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FORMULA
| a(n) = Fibonacci(3^n). - Leroy Quet, March 17, 2002
The first example I know in which a(n) can be expressed as (4/5)^(1/2)*cosh(3^n*argch((5/4)^(1/2)).
a(n+1) = a(n)*A002814(n+1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 16 2003
a(n) = (G^(3^n) - (1 - G)^(3^n))/Sqrt[5] where G = GoldenRatio = (1 + Sqrt[5])/2 [From Artur Jasinski (grafix(AT)csl.pl), Oct 05 2008]
a(n)=(4/5)^(1/2)*cosh((3^n)*arccosh((5/4)^(1/2)). [From Artur Jasinski (grafix(AT)csl.pl), Oct 05 2008]
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MATHEMATICA
| G = (1 + Sqrt[5])/2; Table[Expand[(G^(3^n) - (1 - G)^(3^n))/Sqrt[5]], {n, 1, 7}] [From Artur Jasinski (grafix(AT)csl.pl), Oct 05 2008]
Table[Round[(4/5)^(1/2)*Cosh[3^n*ArcCosh[((5/4)^(1/2))]]], {n, 1, 4}] [From Artur Jasinski (grafix(AT)csl.pl), Oct 05 2008]
RecurrenceTable[{a[0]==1, a[n]==5a[n-1]^3-3a[n-1]}, a[n], {n, 6}] (* From Harvey P. Dale, Oct 24 2011 *)
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