A088306 -id:A088306 - cp's OEIS frontend

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

0, 1, -3, -1, 1, -4, -1, 0, -7, -1, 0, -226, -1, 0, 7, -1, 0, 3, -2, 0, 2, -2, 0, 1, -3, -1, 1, -4, -1, 0, -7, -1, 0, -76, -1, 0, 7, -1, 0, 3, -2, 0, 2, -2, 0, 1, -3, -1, 1, -4, -1, 0, -7, -1, 0, -46, -1, 0, 8, -1, 0, 3, -2, 0, 2, -2, 0, 1, -3, -1, 1, -4, -1, 0, -6, -1, 0, -33, -1, 0, 9, -1, 0, 3, -2, 0, 2, -2, 0, 1, -2, -1, 1, -3, -1, 0, -6, -1, 0, -26

Offset: 0

N. J. A. Sloane

Every integer appears infinitely often. - Charles R Greathouse IV, Aug 06 2012

Does not satisfy Benford's law [Whyman et al., 2016]. - N. J. A. Sloane, Feb 12 2017

T. D. Noe, Table of n, a(n) for n = 0..1000
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. 782-784.
Daniel Forgues and Jon E. Schoenfield, Discussion of A000503
G. Whyman, N. Ohtori, E. Shulzinger, Ed. Bormashenko, Revisiting the Benford law: When the Benford-like distribution of leading digits in sets of numerical data is expectable?, Physica A: Statistical Mechanics and its Applications, 461 (2016), 595-601.
Index entries for sequences related to Benford's law

Cf. A005657, A000493, A000480, A000494, A000484, A088306, A195911, A195910, A037448, A258024.

[Floor(Tan(n)): n in [0..80]]; // Vincenzo Librandi, Jun 13 2015

Maple
```
f := n->floor(evalf(tan(n)));
```

Table[Floor[Tan[n]], {n, 0, 100}] (* Stefan Steinerberger, Apr 09 2006 *)

a(n)=tan(n)\1 \\ Charles R Greathouse IV, Sep 04 2014

More terms from Stefan Steinerberger, Apr 09 2006

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

Original entry on oeis.org

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

Offset: 1

Jacob Vecht, Nov 07 2014

nonn

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

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

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.

Subsequence of positive terms of A088306. Supersequence of A079332.

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

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

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

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

A092328 Solutions of x^2 = ceiling(xrfloor(x/r)) where r=Pi.

Original entry on oeis.org

0, 22, 44, 355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195, 312689, 1146408, 5419351, 10838702

Offset: 1

Benoit Cloitre, Feb 14 2004

Does limit n->infinity log(a(n))/n exist?
Notice that the entries above are either numerators of convergents to Pi (A002485) or multiples thereof. - Robert G. Wilson v, Feb 26 2004
a(23) <= 430010946591069243. - Robert G. Wilson v, Jul 19 2019
From M. F. Hasler, Sep 10 2020: (Start)
Appears to be the same as: n >= 0 such that n*tan(n) < 1, cf. A332095. Is there a counterexample?
Most terms are multiples of a smaller term: 44 = 22*2 and a(4..12) = {355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195} = 355*{1, 2, 3, ..., 9}. See A332095 for more. (End)

Cf. A249836, A088306, A332095.

Do[ If[ n^2 == Ceiling[n*3.1415926535897932346264*Floor[n/3.1415926535897932346264]], Print[n]], {n, 0, 10^8}] (* Robert G. Wilson v, Feb 26 2004 *)

for(x=0,2000000,if(x^2==ceil(Pi*x*floor(x/Pi)),print1(x,",")))

More terms from Robert G. Wilson v, Feb 26 2004

A332095 Numbers m such that 0 <= m*tan(m) < 1, ordered by |m|.

