A283393 -id:A283393 - cp's OEIS frontend

A000034 Period 2: repeat [1, 2]; a(n) = 1 + (n mod 2).

1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 0

N. J. A. Sloane

Also continued fraction for (sqrt(3)+1)/2 (cf. A040001) and base-3 digital root of n+1 (cf. A007089, A010888). - Henry Bottomley, Jul 05 2001
The sequence 1,-2,-1,2,1,-2,-1,2,... with g.f. (1-2x)/(1+x^2) has a(n) = cos(Pi*n/2)-2*sin(Pi*n/2). - Paul Barry, Oct 18 2004
Hankel transform is [1,-3,0,0,0,0,0,0,0,...]. - Philippe Deléham, Mar 29 2007
4/33 = 0.121212... - Eric Desbiaux, Nov 03 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=A[i,i]:=1, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1) = charpoly(A,2). - Milan Janjic, Jan 24 2010
First differences of A032766. - Tom Edgar, Jul 17 2014
Denominator of the harmonic mean of the first n triangular numbers. - Colin Barker, Nov 13 2014
This is the lexicographically earliest sequence of positive integers such that no polynomial of degree d can be fitted to d+2 consecutive terms (equivalently, such that no iterated difference is zero). - Pontus von Brömssen, Dec 26 2021 [See A300002 for the case where not only consecutive terms are considered. - Pontus von Brömssen, Jan 03 2023]
Number of maximum antichains in the power set of {1,2,...,n} partially ordered by set inclusion. For even n, there is a unique maximum antichain formed by all subsets of size n/2; for odd n, there are two maximum antichains, one formed by all subsets of size (n-1)/2 and the other formed by all subsets of size (n+1)/2. See the David Guichard link below for a proof. - Jianing Song, Jun 19 2022

Jozsef Beck, Combinatorial Games, Cambridge University Press, 2008.
J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 545 pages 73 and 260, Ellipses, Paris 2004.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Sean A. Irvine, Table of n, a(n) for n = 0..9999
Paul Barry, On a Central Transform of Integer Sequences, arXiv:2004.04577 [math.CO], 2020.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seqs., Vol. 6, 2003.
David Guichard, Sperner's Theorem
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 383
Agustín Moreno Cañadas, Hernán Giraldo and Robinson Julian Serna Vanegas, Some integer partitions induced by orbits of Dynkin type, Far East Journal of Mathematical Sciences (FJMS), Vol. 101, No. 12 (2017), pp. 2745-2766.
Wikipedia, Collatz conjecture
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A000035, A003945 (binomial transf), A007089, A010693, A010704, A010888, A032766, A040001, A123344, A134451, A300002.

Cf. sequences listed in Comments section of A283393.

List([0..120],n->1+(n mod 2)); # Muniru A Asiru, Feb 01 2019

a000034 = (+ 1) . (`mod` 2)
a000034_list = cycle [1,2]
-- Reinhard Zumkeller, Jul 03 2012, Dec 02 2011 and corrected by James Spahlinger, Oct 08 2012

[1+(n mod 2) : n in [0..100]]; // Wesley Ivan Hurt, Sep 11 2014

(1+2*x)/(1-x^2);
A000034 := proc(n) op((n mod 2)+1,[1,2]) ; end proc: # R. J. Mathar, Feb 05 2011

a[n_] := If[OddQ[n], 2, 1]; Table[a[n], {n, 0, 90}] (* Stefan Steinerberger, Apr 17 2006 *)
Nest[ Flatten[# /. {    0 -> {1}, 1 -> {2}, 2 -> {1, 2, 1}  }] &, {1}, 8] (* Robert G. Wilson v, May 20 2014 *)

