%I #70 Apr 21 2023 16:54:54
%S 1,12,145,1752,21169,255780,3090529,37342128,451196065,5451694908,
%T 65871534961,795910114440,9616792908241,116197425013332,
%U 1403985893068225,16964028141832032,204972323595052609,2476631911282463340,29924555258984612689,361571295019097815608
%N Denominators of continued fraction convergents to sqrt(37).
%C Sqrt(37) = 6.08276253... = 12/2 + 12/145 + 12/(145*21169) + 12/(21169*3090529) + ... - _Gary W. Adamson_, Jun 13 2008
%C For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 12's along the main diagonal and 1's along the superdiagonal and the subdiagonal. - _John M. Campbell_, Jul 08 2011
%C a(n) equals the number of words of length n on alphabet {0,1,...,12} avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015
%C From _Michael A. Allen_, Apr 02 2023: (Start)
%C Also called the 12-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
%C a(n) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 12 kinds of squares available. (End)
%H Nathaniel Johnston, <a href="/A041061/b041061.txt">Table of n, a(n) for n = 0..500</a>
%H Michael A. Allen and Kenneth Edwards, <a href="https://www.fq.math.ca/Papers1/60-5/allen.pdf">Fence tiling derived identities involving the metallonacci numbers squared or cubed</a>, Fib. Q. 60:5 (2022) 5-17.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla and Fausto Jarquín-Zárate, <a href="https://arxiv.org/abs/1904.13002">The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d))</a>, arXiv:1904.13002 [math.NT], 2019.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (12,1).
%F a(n) = F(n, 12), the n-th Fibonacci polynomial evaluated at x=12. - _T. D. Noe_, Jan 19 2006
%F From _Philippe Deléham_, Nov 21 2008: (Start)
%F a(n) = 12*a(n-1) + a(n-2), n>1; a(0)=1, a(1)=12.
%F G.f.: 1/(1 - 12*x - x^2). (End)
%F a(n) = ((6+sqrt(37))^(n+1) - (6-sqrt(37))^(n+1))/(2*sqrt(37)). - _Rolf Pleisch_, May 14 2011
%t Denominator[Convergents[Sqrt[37],30]] (* or *) LinearRecurrence[{12,1},{1,12},30] (* _Harvey P. Dale_, May 26 2014 *)
%o (Sage) [lucas_number1(n,12,-1) for n in range(1, 18)] # _Zerinvary Lajos_, Apr 28 2009
%Y Cf. A010491, A041060.
%Y Cf. A243399.
%Y Row n=12 of A073133, A172236 and A352361 and column k=12 of A157103.
%K nonn,frac,easy
%O 0,2
%A _N. J. A. Sloane_