%I #70 Oct 05 2024 14:33:32
%S 0,1,15,226,3405,51301,772920,11645101,175449435,2643386626,
%T 39826248825,600037119001,9040383033840,136205782626601,
%U 2052127122432855,30918112619119426,465823816409224245,7018275358757483101,105739954197771470760,1593117588325329544501
%N a(n) = 15*a(n-1) + a(n-2) with a(0) = 0, a(1) = 1.
%C Limit_{n -> infinity} a(n)/a(n-1) = (15 + sqrt(229))/2. - _Klaus Brockhaus_, Oct 07 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 For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 15'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 - 1 on alphabet {0,1,...,15} avoiding runs of zeros of odd lengths. - _Milan Janjic_, Jan 28 2015
%C From _Michael A. Allen_, Apr 30 2023: (Start)
%C Also called the 15-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
%C a(n+1) 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 15 kinds of squares available. (End)
%H G. C. Greubel, <a href="/A154597/b154597.txt">Table of n, a(n) for n = 0..840</a> (a(0) = 0 added by Jianing Song)
%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> [From _Johannes W. Meijer_, Jun 12 2010]
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (15,1).
%F G.f.: x/(1 - 15*x - x^2). - _Klaus Brockhaus_, Jan 12 2009, corrected Oct 07 2009
%F a(n) = ((15 + sqrt(229))^n - (15 - sqrt(229))^n)/(2^n*sqrt(229)).
%F From _Johannes W. Meijer_, Jun 12 2010: (Start)
%F Limit_{k -> infinity} a(n+k)/a(k) = (A090301(n) + a(n)*sqrt(229))/2.
%F Limit_{n -> infinity} A090301(n)/a(n) = sqrt(229).
%F a(2n+1) = 15*A098245(n-1).
%F a(3n+1) = A041427(5n), a(3n+2) = A041427(5n+3), a(3n+3) = 2*A041427(5n+4). (End)
%F E.g.f.: (2/sqrt(229))*exp(15*x/2)*sinh(sqrt(229)*x/2). - _G. C. Greubel_, Sep 20 2024
%t LinearRecurrence[{15,1}, {0,1}, 31] (* _Vladimir Joseph Stephan Orlovsky_, Oct 27 2009 *)
%t CoefficientList[Series[x/(1-15*x-x^2), {x,0,50}], x] (* _G. C. Greubel_, Apr 16 2017 *)
%o (Magma) Z<x>:=PolynomialRing(Integers()); N<r>:=NumberField(x^2-229); S:=[ ((15+r)^n-(15-r)^n)/(2^n*r): n in [1..17] ]; [ Integers()!S[j]: j in [1..#S] ]; // _Klaus Brockhaus_, Jan 12 2009
%o (Magma)
%o [n le 2 select n-1 else 15*Self(n-1) +Self(n-2): n in [1..30]]; // _G. C. Greubel_, Sep 20 2024
%o (PARI) my(x='x+O('x^50)); concat([0], Vec(x/(1-15*x-x^2))) \\ _G. C. Greubel_, Apr 16 2017
%o (SageMath)
%o def A154597(n): return (-i)^(n-1)*chebyshev_U(n-1, 15*i/2)
%o [A154597(n) for n in range(31)] # _G. C. Greubel_, Sep 20 2024
%Y Row n=15 of A073133, A172236 and A352361 and column k=15 of A157103.
%Y First bisection is A098247.
%Y Cf. A166125 (decimal expansion of sqrt(229)), A166126 (decimal expansion of (15 + sqrt(229))/2).
%Y Cf. also A041427, A090301, A098245.
%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), this sequence (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
%K nonn,easy
%O 0,3
%A Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
%E Extended beyond a(7) by _Klaus Brockhaus_ and _Philippe Deléham_, Jan 12 2009
%E Name from _Philippe Deléham_, Jan 12 2009
%E Edited by _Klaus Brockhaus_, Oct 07 2009
%E Missing a(0) added by _Jianing Song_, Jan 29 2019