
COMMENTS

Supersequence of A079332, hence 3083975227 and 214112296674652 are members.  Charles R Greathouse IV, Nov 07 2014
This sequence consists of all positivevalued terms of A088306. (Of the first 1000 terms in A088306, 518 are positive.)  Jon E. Schoenfield, Nov 07 2014
From Daniel Forgues, May 27 2015, Jun 12 2005: (Start)
Numbers n for which tanc(n) > 1, where tanc(n) = tan(n)/n, tanc(0) = 1, where n are radians; cf. Weisstein link.
It is an open problem whether tan(n) > n for infinitely many integer n.
Jan Kristian Haugland found a(3) = 37362253, Bob Delaney found a(6) = 3083975227.
For n <= tan(n) < n+1, or floor(tan(n)) = n, we get 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) (End)
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 likely that not only n < tan(n) < n+1 for infinitely many integers n (not just 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 requisite positive delta s.t. n mod Pi = Pi/2  delta.  Daniel Forgues, Jun 15 2015
It appears that we need {n / Pi} = 0.5  delta, with delta < k/n, for some k, where {.} denotes the fractional part: we have, 260515/Pi = 82924.49999917..., 37362253/Pi = 11892774.4999999915... etc.  Daniel Forgues, Jun 18 2015, edited by M. F. Hasler, Aug 19 2015
Indeed, from the graph of the function we see that tan(n) > n for numbers of the form n = (m + 1/2)*Pi  epsilon (i.e., n/Pi = m + 1/2  epsilon/Pi) with small epsilon > 0, for which tan(n) = tan((m + 1/2)*Pi  epsilon) = tan(Pi/2epsilon) ~ 1/epsilon, using tan(Pi/2x) = sin(Pi/2x)/cos(Pi/2x) = cos(x)/sin(x) ~ 1/x as x > 0. Thus tan(n) > n if epsilon < 1/n, or delta = epsilon/Pi < k/n with k = 1/Pi.  M. F. Hasler, Aug 19 2015


LINKS

Table of n, a(n) for n=1..16.
David P. Bellamy, Jeffrey C. Lagarias, Felix Lazebnik, Proposed Problem: Large Values of Tan n
David P. Bellamy, Jeffrey C. Lagarias, Felix Lazebnik and Stephen M. Gagola, Jr., Large Values of Tangent: 10656, The American Mathematical Monthly, Vol. 106, No. 8 (Oct. 1999), pp. 782784.
Eric W. Weisstein, Tanc Function. From MathWorldA Wolfram Web Resource.
