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A081704
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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.
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5
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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
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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a(n+1) = (a(n)^2 + 3^n) / a(n-1).
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
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
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MAPLE
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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));
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MATHEMATICA
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a[0]=1; a[1]=3; a[n_] := a[n]=5a[n-1]-3a[n-2]; Array[a, 25, 0]
LinearRecurrence[{5, -3}, {1, 3}, 30] (* Harvey P. Dale, Jul 28 2013 *)
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PROG
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(PARI) Vec((1-2*x)/(1-5*x+3*x^2) + O(x^30)) \\ Colin Barker, Nov 26 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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Victor Ufnarovski (ufn(AT)maths.lth.se), Apr 02 2003
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STATUS
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approved
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