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%I #13 Jun 05 2015 03:51:21
%S 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,
%T 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,
%U 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
%N Number of steps to reach a fixed point with map x -> floor(tan(x)) when starting the iteration with the initial value x = n.
%C Note that this sequence lists such values only for nonnegative integers, although the function is defined in all Z.
%H Antti Karttunen, <a href="/A258020/b258020.txt">Table of n, a(n) for n = 0..10000</a>
%F 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.]
%e 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.
%e 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.
%o (Scheme) (define (A258020 n) (if (= n (floor->exact (tan n))) 0 (+ 1 (A258020 (floor->exact (tan n))))))
%Y Cf. A000503, A258021, A258022, A258024, A258201.
%K nonn
%O 0,3
%A _Antti Karttunen_, May 24 2015
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