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

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

Table of n, a(n) for n=1..78.

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,
- f(4/5) = 2/5,
- hence a(5) = 2.

Prog

(PARI) a(n, f=x -> min(2*x, 2-2*x)) = my (x=f(1/n), y=f(x)); while (x!=y, x=f(x); y=f(f(y))); for (k=1, oo, if (x==y=f(y), return (k)))

Crossrefs

Cf. A003558, A020714, A029744.

Sequence in context: A007735 A002616 A046073 * A162912 A230070 A224762

Adjacent sequences: A309783 A309784 A309785 * A309787 A309788 A309789

Keyword

nonn

Author

Rémy Sigrist, Aug 17 2019

Status

approved