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A081704 Let f(0)=1, f(1)=t, f(n+1) = (f(n)^2+t^n)/f(n-1). f(t) is a polynomial with integer coefficients. Then a(n) = f(n) when t=3.

%I A081704 #26 Aug 23 2024 22:05:33
%S A081704 1,3,12,51,219,942,4053,17439,75036,322863,1389207,5977446,25719609,
%T A081704 110665707,476169708,2048851419,8815747971,37932185598,163213684077,
%U A081704 702271863591,3021718265724,13001775737847,55943723892063,240713292246774,1035735289557681
%N A081704 Let f(0)=1, f(1)=t, f(n+1) = (f(n)^2+t^n)/f(n-1). f(t) is a polynomial with integer coefficients. Then a(n) = f(n) when t=3.
%C A081704 f satisfies the linear recursion f(n+1) = (t+2)*f(n)-t*f(n-1). For t=3 this gives a(n+1) = 5*a(n)-3*a(n-1).
%C A081704 Given the 3 X 3 matrix [1,1,1; 1,1,2; 1,1,3] = M, a(n) = term (1,1) in M^(n+1). - _Gary W. Adamson_, Aug 06 2010
%H A081704 Colin Barker, <a href="/A081704/b081704.txt">Table of n, a(n) for n = 0..1000</a>
%H A081704 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (5,-3).
%F A081704 a(n+1) = (a(n)^2 + 3^n) / a(n-1).
%F A081704 From _Philippe Deléham_, Nov 14 2008: (Start)
%F A081704 G.f.: (1-2*x)/(1-5*x+3*x^2).
%F A081704 a(n) = Sum_{k, 0<=k<=n} A147703(n,k)*2^k. (End)
%F A081704 a(n) = (2^(-1-n)*((5-sqrt(13))^n*(-1+sqrt(13)) + (1+sqrt(13))*(5+sqrt(13))^n))/sqrt(13). - _Colin Barker_, Nov 26 2016
%F A081704 E.g.f.: exp(5*x/2)*(sqrt(13)*cosh(sqrt(13)*x/2) + sinh(sqrt(13)*x/2))/sqrt(13). - _Stefano Spezia_, Jul 09 2022
%p A081704 f := proc(n) if n=0 then 1 elif n=1 then t else sort(simplify((f(n-1)^2+t^(n-1))/f(n-2)),t) fi end; a := i->subs(t=3,f(i));
%t A081704 a[0]=1; a[1]=3; a[n_] := a[n]=5a[n-1]-3a[n-2]; Array[a,25,0]
%t A081704 LinearRecurrence[{5,-3},{1,3},30] (* _Harvey P. Dale_, Jul 28 2013 *)
%o A081704 (PARI) Vec((1-2*x)/(1-5*x+3*x^2) + O(x^30)) \\ _Colin Barker_, Nov 26 2016
%Y A081704 Cf. A006012, A001519.
%Y A081704 Equals 3*A018902(n-1) for n>0.
%K A081704 nonn,easy
%O A081704 0,2
%A A081704 Victor Ufnarovski (ufn(AT)maths.lth.se), Apr 02 2003