A286948 a(n) is the number of distinct products p of Fibonacci numbers such that Fibonacci(n) < p <= Fibonacci(n + 1).

0, 1, 1, 2, 2, 4, 5, 7, 10, 12, 17, 21, 29, 36, 48, 60, 78, 96, 124, 151, 190, 234, 290, 354, 436, 529, 648, 784, 952, 1141, 1382, 1651, 1984, 2367, 2831, 3359, 3999, 4733, 5608, 6614, 7816, 9178, 10802, 12667, 14850, 17356, 20297, 23653, 27579, 32062, 37277, 43235, 50139

Offset: 1

Author

David A. Corneth, Jun 11 2017

Keywords

• nonn

Comments

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).

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?

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.

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.

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.

Examples

			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.
		

Crossrefs

Cf. A000041, A000045, A065108, A235383.