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A041007

Denominators of continued fraction convergents to sqrt(6).

1, 2, 9, 20, 89, 198, 881, 1960, 8721, 19402, 86329, 192060, 854569, 1901198, 8459361, 18819920, 83739041, 186298002, 828931049, 1844160100, 8205571449, 18255302998, 81226783441, 180708869880

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OFFSET

0,2

COMMENTS

sqrt(6) = 4/2 + 4/9 + 4/(9*89) + 4/(89*881) + 4/(881*8721), ...; where sqrt(6) = 2.4494897427... and the sum of the first 5 terms of this series = 2.449489737... - Gary W. Adamson, Dec 21 2007

sqrt(6) = 2 + continued fraction [2, 4, 2, 4, 2, 4, ...] = 4/2 + 4/9 + 4/(9*89) + 4/(89*881) + 4/(881*8721) + ... - Gary W. Adamson, Dec 21 2007

Interspersion of 2 sequences, A072256 and 2*A004189. - Gerry Martens, Jun 10 2015

For n > 0, a(n) equals the permanent of the n X n tridiagonal matrix with the main diagonal alternating sequence [2, 4, 2, 4, ...] and 1's along the superdiagonal and the subdiagonal. - Rogério Serôdio, Apr 01 2018

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,10,0,-1).

FORMULA

G.f.: (1+2*x-x^2)/(1-10*x^2+x^4). - Colin Barker, Dec 31 2011

From Rogério Serôdio, Apr 01 2018: (Start)

Recurrence formula: a(n) = (3 + (-1)^n)*a(n-1) + a(n-2), a(0) = 1, a(1) = 2.

Some properties:

(1) a(n)^2 - a(n-2)^2 = (3+(-1)^n)*a(2*n-1), for n > 1;

(2) a(2*n+1) = a(n)*(a(n+1) + a(n-1)), for n > 0;

(3) a(2*n) = A142239(2*n), for n >= 0;

(4) a(2*n+1) = A041007(2*n+1)/2, for n >= 0;