

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.


5



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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 A077382 A046527
Adjacent sequences: A258017 A258018 A258019 * A258021 A258022 A258023


KEYWORD

nonn


AUTHOR

Antti Karttunen, May 24 2015


STATUS

approved



