A249836 - OEIS

%I #113 Jan 10 2022 11:03:00

%S 1,260515,37362253,122925461,534483448,3083975227,902209779836,

%T 74357078147863,214112296674652,642336890023956,18190586279576483,

%U 248319196091979065,1108341089274117551,118554299812338354516058,1428599129020608582548671,4285797387061825747646013

%N Numbers n for which tan(n) > n.

%C Supersequence of A079332, hence 3083975227 and 214112296674652 are members. - _Charles R Greathouse IV_, Nov 07 2014

%C This sequence consists of all positive-valued terms of A088306. (Of the first 1000 terms in A088306, 518 are positive.) - _Jon E. Schoenfield_, Nov 07 2014

%C From _Daniel Forgues_, May 27 2015, Jun 12 2005: (Start)

%C Numbers n for which tanc(n) > 1, where tanc(n) = tan(n)/n, tanc(0) = 1, where n are radians; cf. Weisstein link.

%C It is an open problem whether tan(n) > n for infinitely many integer n.

%C _Jan Kristian Haugland_ found a(3) = 37362253, Bob Delaney found a(6) = 3083975227.

%C 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.)

%C 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)

%C 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 such that n mod Pi = Pi/2 - delta. - _Daniel Forgues_, Jun 15 2015

%C 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

%C 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/2-epsilon) ~ 1/epsilon, using tan(Pi/2-x) = sin(Pi/2-x)/cos(Pi/2-x) = 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

%C The first prime term is a(28). - _Jacob Vecht_, Aug 09 2020

%H Michel Marcus, <a href="/A249836/b249836.txt">Table of n, a(n) for n = 1..518</a> (from A088306 b-file).

%H David P. Bellamy, Jeffrey C. Lagarias, and Felix Lazebnik, <a href="http://math.udel.edu/~lazebnik/papers/tan_n.pdf">Proposed Problem: Large Values of Tan n</a>

%H David P. Bellamy, Jeffrey C. Lagarias, Felix Lazebnik and Stephen M. Gagola, Jr., <a href="http://www.jstor.org/stable/2589040">Large Values of Tangent: 10656</a>, The American Mathematical Monthly, Vol. 106, No. 8 (Oct. 1999), pp. 782-784.

%H Matt Parker, <a href="https://youtu.be/A7eJb8n8zAw">What is the biggest tangent of a prime?</a>, Channel Stand-up Maths, YouTube, Aug 19 2020.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TancFunction.html">Tanc Function</a>.

%F log(a(n)) / n ~ Pi, conjectured. - _M. F. Hasler_, Sep 10 2020 [corrected thanks to _Vaclav Kotesovec_, Feb 22 2021]

%e tan(1) = 1.557... > 1 so 1 is a member.

%t a249836[n_Integer] := Select[Range[n], Tan[#] > # &]; a249836[270000] (* _Michael De Vlieger_, Nov 23 2014 *)

%o (PARI) is(n)=tan(n)>n \\ _Charles R Greathouse IV_, Nov 07 2014

%Y Subsequence of positive terms of A088306. Supersequence of A079332.

%Y Cf. A000503(n) = floor(tan(n)).

%K nonn

%O 1,2

%A _Jacob Vecht_, Nov 07 2014