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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: A191332 A229982 A289848 * A021803 A077382 A046527

Adjacent sequences:  A258017 A258018 A258019 * A258021 A258022 A258023

KEYWORD

nonn

AUTHOR

Antti Karttunen, May 24 2015

STATUS

approved

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Last modified July 21 17:43 EDT 2019. Contains 325198 sequences. (Running on oeis4.)