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a(0)=1, a(1)=8, a(n)=17*a(n-1)-64*a(n-2) for n>1.
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%I #15 Jun 18 2022 15:47:50

%S 1,8,72,712,7496,81864,911944,10263752,116119368,1317149128,

%T 14959895624,170020681416,1932918264136,21978286879688,

%U 249924108049992,2842099476549832,32320548186147656,367554952665320904

%N a(0)=1, a(1)=8, a(n)=17*a(n-1)-64*a(n-2) for n>1.

%C a(n)/a(n-1) tends to (17+sqrt(33))/2 = 11.3722813...

%C For n>=2, a(n) equals 8^n times the permanent of the (2n-2) X (2n-2) tridiagonal matrix with 1/sqrt(8)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (17,-64).

%F G.f.: (1-9*x)/(1-17*x+64*x^2).

%F a(n) = Sum_{k=0..n} A165253(n,k)*8^(n-k).

%F a(n) = ((33-sqrt(33))*(17+sqrt(33))^n+(33+sqrt(33))*(17-sqrt(33))^n)/(66*2^n). [_Klaus Brockhaus_, Sep 28 2009]

%t LinearRecurrence[{17,-64},{1,8},20] (* _Harvey P. Dale_, Jun 08 2018 *)

%Y Cf. A165253.

%K nonn

%O 0,2

%A _Philippe Deléham_, Sep 14 2009