Original entry on oeis.org

0, -3, 22, 44, 355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195, 312689, 1146408, 5419351, 10838702, -6167950454, -21053343141, -42106686282, -63160029423, -84213372564, -105266715705, -8958937768937, -17917875537874, -428224593349304, -856449186698608, -6134899525417045

Offset: 1

M. F. Hasler, Sep 10 2020

sign

Equivalently, 0 together with integers m such that |tan(m)| < 1/m, multiplied by sign(tan(m)).
The term a(2) = 3 is up to 10^7 the only term m for which tan(m) < 0.
A092328 appears to be a subsequence. Does it contain all terms with tan(m) > 0?
Many terms are multiples of a smaller term: 44 = 22*2 and a(4..12) = {355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195} = 355*{1, 2, 3, ..., 9}.
Indeed, if |m*tan(m)| < 1/k^2 for some k = 1, 2, 3..., then (k*m)*tan(k*m) ~ k^2*m*tan(m) < 1. (E.g., for m = 355, m*tan(m) ~ 0.01.)
The "seeds" for which |m*tan(m)| is particularly small are numerators of convergents of continued fractions for Pi (A002485) (and/or Pi/2: A096456), e.g., a(3) = numerator(22/7), a(5) = numerator(355/113), ...
Other terms in the sequence include: -21053343141*{1, 2, 3, 4, 5}, -8958937768937*{1, 2}, -6134899525417045, -66627445592888887, 430010946591069243, -2646693125139304345*{1, 2, 3, 4, 5}, ...
The absolute values of nonzero terms are a subsequence of A337371. - R. J. Mathar, Sep 24 2020
Can someone find a counterexample for which |sin(m)| < 1/m and |m*tan(m)| > 1? - M. F. Hasler, Oct 09 2020

Cf. A092328, A088306, A337371 (similar, with sin, a superset except for the initial term).

is_A332095(n)={tan(n)*n < 1 && n*tan(n) >= 0}
for(n=0,oo, n*abs(tan(n))<1 && print1(sign(tan(n))*n", "))
/* Much faster: apply to numerators of convergents of Pi the function check(n) which prints all nonzero k*n in the sequence and returns the largest such k, largest in magnitude, possibly negative. N.B.: stops when (k+1)n is not in the sequence, so e.g., n = 11 (in convergents of Pi/2) does not give 22 and 44! */
print1(0); apply( {check(n)=for(i=1,oo,abs(i*n*tan(i*n))<1||return(sign(tan(n))*(i-1)); print1(", "sign(tan(n*i))*i*n))}, contfracpnqn(c=contfrac(Pi),#c)[1,]) \\ M. F. Hasler, Oct 09 2020

A337371 Integers k with abs(sin(k)) < 1/k.

Original entry on oeis.org

1, 3, 22, 44, 355, 710, 1065, 1420, 1775, 2130, 2485, 2840, 3195, 312689, 1146408, 5419351, 10838702, 6167950454, 21053343141, 42106686282, 63160029423, 84213372564, 105266715705, 8958937768937, 17917875537874, 428224593349304, 856449186698608, 6134899525417045

Offset: 1

Anian Brosig, Aug 25 2020

nonn

The values > 1 appear to be a subset of the numerators of continued fractions of Pi (A002485) (and/or Pi/2: A096456) and their multiples. Is it possible to find a term k here but not in |A332095| (k |tan k| < 1)? - M. F. Hasler, Oct 09 2020

Cf. A092328, A332095, A046965, A088306, A337249, A332095.

Select[Range[3200], Abs[Sin[#]] < 1/# &] (* Amiram Eldar, Aug 25 2020 *)

print1(1);apply( n-> forstep(n=n,oo,n,abs(sin(n))<1/n||return; print1(","n)), contfracpnqn(c=contfrac(Pi),#c)[1,]); \\ M. F. Hasler, Oct 09 2020

import numpy as np
for x in range(1, 10**9):
    if np.abs(np.sin(x)) < 1/x:
        print(x, end=", ")

More terms from M. F. Hasler, Oct 09 2020

A051512 a(n) = floor(tan(prime(n))).

