The sequence of numbers {Q(m)} is defined recursively by the following relationships:
Q(1) = 1, and:
Q(m) = 1 + Q(m/2), whenever m is ≥ 2 and even, and:
Q(m) = 1/Q(m-1), whenever m is ≥ 3 and odd.
Determine the value of d, given that Q(d) = 19/87
(In reply to
No Subject by Dej Mar)
In other words, find the continued fraction:
19/87 = [0; 4, 1, 1, 2, 1, 2],
take the partial sums, and decrement the last number by 1:
{0, 4, 5, 6, 8, 9, 11-1}
These are the positions of the 1s in the binary expansion of
11101110001 = 1905.
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Posted by Eigenray
on 2008-02-16 03:41:53 |