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Numbers of the form Fibonacci(2^c)/Fibonacci(2^b), 1 <= b < c.
2

%I #7 Jan 21 2020 00:14:40

%S 3,7,21,47,329,987,2207,103729,726103,2178309,4870847,10610209857723,

%T 23725150497407,10749959329,505248088463,3536736619241

%N Numbers of the form Fibonacci(2^c)/Fibonacci(2^b), 1 <= b < c.

%C 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.

%F 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).

%e If k=3, m=1, by the latter formula, we have a(8) = A001566(2)*A001566(3) = 47*2207 = 103729.

%Y Cf. A000045, A001566.

%K nonn

%O 1,1

%A _Vladimir Shevelev_, Oct 17 2010