A249836 Numbers n for which tan(n) > n.

1, 260515, 37362253, 122925461, 534483448, 3083975227, 902209779836, 74357078147863, 214112296674652, 642336890023956, 18190586279576483, 248319196091979065, 1108341089274117551, 118554299812338354516058, 1428599129020608582548671, 4285797387061825747646013

Offset

1,2

Comments

Supersequence of A079332, hence 3083975227 and 214112296674652 are members. - Charles R Greathouse IV, Nov 07 2014

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

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 such that 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/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

The first prime term is a(28). - Jacob Vecht, Aug 09 2020

Links

Michel Marcus, Table of n, a(n) for n = 1..518 (from A088306 b-file).

David P. Bellamy, Jeffrey C. Lagarias, and 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. 782-784.

Matt Parker, What is the biggest tangent of a prime?, Channel Stand-up Maths, YouTube, Aug 19 2020.

Eric Weisstein's World of Mathematics, Tanc Function.

Formula

log(a(n)) / n ~ Pi, conjectured. - M. F. Hasler, Sep 10 2020 [corrected thanks to Vaclav Kotesovec, Feb 22 2021]

Example

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

Mathematica

a249836[n_Integer] := Select[Range[n], Tan[#] > # &]; a249836[270000] (* Michael De Vlieger, Nov 23 2014 *)

Prog

(PARI) is(n)=tan(n)>n \\ Charles R Greathouse IV, Nov 07 2014

Crossrefs

Subsequence of positive terms of A088306. Supersequence of A079332.

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

Sequence in context: A252445 A184781 A146897 * A226111 A002272 A172850

Adjacent sequences: A249833 A249834 A249835 * A249837 A249838 A249839

Keyword

nonn

Author

Jacob Vecht, Nov 07 2014

Status

approved