%I #46 May 04 2023 02:27:11
%S 1,19,362,6897,131405,2503592,47699653,908796999,17314842634,
%T 329890807045,6285240176489,119749454160336,2281524869222873,
%U 43468721969394923,828187242287726410,15779026325436196713,300629687425575463957,5727743087411370011896
%N a(0) = 1, a(1) = 19; for n > 1, a(n) = 19*a(n-1) + a(n-2).
%C a(n+1)/a(n) tends to (19 + sqrt(365))/2.
%C a(n) equals the number of words of length n on alphabet {0,1,...,19} avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015
%C From _Michael A. Allen_, May 03 2023: (Start)
%C Also called the 19-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 19 kinds of squares available. (End)
%H Bruno Berselli, <a href="/A243399/b243399.txt">Table of n, a(n) for n = 0..200</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#dPlusOne">Recursive Sequences</a>.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>.
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (19,1).
%F G.f.: 1/(1 - 19*x - x^2).
%F a(n) = (-1)^n*a(-n-2) = ((19 + sqrt(365))^(n+1)-(19 - sqrt(365))^(n+1))/(2^(n+1)*sqrt(365)).
%F a(n) = F(n+1, 19), the (n+1)-th Fibonacci polynomial evaluated at x = 19.
%F a(n)*a(n-2) - a(n-1)^2 = (-1)^n, with a(-2)=1, a(-1)=0.
%t RecurrenceTable[{a[n] == 19 a[n - 1] + a[n - 2], a[0] == 1, a[1] == 19}, a, {n, 0, 20}]
%o (PARI) v=vector(20); v[1]=1; v[2]=19; for(i=3, #v, v[i]=19*v[i-1]+v[i-2]); v
%o (Magma) [n le 2 select 19^(n-1) else 19*Self(n-1)+Self(n-2): n in [1..20]];
%o (Maxima) a[0]:1$ a[1]:19$ a[n]:=19*a[n-1]+a[n-2]$ makelist(a[n], n, 0, 20);
%o (Sage)
%o from sage.combinat.sloane_functions import recur_gen2
%o a = recur_gen2(1,19,19,1)
%o [next(a) for i in (0..20)]
%Y Row n=19 of A073133, A172236 and A352361 and column k=19 of A157103.
%Y Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), this sequence (k=19), A041181 (k=20). Also, many other sequences are in the OEIS with even k greater than 20 (denominators of continued fraction convergents to sqrt((k/2)^2+1)).
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Jun 04 2014