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a(0)=1, a(1)=7, a(n)=15*a(n-1)-49*a(n-2) for n>1.
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%I #13 Sep 02 2021 04:01:58

%S 1,7,56,497,4711,46312,463841,4688327,47596696,484222417,4931098151,

%T 50239573832,511969798081,5217807853447,53180597695736,

%U 542036380617137,5524696422165991,56310663682250152,573949830547618721

%N a(0)=1, a(1)=7, a(n)=15*a(n-1)-49*a(n-2) for n>1.

%C a(n)/a(n-1) tends to (15+sqrt(29))/2=10,192582...

%C For n>=2, a(n) equals 7^n times the permanent of the (2n-2)X(2n-2) tridiagonal matrix with 1/sqrt(7)'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 (15,-49).

%F G.f.: (1-8x)/(1-15x+49x^2).

%F a(n) = Sum_{k=0..n} A165253(n,k)*7^(n-k).

%F a(n) = ((29-sqrt(29))*(15+sqrt(29))^n+(29+sqrt(29))*(15-sqrt(29))^n )/(58*2^n). [_Klaus Brockhaus_, Sep 26 2009]

%t LinearRecurrence[{15,-49},{1,7},20] (* _Harvey P. Dale_, Jun 04 2021 *)

%Y Cf. A165253.

%K nonn

%O 0,2

%A _Philippe Deléham_, Sep 14 2009