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%I #31 Dec 31 2023 10:12:54
%S 1,9,25,113,401,1593,5993,23137,88225,338409,1294393,4957649,18976049,
%T 72655641,278143625,1064876737,4076758849,15607654857,59752621657,
%U 228758827313,875786006225,3352883803641,12836269650857,49142725927201
%N a(n) = 2*a(n-1) + 7*a(n-2).
%C a(n) = a(n-1) + 8*A015519(n). a(n)/A015519(n+1) converges to sqrt(8).
%C a(n-1) is the number of compositions of n when there is 1 type of 1 and 8 types of other natural numbers. - _Milan Janjic_, Aug 13 2010
%H G. C. Greubel, <a href="/A083100/b083100.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2, 7).
%F G.f.: (1+7*x)/(1-2*x-7*x^2).
%F a(n) = binomial transform of 1,8,8,64,64,512. - Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009
%F If p[1]=1, and p[i]=8,(i>1), and if A is Hessenberg matrix of order n defined by: A[i,j]=p[j-i+1], (i<=j), A[i,j]=-1, (i=j+1), and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=det A. - _Milan Janjic_, Apr 29 2010
%F G.f.: G(0)/(2*x) - 1/x, where G(k)= 1 + 1/(1 - x*(8*k-1)/(x*(8*k+7) - 1/G(k+1))); (continued fraction). - _Sergei N. Gladkovskii_, May 28 2013
%t CoefficientList[Series[(1 + 7 x)/(1 - 2 x - 7 x^2), {x, 0, 25}], x] (* Or *) a[n_] := Simplify[((1 + Sqrt[8])^n + (1 - Sqrt[8])^n)/2]; Array[a, 25, 0] (* Or *) LinearRecurrence[{2, 7}, {1, 1}, 28] (* Or *) Table[ MatrixPower[{{1, 2}, {4, 1}}, n][[1, 1]], {n, 0, 25}] (* _Robert G. Wilson v_, Sep 18 2013 *)
%o (PARI) a(n)=([0,1; 7,2]^n*[1;9])[1,1] \\ _Charles R Greathouse IV_, Apr 06 2016
%o (PARI) x='x+O('x^30); Vec((1+7*x)/(1-2*x-7*x^2)) \\ _G. C. Greubel_, Jan 08 2018
%o (Magma) I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) + 7*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Jan 08 2018
%Y Essentially a duplicate of A084058.
%Y The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519.
%K easy,nonn
%O 0,2
%A Mario Catalani (mario.catalani(AT)unito.it), Apr 23 2003
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