A258020 Number of steps to reach a fixed point with map x -> floor(tan(x)) when starting the iteration with the initial value x = n.

0, 0, 2, 5, 1, 5, 5, 1, 6, 5, 1, 2, 5, 1, 2, 5, 1, 6, 4, 1, 3, 4, 1, 1, 2, 5, 1, 5, 5, 1, 6, 5, 1, 6, 5, 1, 2, 5, 1, 6, 4, 1, 3, 4, 1, 1, 2, 5, 1, 5, 5, 1, 6, 5, 1, 4, 5, 1, 7, 5, 1, 6, 4, 1, 3, 4, 1, 1, 2, 5, 1, 5, 5, 1, 2, 5, 1, 7, 5, 1, 6, 5, 1, 6, 4, 1, 3, 4, 1, 1, 4, 5, 1, 2, 5, 1, 2, 5, 1, 5, 5, 1, 6, 5, 1, 2, 4, 1, 3, 4, 1, 1, 4, 5, 1, 2, 5, 1, 2, 5, 1

Offset

0,3

Comments

Note that this sequence lists such values only for nonnegative integers, although the function is defined in all Z.

Links

Antti Karttunen, Table of n, a(n) for n = 0..10000

Formula

If n is equal to floor(tan(n)) then a(n) = 0; for any other n (positive or negative): a(n) = 1 + a(floor(tan(n))). [The domain of the recurrence is whole Z.]

Example

The only known fixed points of function x -> floor(tan(x)) are 0 and 1 (and it is conjectured there are no others), thus a(0) = a(1) = 0.
For n=2, we get tan(2) = -2.185, thus floor(tan(2)) = -3. tan(-3) = 0.1425, thus floor(tan(-3)) = 0, and we have reached a fixed point in two steps, thus a(2) = 2.

Prog

(Scheme) (define (A258020 n) (if (= n (floor->exact (tan n))) 0 (+ 1 (A258020 (floor->exact (tan n))))))

Crossrefs

Cf. A000503, A258021, A258022, A258024, A258201.

Sequence in context: A229982 A327123 A289848 * A021803 A356273 A077382

Adjacent sequences: A258017 A258018 A258019 * A258021 A258022 A258023

Keyword

nonn

Author

Antti Karttunen, May 24 2015

Status

approved