Discussion of A000503 from Daniel Forgues and Jon E. Schoenfield, May 27 2015 - June 18 2015

DF: It is an open problem whether tan(n) > n for infinitely many integers n; thus a fortiori whether n < tan(n) < n+1 for infinitely many integers n;
  and tan(n) = n for integer n only when n = 0.
J. K. Haugland found n = 37362253 s.t. tan(n) > n.
Bob Delaney found n = 3083975227 s.t. tan(n) > n.
For n where tan(n) > n, see A249836.
Floor(tan(n)) = n yields a fixed point of the iterated Floor(tan(n)).
Currently, the only known fixed points are 0 and 1. (Cf. A258024.)
It is proved that |tan n| > n for infinitely many n, and that tan n > n/4 for infinitely many n. (Bellamy, Lagarias, Lazebnik) 

JES: Is it really an open problem whether tan(n) > n for infinitely many integers n? Since |tan n| > n for infinitely many n, it seems very unlikely that tan(n) > n for only finitely many n, since this would require the existence of a largest number n for which tan(n)/n > 1, even though there would remain an infinite number of larger numbers n for which |tan n| / n > 1; for every one of them, tan(n) would have to be negative. The behavior of the first 1000 terms of A088306, including the nearly equal frequency of positive and negative values of tan(n)/n among them, would be consistent with the seemingly far more likely situation that, among positive values of n where |tan n| > n, tan(n) shows no bias toward negative values, much less any indication that only the positive values will come to an end.

DF: Since tan(n) has a transcendental period, namely Pi, it seems very likely that not only tan(n) > n for infinitely many integers n, but also that tan(n) > kn for infinitely many integers n, for any integer k. It even seems very likely that not only n < tan(n) < n+1 for infinitely many integers n (not only for n = 1), but also that kn < tan(n) < kn + 1, for infinitely many integers n, for any integer k. It seems that we are bound to stumble upon the required (to satisfy any condition) positive delta s.t. n mod Pi = Pi/2 - delta. 

For n=0 to 1000, the 1st convergent of Pi, 3/1, yields:
(poor convergent of Pi: a(n) drifts rapidly...)
For n=0 to 1000, the 2nd convergent of Pi, 22/7, yields:
n = 11 = 11 + 0*22: a(n) = -226;
n = 33 = 11 + 1*22: a(n) = -76;
n = 55 = 11 + 2*22: a(n) = -46; (...)
For n=0 to 1000, the 3rd convergent of Pi, 333/106, yields:
n = 322 = -11 + 1*333: a(n) = 75;
n = 655 = -11 + 2*333: a(n) = 45;
n = 988 = -11 + 3*333: a(n) = 32;
and:
n = 344 = 11 + 1*333: a(n) = 227;
n = 677 = 11 + 2*333: a(n) = 75;
For n=0 to 1000, the 4th convergent of Pi, 355/113, yields:
n = 11 = 11 + 0*355: a(n) = -226;
n = 366 = 11 + 1*355: a(n) = -225;
n = 721 = 11 + 2*355: a(n) = -223;
and:
n = 344 = -11 + 1*355: a(n) = 227;
n = 699 = -11 + 2*355: a(n) = 229.
11 mod Pi = Pi/2 - 0.0044257..., -11 mod Pi = -Pi/2 + 0.0044257... and 355/113 = 3.1415929203539... is a much better approximation of Pi than the lower convergents (of the simple continued fraction) of Pi.