%I #16 Sep 02 2021 04:02:07

%S 1,5,30,205,1505,11430,88105,683405,5314830,41378005,322287305,

%T 2510710230,19560629905,152399173205,1187375157630,9251147403805,

%U 72078242501105,561581982417030,4375445744059705,34090353624231005

%N a(0)=1, a(1)=5, a(n)=11*a(n-1)-25*a(n-2) for n>1.

%C a(n)/a(n-1) tends to (11+sqrt(21))/2 = 7.79128784...

%C For n>=2, a(n) equals 5^n times the permanent of the (2n-2)X(2n-2) tridiagonal matrix with 1/sqrt(5)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [_John M. Campbell_, Jul 08 2011]

%H Indranil Ghosh, <a href="/A165312/b165312.txt">Table of n, a(n) for n = 0..1119</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-25).

%F G.f.: (1-6x)/(1-11x+25x^2).

%F a(n) = Sum_{k=0..n} A165253(n,k)*5^(n-k).

%F a(n) = ((21-sqrt(21))*(11+sqrt(21))^n+(21+sqrt(21))*(11-sqrt(21))^n )/(42*2^n). [_Klaus Brockhaus_, Sep 26 2009]

%t LinearRecurrence[{11,-25},{1,5},30] (* _Harvey P. Dale_, Oct 02 2016 *)

%Y Cf. A165253.

%K nonn

%O 0,2

%A _Philippe Deléham_, Sep 14 2009