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A181393
Numbers of the form Fibonacci(2^c)/Fibonacci(2^b), 1 <= b < c.
2
3, 7, 21, 47, 329, 987, 2207, 103729, 726103, 2178309, 4870847, 10610209857723, 23725150497407, 10749959329, 505248088463, 3536736619241
OFFSET
1,1
COMMENTS
Using an Eratosthenes-like sieve, we find "primes" of the form P_k = Fibonacci(2^(k+1)) / Fibonacci(2^k) = A001566(k-1), k=1,2,..., such that every term has a unique "prime" factorization.
FORMULA
For n >= 1, a((n^2-n+2)/2) = P_n = A001566(n-1); for 1 <= m < k, a((k^2+3*k)/2-m)/2) = Product_{i=m+1..k} A001566(i).
EXAMPLE
If k=3, m=1, by the latter formula, we have a(8) = A001566(2)*A001566(3) = 47*2207 = 103729.
CROSSREFS
Sequence in context: A092203 A018760 A050614 * A036569 A018303 A098545
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Oct 17 2010
STATUS
approved