A000484 -id:A000484 - 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

A057682 a(n) = Sum_{j=0..floor(n/3)} (-1)^jbinomial(n,3j+1).

Original entry on oeis.org

0, 1, 2, 3, 3, 0, -9, -27, -54, -81, -81, 0, 243, 729, 1458, 2187, 2187, 0, -6561, -19683, -39366, -59049, -59049, 0, 177147, 531441, 1062882, 1594323, 1594323, 0, -4782969, -14348907, -28697814, -43046721, -43046721, 0, 129140163, 387420489, 774840978

Offset: 0

N. J. A. Sloane, Oct 20 2000

Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n)-a(n)*M+z(n)*M^2, where z(0)=z(1)=0 and, apparently, z(n+2)=A057083(n). - Stanislav Sykora, Jun 10 2012
From Tom Copeland, Nov 09 2014: (Start)
This array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interp. (here t=-2) o.g.f. G(x,t) = x(1-x)/[1+(t-1)x(1-x)] and inverse o.g.f. Ginv(x,t) =  [1-sqrt(1-4x/(1+(1-t)x))]/2 (Cf. A005773 and A091867 and A030528 for more info on this family). (End)
{A057681, A057682, A*}, where A* is A057083 prefixed by two 0's, is the difference analog of the trigonometric functions {k_1(x), k_2(x), k_3(x)} of order 3. For the definitions of {k_i(x)} and the difference analog {K_i (n)} see [Erdelyi] and the Shevelev link respectively. - Vladimir Shevelev, Jul 31 2017

			G.f. = x + 2*x^2 + 3*x^3 + 3*x^4 - 9*x^6 - 27*x^7 - 54*x^8 - 81*x^9 + ...
If M^3=1 then (1-M)^6 = A057681(6) - a(6)*M + A057083(4)*M^2 = -18 + 9*M + 9*M^2. - _Stanislav Sykora_, Jun 10 2012

A. Erdelyi, Higher Transcendental Functions, McGraw-Hill, 1955, Vol. 3, Chapter XVIII.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Mark W. Coffey, Reductions of particular hypergeometric functions 3F2 (a, a+1/3, a+2/3; p/3, q/3; +-1), arXiv preprint arXiv:1506.09160 [math.CA], 2015.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for linear recurrences with constant coefficients, signature (3,-3).

Alternating row sums of triangle A030523.

Cf. A000108, A000484, A000748, A005043, A005773, A057083, A057681, A091867.

I:=[0,1,2]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2): n in [1..45]]; // Vincenzo Librandi, Nov 10 2014

A057682:=n->add((-1)^j*binomial(n,3*j+1), j=0..floor(n/3)):
seq(A057682(n), n=0..50); # Wesley Ivan Hurt, Nov 11 2014

A[n_] := Array[KroneckerDelta[#1, #2 + 1] - KroneckerDelta[#1, #2] + Sum[KroneckerDelta[#1, #2 -q], {q, n}] &, {n, n}];
Join[{0,1}, Table[(-1)^(n-1)*Total[CoefficientList[ CharacteristicPolynomial[A[(n-1)], x], x]], {n,2,30}]] (* John M. Campbell, Mar 16 2012 *)
Join[{0}, LinearRecurrence[{3,-3}, {1,2}, 40]] (* Jean-François Alcover, Jan 08 2019 *)

{a(n) = sum( j=0, n\3, (-1)^j * binomial(n, 3*j + 1))} /* Michael Somos, May 26 2004 */

{a(n) = if( n<2, n>0, n-=2; polsym(x^2 - 3*x + 3, n)[n + 1])} /* Michael Somos, May 26 2004 */

b=BinaryRecurrenceSequence(3,-3,1,2)
def A057682(n): return 0 if n==0 else b(n-1)
[A057682(n) for n in range(41)] # G. C. Greubel, Jul 14 2023