Original entry on oeis.org

-3, -1, -4, 0, -226, 0, 3, 0, 1, 0, -1, -1, 0, -2, -1, -1, -1, 3, 1, -4, 0, 0, 3, 1, -1, 0, -1, 0, -2, -1, 4, -2, -3, 0, 4, 0, -1, -1, 0, 0, -1, -3, -1, 4, -2, 1, 0, -1, 1, -1, 0, 0, -2, -1, -1, -2, -3, 1, 0, 5, 0, 1, -2, -1, -3, -1, 2, 1, 6, 0, 2, 1, -1, -2, -3, -1, -1, 2, -3, 0, 2, 0, 0, -1, -2, 0, -1, 9, -2, 2, -2, 10

Offset: 1

N. J. A. Sloane

sign

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

Cf. A000503, A000040, A088306, A249836.

Table[Floor[Tan[Prime[n]]], {n, 100}] (* Wesley Ivan Hurt, Mar 28 2015 *)

apply( A051512(n)=tan(prime(n))\1, [1..77]) \\ M. F. Hasler, Sep 10 2020

a(n) = A000503(prime(n)), i.e., this A051512 = A000503 o A000040. - M. F. Hasler, Sep 10 2020

A337248 Numbers k for which sec(k) > k.

Original entry on oeis.org

1, 11, 52174, 260515, 37362253, 42781604, 2685575996367

Offset: 1

Joseph C. Y. Wong, Aug 21 2020

This sequence includes abs(m) for many terms m from A088306, including 1, 11, 52174, 260515, 37362253, 42781604, 2685575996367, 65398140378926, 214112296674652, 12055686754159438, 18190586279576483, 1538352035865186794, 1428599129020608582548671, 103177264599407569664999125, 9322105473781932574489648896, .... - Jon E. Schoenfield, Feb 12 2021
From Wolfe Padawer, Jan 05 2023: (Start)
For any given value in this sequence, it is extremely unlikely that it or its negation is not also in A088306. Take the following facts:
[1] |sec(x)| > |tan(x)| for any finite value of sec(x) and tan(x).
[2] |sec(x)| - |tan(x)| approaches 0, and |sec(x)| and |tan(x)| approach infinity, as x approaches (0.5 + n)*Pi where n is any integer.
[3] Any integer k where |sec(k)| > k or |tan(k)| > k must be close to some value of (0.5 + n)*Pi, increasingly so with larger k.
[4] sec(2685575996367) - |tan(2685575996367)| is approximately 8.437*10^-14.
Therefore, for any integer k > 2685575996367 where sec(k) > k, it must be that sec(k) - |tan(k)| < 8.437*10^-14. In order for sec(k) > k but |tan(k)| < k, it must be that k + 8.437*10^-14 > sec(k) > k, a very small interval that only gets smaller as k increases.
It is thus extremely likely, but not yet explicitly proven, that a(8) = 65398140378926, a(9) = 214112296674652, and a(10) = 12055686754159438. Assuming it exists, the smallest k for which sec(k) > k but |tan(k)| < k is probably very large, and it is unknown whether it is currently computable. (End)

			sec(1) = 1.8508... so 1 is a term.

Subsequence of A337371.

Cf. A337249, A092328, A249836.

Select[Range[10^6], Sec[#] > # &] (* Amiram Eldar, Aug 21 2020 *)

isok(m) = 1/cos(m) > m; \\ Michel Marcus, Aug 27 2020

import math
i = 1
while True:
  if 1 / math.cos(i) > i:
    print(i)
  i += 1

a(7) from Wolfe Padawer, Jan 05 2023

cp's OEIS Frontend

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

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

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A092328 Solutions of x^2 = ceiling(x*r*floor(x/r)) where r=Pi.

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A332095 Numbers m such that 0 <= m*tan(m) < 1, ordered by |m|.

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A337371 Integers k with abs(sin(k)) < 1/k.

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A051512 a(n) = floor(tan(prime(n))).

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A337248 Numbers k for which sec(k) > k.

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A092328 Solutions of x^2 = ceiling(xrfloor(x/r)) where r=Pi.