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A083697
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2^(2^n-1) * Fibonacci(2^n).
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1
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OFFSET
| 0,2
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COMMENTS
| A083696(n)/a(n) converges to sqrt(5). a(n)=2*a(n-1)*A083696(n-1). Similar to A081460: a(n) is the denominator of the same mapping f(r)=(1/2)(r+5/r) but with initial value r=1.
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FORMULA
| a(n)=Sum[Prod(2^n-k), (k=0, 2r)]5^r/(2r+1)!, (r=0, 2^n-1).
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MATHEMATICA
| Table[Sum[Product[2^n - k, {k, 0, 2*r}]k^r/(2*r + 1)!, {r, 0, 2^n - 1}], {n, 0, 8}]
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CROSSREFS
| Equals A058635(n) * A058891(n).
Sequence in context: A053995 A184595 A007079 * A110131 A112332 A101339
Adjacent sequences: A083694 A083695 A083696 * A083698 A083699 A083700
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KEYWORD
| easy,nonn
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AUTHOR
| Mario Catalani (mario.catalani(AT)unito.it), May 22 2003
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EXTENSIONS
| The next term is too large to include.
Better description from Ralf Stephan, Aug 29 2004
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