G.f.: (x - x^2) / (1 - 3*x + 3*x^2).
a(n) = 3*a(n-1) - 3*a(n-2), if n>1.
Starting at 1, the binomial transform of A000484. - Paul Barry, Jul 21 2003
It appears that abs(a(n)) = floor(abs(A000748(n))/3). - John W. Layman, Sep 05 2003
a(n) = ((3+i*sqrt(3))/2)^(n-2) + ((3-i*sqrt(3))/2)^(n-2). - Benoit Cloitre, Oct 27 2003
a(n) = n*3F2(1/3-n/3,2/3-n/3,1-n/3 ; 2/3,4/3 ; 1) for n>=1. - John M. Campbell, Jun 01 2011
Let A(n) be the n X n matrix with -1's along the main diagonal, 1's everywhere above the main diagonal, and 1's along the subdiagonal. Then a(n) equals (-1)^(n-1) times the sum of the coefficients of the characteristic polynomial of A(n-1), for all n>1 (see Mathematica code below). - John M. Campbell, Mar 16 2012
Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) - z(n), y(n+1) = y(n) - x(n), z(n+1) = z(n) - y(n). Then a(n) = -y(n). But this recurrence falls into a repetitive cycle of length 6 and multiplicative factor -27, so that a(n) = -27*a(n-6) for any n>6. - Stanislav Sykora, Jun 10 2012
a(n) = A057083(n-1) - A057083(n-2). - R. J. Mathar, Oct 25 2012
G.f.: 3*x - 1/3 + 3*x/(G(0) - 1) where G(k)= 1 + 3*(2*k+3)*x/(2*k+1 - 3*x*(k+2)*(2*k+1)/(3*x*(k+2) + (k+1)/G(k+1)));(continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Nov 23 2012
G.f.: Q(0,u) -1, where u=x/(1-x), Q(k,u) = 1 - u^2 + (k+2)*u - u*(k+1 - u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
From Vladimir Shevelev, Jul 31 2017: (Start)
For n>=1, a(n) = 2*3^((n-2)/2)*cos(Pi*(n-2)/6);
For n>=2, a(n) = K_1(n) + K_3(n-2);
For m,n>=2, a(n+m) = a(n)*K_1(m) + K_1(n)*a(m) - K_3(n-2)*K_3(m-2), where
K_1 = A057681, K_3 = A057083. (End)

A037448 a(n) = floor(cot(n)).

Original entry on oeis.org

0, -1, -8, 0, -1, -4, 1, -1, -3, 1, -1, -2, 2, 0, -2, 3, 0, -1, 6, 0, -1, 112, 0, -1, -8, 0, -1, -4, 1, -1, -3, 1, -1, -2, 2, 0, -2, 3, 0, -1, 6, 0, -1, 56, 0, -1, -9, 0, -1, -4, 1, -1, -3, 1, -1, -2, 2, 0, -2, 3, 0, -1, 5, 0, -1, 37, 0, -1, -9, 0, -1, -4, 1, -1, -3, 1, -1, -2, 2, 0, -2, 3, 0, -1, 5, 0, -1, 28, 0, -1, -10, 0, -1, -4, 1, -1, -3

Offset: 1

Jason Earls, Jun 30 2001

sign

Contains all integers infinitely often. - Charles R Greathouse IV, Aug 06 2012

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A195911, A195910, A000503, A005657, A000493, A000480, A000494, A000484.

[Floor(Cot(n)): n in [1..100]]; // Vincenzo Librandi, Jun 15 2015

Floor[Cot[Range[100]]] (* Harvey P. Dale, Dec 26 2024 *)

v=[]; for(n=1,260,v=concat(v,floor(cotan(n)))); v

a(44) corrected by T. D. Noe, Jan 21 2008

A195911 a(n) = ceiling(cot(n)).

Original entry on oeis.org

1, 0, -7, 1, 0, -3, 2, 0, -2, 2, 0, -1, 3, 1, -1, 4, 1, 0, 7, 1, 0, 113, 1, 0, -7, 1, 0, -3, 2, 0, -2, 2, 0, -1, 3, 1, -1, 4, 1, 0, 7, 1, 0, 57, 1, 0, -8, 1, 0, -3, 2, 0, -2, 2, 0, -1, 3, 1, -1, 4, 1, 0, 6, 1, 0, 38, 1, 0, -8, 1, 0, -3, 2, 0, -2, 2, 0, -1, 3

Offset: 1

Mohammad K. Azarian, Mar 15 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A195910, A000503, A037448, A005657, A000493, A000480, A000494, A000484.

