A309786 - OEIS

A309786

a(n) is the length of the cycle of the trajectory of 1/n under the map f(x) = min(2*x, 2-2*x).

1, 1, 1, 1, 2, 1, 3, 1, 3, 2, 5, 1, 6, 3, 4, 1, 4, 3, 9, 2, 6, 5, 11, 1, 10, 6, 9, 3, 14, 4, 5, 1, 5, 4, 12, 3, 18, 9, 12, 2, 10, 6, 7, 5, 12, 11, 23, 1, 21, 10, 8, 6, 26, 9, 20, 3, 9, 14, 29, 4, 30, 5, 6, 1, 6, 5, 33, 4, 22, 12, 35, 3, 9, 18, 20, 9, 30, 12

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OFFSET

1,5

COMMENTS

For any rational number q in the interval [0, 1]:

- f(q) is rational and lies in the interval [0, 1],

- denominator(f(q)) <= denominator(q),

- as there are only finitely many rational numbers whose denominator is less than or equal to that of q in the interval [0, 1],

- iteratively applying f to q eventually leads to a cycle,

- and the sequence is well defined.

For any irrational number x in the interval [0, 1]:

- f(x) is irrational and lies in the interval [0, 1],

- for any k > 0, as fixed points of the k-th iterate of f are rational,

- iteratively applying f to x never leads to a cycle.

As f is continuous on the interval [0, 1] and 1/7 has least period 3, according to the period three theorem, f has points of any period length, as well as chaotic points.

LINKS

Roman Khrabrov, Table of n, a(n) for n = 1..10000

Eric Weisstein's World of Mathematics, Period Three Theorem

FORMULA

a(2*n-1) = A003558(n-1).

a(2*n) = a(n).

a(n) = 1 iff n belongs to A029744.

a(n) = 2 iff n belongs to A020714.

EXAMPLE

For n = 5:

- f(1/5) = 2/5,

- f(2/5) = 4/5,