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%I #123 Mar 31 2023 05:10:51
%S 1,5,26,135,701,3640,18901,98145,509626,2646275,13741001,71351280,
%T 370497401,1923838285,9989688826,51872282415,269351100901,
%U 1398627786920,7262490035501,37711077964425,195817879857626
%N a(0) = 1, a(1) = 5, a(n+1) = 5*a(n) + a(n-1).
%C A087130(n)^2 - 29*a(n-1)^2 = 4*(-1)^n, n >= 1. - _Gary W. Adamson_, Jul 01 2003, corrected Oct 07 2008, corrected by _Jianing Song_, Feb 01 2019
%C a(p) == 29^((p-1)/2)) (mod p), for odd primes p. - _Gary W. Adamson_, Feb 22 2009
%C For more information about this type of recurrence, follow the Khovanova link and see A054413, A086902 and A178765. - _Johannes W. Meijer_, Jun 12 2010
%C Binomial transform of A015523. - _Johannes W. Meijer_, Aug 01 2010
%C For positive n, a(n) equals the permanent of the n X n tridiagonal matrix with 5'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,...,5} avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015
%C From _Michael A. Allen_, Feb 15 2023: (Start)
%C Also called the 5-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 5 kinds of squares available. (End)
%H Vincenzo Librandi, <a href="/A052918/b052918.txt">Table of n, a(n) for n = 0..1000</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 D. Birmajer, J. B. Gil, and M. D. Weiner, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL19/Gil/gil6.html">On the Enumeration of Restricted Words over a Finite Alphabet</a>, J. Int. Seq. 19 (2016) # 16.1.3, Example 8.
%H INRIA Algorithms Project, <a href="http://ecs.inria.fr/services/structure?nbr=901">Encyclopedia of Combinatorial Structures 901</a>
%H Milan Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL13/Janjic/janjic33.html">Hessenberg Matrices and Integer Sequences </a>, J. Int. Seq. 13 (2010) # 10.7.8, section 3.
%H Milan Janjic, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL18/Janjic/janjic63.html">On Linear Recurrence Equations Arising from Compositions of Positive Integers</a>, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Kai Wang, <a href="https://www.researchgate.net/publication/339487198_On_k-Fibonacci_Sequences_And_Infinite_Series_List_of_Results_and_Examples">On k-Fibonacci Sequences And Infinite Series List of Results and Examples</a>, 2020.
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,1).
%F G.f.: 1/(1 - 5*x - x^2).
%F a(3n) = A041047(5n), a(3n+1) = A041047(5n+3), a(3n+2) = 2*A041047(5n+4). - _Henry Bottomley_, May 10 2000
%F a(n) = Sum_{alpha=RootOf(-1+5*z+z^2)} (1/29)*(5+2*alpha)*alpha^(-1-n).
%F a(n-1) = (((5 + sqrt(29))/2)^n - ((5 - sqrt(29))/2)^n)/sqrt(29). - _Gary W. Adamson_, Jul 01 2003
%F a(n) = U(n, 5*i/2)*(-i)^n with i^2 = -1 and Chebyshev's U(n, x/2) = S(n, x) polynomials. See triangle A049310.
%F Let M = {{0, 1}, {1, 5}}, then a(n) is the lower-right term of M^n. - _Roger L. Bagula_, May 29 2005
%F a(n) = F(n, 5), the n-th Fibonacci polynomial evaluated at x = 5. - _T. D. Noe_, Jan 19 2006
%F a(n) = denominator of n-th convergent to [1, 4, 5, 5, 5, ...], for n > 0. Continued fraction [1, 4, 5, 5, 5, ...] = 0.807417596..., the inradius of a right triangle with legs 2 and 5. n-th convergent = A100237(n)/A052918(n), the first few being: 1/1, 4/5, 21/26, 109/135, 566/701, ... - _Gary W. Adamson_, Dec 21 2007
%F From _Johannes W. Meijer_, Jun 12 2010: (Start)
%F a(2n+1) = 5*A097781(n), a(2n) = A097835(n).
%F Limit_{k->infinity} a(n+k)/a(k) = (A087130(n) + a(n-1)*sqrt(29))/2.
%F Limit_{n->infinity} A087130(n)/a(n-1) = sqrt(29).
%F (End)
%F From _L. Edson Jeffery_, Jan 07 2012: (Start)
%F Define the 2 X 2 matrix A = {{1, 1}, {5, 4}}. Then:
%F a(n) is the upper-left term of (1/5)*(A^(n+2) - A^(n+1));
%F a(n) is the upper-right term of A^(n+1);
%F a(n) is the lower-left term of (1/5)*A^(n+1);
%F a(n) is the lower-right term of (Sum_{k=0..n} A^k). (End)
%F Sum_{n>=0} (-1)^n/(a(n)*a(n+1)) = (sqrt(29) - 5)/2. - _Vladimir Shevelev_, Feb 23 2013
%p spec := [S,{S=Sequence(Union(Z,Z,Z,Z,Z,Prod(Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..30);
%p a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=5*a[n-1]+a[n-2] od: seq(a[n], n=0..30); # _Zerinvary Lajos_, Jul 26 2006
%p with(combinat):a:=n->fibonacci(n,5):seq(a(n),n=1..30); # _Zerinvary Lajos_, Dec 07 2008
%t LinearRecurrence[{5, 1}, {1, 5}, 30] (* _Vincenzo Librandi_, Feb 23 2013 *)
%t Table[Fibonacci[n+1, 5], {n,0,30}] (* _Vladimir Reshetnikov_, May 08 2016 *)
%o (Sage) [lucas_number1(n,5,-1) for n in range(1, 22)] # _Zerinvary Lajos_, Apr 24 2009
%o (PARI) Vec(1/(1-5*x-x^2)+O(x^30)) \\ _Charles R Greathouse IV_, Nov 20 2011
%o (Magma) I:=[1, 5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // _Vincenzo Librandi_, Feb 23 2013
%o (Magma) R<x>:=PowerSeriesRing(Integers(), 22); Coefficients(R!( 1/(1 - 5*x - x^2) )); // _Marius A. Burtea_, Oct 16 2019
%o (GAP) a:=[1,5];; for n in [3..30] do a[n]:=5*a[n-1]+a[n-2]; od; a; # _G. C. Greubel_, Oct 16 2019
%Y Row 5 of A073133, A172236, and A352361.
%Y Cf. A087130, A099365 (squares), A100237, A175184 (Pisano periods), A201005 (prime subsequence).
%K easy,nonn
%O 0,2
%A encyclopedia(AT)pommard.inria.fr, Jan 25 2000
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