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

%H G. C. Greubel, <a href="/A135829/b135829.txt">Table of n, a(n) for n = 0..98</a>

%F 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

%F a(n) ~ c * ((1 + sqrt(5))/2)^(n*(n+1)/2) / 5^(n/2), where c = 2.25240516839867905756631574518868900987391688308922490621152619277084562178... - _Vaclav Kotesovec_, Dec 29 2019

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

%p a:= proc(n) option remember; `if`(n<2, n,

%p combinat[fibonacci](n)*a(n-1)+a(n-2))

%p end:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, Jan 24 2021

%t 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 *)

%Y Cf. A000045.

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

%K nonn,easy

%O 0,4

%A _Gary W. Adamson_, Nov 29 2007

%E More terms from _Michel Lagneau_, Apr 12 2010

%E Offset changed by _N. J. A. Sloane_, Apr 21 2010

%E Replaced n with n+1 where needed. - _Seung Ju Lee_, Aug 30 2020

%E Incorrect program removed by _Alois P. Heinz_, Jan 24 2021