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Search: id:A081704
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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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+0
2
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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
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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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)=5a(n)-3a(n-1).
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FORMULA
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a(n+1)=(a(n)^2+3^n)/a(n-1)
G.f.: (1-2x)/(1-5x+3*x^2). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 14 2008]
a(n)=Sum_{k, 0<=k<=n}A147703(n,k)*2^k. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 14 2008]
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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]
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CROSSREFS
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Cf. A006012, A001519.
Equals 3*A018902(n-1), n>0.
Sequence in context: A135343 A083314 A104268 this_sequence A007854 A151182 A151316
Adjacent sequences: A081701 A081702 A081703 this_sequence A081705 A081706 A081707
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KEYWORD
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nonn,new
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AUTHOR
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Victor Ufnarovski (ufn(AT)maths.lth.se), Apr 02 2003
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