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a(n) = ((1+sqrt(13))^n + (1-sqrt(13))^n)/2.
8

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

%S 1,1,14,40,248,976,4928,21568,102272,463360,2153984,9868288,45584384,

%T 209588224,966189056,4447436800,20489142272,94347526144,434564759552,

%U 2001299832832,9217376780288,42450351554560,195509224472576

%N a(n) = ((1+sqrt(13))^n + (1-sqrt(13))^n)/2.

%C Binomial transform of A001022(powers of 13), with interpolated zeros. - _Philippe Deléham_, Dec 20 2007

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

%H G. C. Greubel, <a href="/A125816/b125816.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,12).

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

%F a(n) = 2*a(n-1) + 12*a(n-2), with a(0)=a(1)=1.

%F G.f.: (1-x)/(1-2*x-12*x^2). (End)

%F a(n) = Sum_{k=0..n} A098158(n,k)*13^(n-k). - _Philippe Deléham_, Dec 20 2007

%F If p[1]=1, and p[i]=13, (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+Sqrt[13])^n +(1-Sqrt[13])^n)/(2), {n,0,30}]] (* Artur Jasinski *)

%t LinearRecurrence[{2,12}, {1,1}, 30] (* _G. C. Greubel_, Aug 02 2019 *)

%o (PARI) my(x='x+O('x^30)); Vec((1-x)/(1-2*x-12*x^2)) \\ _G. C. Greubel_, Aug 02 2019

%o (Magma) I:=[1,1]; [n le 2 select I[n] else 2*Self(n-1) +12*Self(n-2): n in [1..30]]; // _G. C. Greubel_, Aug 02 2019

%o (Sage) ((1-x)/(1-2*x-12*x^2)).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Aug 02 2019

%o (GAP) a:=[1,1];; for n in [3..30] do a[n]:=2*a[n-1]+12*a[n-2]; od; a; # _G. C. Greubel_, Aug 02 2019

%Y Cf. A127262. First differences of A091914.

%K nonn

%O 1,3

%A _Artur Jasinski_, Dec 10 2006