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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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4
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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; refs; listen; history; internal format)
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OFFSET
| 0,2
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COMMENTS
| 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).
Given the 3x3 matrix [1,1,1; 1,1,2; 1,1,3] = M, a(n) = term (1,1) in M^(n+1). [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 06 2010]
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FORMULA
| 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
| 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
| a[0]=1; a[1]=3; a[n_] := a[n]=5a[n-1]-3a[n-2]
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CROSSREFS
| Cf. A006012, A001519.
Equals 3*A018902(n-1), n>0.
Sequence in context: A083314 A155179 A104268 * A166482 A007854 A151182
Adjacent sequences: A081701 A081702 A081703 * A081705 A081706 A081707
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KEYWORD
| nonn
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AUTHOR
| Victor Ufnarovski (ufn(AT)maths.lth.se), Apr 02 2003
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