[Ceiling(1/Tan(n)): n in [1..80]]; // Vincenzo Librandi, Feb 16 2013

Table[Ceiling[Cot[n]], {n, 100}] (* T. D. Noe, Mar 16 2012 *)

a(n)=ceil(1/tan(n)) \\ Charles R Greathouse IV, Feb 07 2013

A099494 A Chebyshev transform of Fibonacci(n)+(-1)^n.

Original entry on oeis.org

1, 0, 1, 1, -1, 0, 0, -2, 0, 1, -1, 1, 2, -1, 0, 1, -2, -1, 1, -1, 0, 2, 0, 0, 1, -1, -1, 0, -1, 0, 1, 0, 1, 1, -1, 0, 0, -2, 0, 1, -1, 1, 2, -1, 0, 1, -2, -1, 1, -1, 0, 2, 0, 0, 1, -1, -1, 0, -1, 0, 1, 0, 1, 1, -1, 0, 0, -2, 0, 1, -1, 1, 2, -1, 0, 1, -2, -1, 1, -1, 0, 2, 0, 0, 1, -1, -1

Offset: 0

Paul Barry, Oct 19 2004

A Chebyshev transform of A008346, which has g.f. 1/(1-2x^2-x^3). The image of G(x) under the Chebyshev transform is (1/(1+x^2))*G(x/(1+x^2)).

Periodic with period length 30. - Ray Chandler, Sep 08 2015

Index entries for linear recurrences with constant coefficients, signature (0,-1,1,-1,0,-1).

Cf. A000484, A008346, A014019

G.f.: (1+x^2)^2/(1+x^2-x^3+x^4+x^6).
a(n) = -a(n-2)+a(n-3)-a(n-4)-a(n-6).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*(F(n-2*k)+(-1)^(n-2*k)).
a(n) = A014019(n-1) + A000484(n).

A249988 Integer part of sine of n degrees multiplied by 1000.

Original entry on oeis.org

0, 17, 34, 52, 69, 87, 104, 121, 139, 156, 173, 190, 207, 224, 241, 258, 275, 292, 309, 325, 342, 358, 374, 390, 406, 422, 438, 453, 469, 484, 500, 515, 529, 544, 559, 573, 587, 601, 615, 629, 642, 656, 669, 681, 694, 707, 719, 731, 743, 754, 766, 777, 788, 798, 809, 819, 829, 838, 848, 857, 866

Offset: 0

Alex Ratushnyak, Nov 11 2014

nonn

Due to multiplication by 1000 almost all a(n) are different for n<90, the only exception is a(89)=a(88)=999.

			sin(Pi/2) = sin(90 deg.) = 1, so a(90)=1000.

Cf. A000493, A000494, A000484, A000480, A249989.

Table[IntegerPart[1000*Sin[n Degree]],{n,0,60}] (* Harvey P. Dale, May 22 2015 *)

import math
for n in range(381): print(int(1000*math.sin(math.pi*n/180)), end=', ')

a(n) = round(1000*sin(Pi*n/180)), where round() is the rounding towards zero function.
a(360+n) = a(n).
a(180+n) = -a(n).

Corrected and extended by Harvey P. Dale, May 22 2015

cp's OEIS Frontend

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

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A057682 a(n) = Sum_{j=0..floor(n/3)} (-1)^j*binomial(n,3*j+1).

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A037448 a(n) = floor(cot(n)).

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Magma

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A195911 a(n) = ceiling(cot(n)).

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Magma

Mathematica

PARI

A099494 A Chebyshev transform of Fibonacci(n)+(-1)^n.

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Crossrefs

Formula

A249988 Integer part of sine of n degrees multiplied by 1000.

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Examples

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Programs

Mathematica

Python

Formula

Extensions

A057682 a(n) = Sum_{j=0..floor(n/3)} (-1)^jbinomial(n,3j+1).