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a(n) = ((1 + 3*sqrt(2))^n + (1 - 3*sqrt(2))^n)/2.
8

%I #22 Sep 08 2022 08:45:28

%S 1,1,19,55,433,1801,10963,52543,291457,1476145,7907059,40908583,

%T 216237169,1127920249,5931872371,31038388975,162918608257,

%U 853489829089,4476595998547,23462519091607,123027170158513,644917164874345,3381296222443411,17726184247750687

%N a(n) = ((1 + 3*sqrt(2))^n + (1 - 3*sqrt(2))^n)/2.

%C Binomial transform of [1, 0, 18, 0, 324, 0, 5832, 0, 104976, 0, ...] =: powers of 18 (A001027) with interpolated zeros. - _Philippe Deléham_, Dec 02 2008

%C a(n-1) is the number of compositions of n when there are 1 type of 1 and 18 types of other natural numbers. - _Milan Janjic_, Aug 13 2010

%H T. D. Noe, <a href="/A125818/b125818.txt">Table of n, a(n) for n = 1..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2, 17).

%F From _Philippe Deléham_, Dec 12 2006: (Start)

%F a(n) = 2*a(n-1) + 17*a(n-2), with a(0)=a(1)=1.

%F G.f.: (1-x)/(1-2*x-17*x^2). (End)

%F a(n) = Sum_{k=0..n} A098158(n,k)*18^(n-k). - _Philippe Deléham_, Dec 26 2007

%F If p[1]=1, and p[i]=18, (i>1), and if A is Hessenberg matrix of order n If p[1]=1, and p[i]=18, (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

%t Expand[Table[((1+3*Sqrt[2])^n +(1-3*Sqrt[2])^n)/2, {n,0,30}]]

%t (* alternate program *)

%t LinearRecurrence[{2, 17}, {1, 1}, 30] (* _T. D. Noe_, Mar 28 2012 *)

%o (PARI) my(x='x+O('x^30)); Vec((1-x)/(1-2*x-17*x^2)) \\ _G. C. Greubel_, Aug 03 2019

%o (Magma) I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) +17*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Aug 03 2019

%o (Sage) ((1-x)/(1-2*x-17*x^2)).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Aug 03 2019

%o (GAP) a:=[1,1];; for n in [3..30] do a[n]:=2*a[n-1]+17*a[n-2]; od; a; # _G. C. Greubel_, Aug 03 2019

%Y Cf. A125817.

%K nonn

%O 1,3

%A _Artur Jasinski_, Dec 10 2006