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%I A286948 #23 Jun 21 2017 13:08:13
%S A286948 0,1,1,2,2,4,5,7,10,12,17,21,29,36,48,60,78,96,124,151,190,234,290,
%T A286948 354,436,529,648,784,952,1141,1382,1651,1984,2367,2831,3359,3999,4733,
%U A286948 5608,6614,7816,9178,10802,12667,14850,17356,20297,23653,27579,32062,37277,43235,50139
%N A286948 a(n) is the number of distinct products p of Fibonacci numbers such that Fibonacci(n) < p <= Fibonacci(n + 1).
%C A286948 The products of Fibonacci numbers are not all distinct. For example, 144 is a product of Fibonacci numbers in more than 1 way but 144 still counts as one product. 8 and 144 are the only Fibonacci numbers with this property (see A235383).
%C A286948 a(n+1)/a(n) seems to converge to about 1.1; up to n = 70, the value drops slightly. Maybe to sqrt(5)/2?
%C A286948 Fibonacci(n) ~= phi^n / sqrt(5) where phi = (1 + sqrt(5)) / 2. If m is a product of k Fibonacci numbers, m is of the form Fibonacci(n_1) *...* Fibonacci(n_k). To count the numbers just once, we restrict n_i for 1 <= i <= k.
%C A286948 Fibonacci(1) = Fibonacci(2) = 1 isn't counted, products with factor Fibonacci(6) = 8 aren't counted and products with the factor Fibonacci(12) = 144 aren't counted. I.e., n_i >= 3, n_i != 6 and n_i != 12.
%C A286948 We can write m uniquely as m = Product_{i=1..k} Fibonacci(n_i) ~= Product_{i=1..k} (phi^(n_i) / sqrt(5)) = phi^(Sum_{i=1..k} n_i) / sqrt(5)^k. To determine the number of such products up to f = Fibonacci(x) of k such Fibonacci factors, we can find an upper bound on Sum_{i=1..k} n_i of about (k*log(5)/2 + log(x)) / log(phi). This somewhat relates this sequence to the partitions.
%e A286948 The products of Fibonacci numbers larger than Fibonacci(7) = 13 and smaller than or equal to Fibonacci(7 + 1) = 21 are the five numbers 15, 16, 18, 20 and 21. Therefore a(7) = 5.
%Y A286948 Cf. A000041, A000045, A065108, A235383.
%K A286948 nonn
%O A286948 1,4
%A A286948 _David A. Corneth_, Jun 11 2017

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