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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. 5
 1, 3, 12, 51, 219, 942, 4053, 17439, 75036, 322863, 1389207, 5977446, 25719609, 110665707, 476169708, 2048851419, 8815747971, 37932185598, 163213684077, 702271863591, 3021718265724, 13001775737847, 55943723892063, 240713292246774, 1035735289557681 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS f satisfies the linear recursion f(n+1) = (t+2)f(n)-tf(n-1)). For t=3 this gives a(n+1) = 5*a(n)-3*a(n-1). 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 LINKS Colin Barker, Table of n, a(n) for n = 0..1000 Index entries for linear recurrences with constant coefficients, signature (5,-3). FORMULA a(n+1) = (a(n)^2 + 3^n) / a(n-1). From Philippe Deléham, Nov 14 2008: (Start) G.f.: (1-2*x)/(1-5*x+3*x^2). a(n) = Sum_{k, 0<=k<=n} A147703(n,k)*2^k. (End) 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 MAPLE 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)); MATHEMATICA a[0]=1; a[1]=3; a[n_] := a[n]=5a[n-1]-3a[n-2] LinearRecurrence[{5, -3}, {1, 3}, 30] (* Harvey P. Dale, Jul 28 2013 *) PROG (PARI) Vec((1-2*x)/(1-5*x+3*x^2) + O(x^30)) \\ Colin Barker, Nov 26 2016 CROSSREFS Cf. A006012, A001519. Equals 3*A018902(n-1) for n>0. Sequence in context: A155179 A228770 A104268 * A166482 A007854 A151182 Adjacent sequences:  A081701 A081702 A081703 * A081705 A081706 A081707 KEYWORD nonn,easy AUTHOR Victor Ufnarovski (ufn(AT)maths.lth.se), Apr 02 2003 STATUS approved

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Last modified December 16 14:52 EST 2018. Contains 318167 sequences. (Running on oeis4.)