1, 2, 2, 1, 2, 2, 4, 2, 8, 4, 32, 8, 256, ?, ?, ?, ...
(In reply to
Solution - no spoiler by Lisa)
f(1) = 0
f(2) = 1
For n odd, f(n) = 2 ^ [log˛{f(n-2)} + log˛{f(n-1)}]
For n even, f(n) = 2 ^ | [log˛{f(n-2)} - log˛{f(n-1)}] |
I'm posting this wondering if there's a better way to express this. The
solution is so uncomplicated, yet the formula seems rather unwieldy.
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Posted by Charley
on 2005-06-20 14:26:16 |
Spoiler