PARI
```
a(n)=1+n%2
```

a(n)=1+bittest(n,0) \\ M. F. Hasler, Jan 13 2012

def A000034(n): return 1 + (n & 1) # Chai Wah Wu, May 25 2022

G.f.: (1+2*x)/(1-x^2).
a(n) = 2^((1-(-1)^n)/2) = 2^(ceiling(n/2) - floor(n/2)). - Paul Barry, Jun 03 2003
a(n) = (3-(-1)^n)/2; a(n) = 1 + (n mod 2) = 3-a(n-1) = a(n-2) = a(-n).
a(n) = gcd(n-1, n+1). - Paul Barry, Sep 16 2004
Binomial transform of A123344, inverse binomial transform of A003945. - Philippe Deléham, Jun 04 2007
a(n) = A134451(n+1). - Reinhard Zumkeller, Oct 27 2007
a(n) = if(n=0,1,if(mod(a(n-1),2)=0,a(n-1)/2,(3*a(n-1)+1)/2)). See Collatz conjecture. - Paul Barry, Mar 31 2008
a(n) = 2^n (mod 3). - Vincenzo Librandi, Feb 05 2011
a(n) = A000035(n) + 1. - M. F. Hasler, Jan 13 2012
a(n) = abs(sin(n*Pi/2) - 2*cos(n*Pi/2)). - Mohammad K. Azarian, Mar 12 2012
a(n) = A010704(n) / 3. - Reinhard Zumkeller, Jul 03 2012
a(n) = floor((4/33)*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((5/8)*3^(n+1)) mod 3. - Hieronymus Fischer, Jan 03 2013
a(n) = floor((n+1)*3/2) - floor((n)*3/2). - Hailey R. Olafson, Jul 23 2014
a(n) = denominator(n/2). - Wesley Ivan Hurt, Sep 11 2014
Dirichlet g.f.: zeta(s)*(1 + 1/2^s). - Mats Granvik, Jul 18 2016
E.g.f.: 2*sinh(x) + cosh(x). - Ilya Gutkovskiy, Jul 18 2016
a(n) = A010693(n) - 1. - Filip Zaludek, Oct 29 2016
a(n) = n + 1 - 2*floor(n/2). - Lorenzo Sauras Altuzarra, Jun 28 2019
Limit_{n->oo} (1/n)*Sum_{k=1..n} a(k) = 3/2 (De Koninck reference). - Bernard Schott, Nov 09 2021

Better definition from M. F. Hasler, Jan 13 2012

A010685 Period 2: repeat (1,4).

Original entry on oeis.org

1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1

Offset: 0

N. J. A. Sloane

Continued fraction of (1 + sqrt(2))/2. - R. J. Mathar, Nov 21 2011

This sequence can be generated by an infinite number of formulas all having the form a^(b*n) mod c subject to the following conditions. The number a can be congruent to either 2,3, or 4 mod 5 (A047202). If a is congruent to 2 or 3 mod 5, then b can be any number of the form 4k+2 and c = 5 or 15. If a is congruent to 4 mod 5, then b can be any number of the form 2k+1 and c = 5. For example: a(n) = 29^(13*n) mod 5, a(n) = 24^(11*n) mod 5, and a(n) = 22^(10*n) mod 15. - Gary Detlefs, May 19 2014

Matthew House, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. sequences listed in Comments section of A283393.

Cf. A047202.

[Modexp(4,n,5): n in [0..100]]; // G. C. Greubel, Nov 22 2021

A010685 := proc(n)
    if type(n,'even') then
        1 ;
    else
        4;
    end if;
end proc: # R. J. Mathar, Aug 03 2015

Table[(5-3(-1)^n)/2, {n, 0, 100}] (* Wesley Ivan Hurt, Mar 26 2014 *)
PadRight[{},120,{1,4}] (* Harvey P. Dale, Aug 08 2022 *)

values(m)=my(v=[]);for(i=1,m,v=concat([1,4],v));v; /* Anders Hellström, Aug 03 2015 */

[power_mod(4,n,5)for n in range(0,81)] # Zerinvary Lajos, Nov 26 2009

a(2n) = 1, a(2n+1) = 4.
From Paul Barry, Jun 03 2003: (Start)
G.f.: (1+4*x)/((1-x)*(1+x)).
E.g.f.: (5*exp(x) - 3*exp(-x))/2.
a(n) = (5 - 3*(-1)^n)/2.
a(n) = 4^((1-(-1)^n)/2) = 2^(1-(-1)^n) = 2/(2^((-1)^n)).
a(n) = 4^(ceiling(n/2) - floor(n/2)). (End)
a(n) = gcd((n-1)^2, (n+1)^2). - Paul Barry, Sep 16 2004
a(n) = A160700(A000302(n)). - Reinhard Zumkeller, Jun 10 2009
a(n) = 4^n mod 5. - Zerinvary Lajos, Nov 26 2009
a(n) = 4^(n mod 2). - Wesley Ivan Hurt, Mar 29 2014

A010689 Periodic sequence: Repeat 1, 8.

Original entry on oeis.org

1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1

Offset: 0

N. J. A. Sloane

Also the digital root of 8^n. Also the decimal expansion of 2/11 = 0.181818181818... - Cino Hilliard, Dec 31 2004
Interleaving of A000012 and A010731. - Klaus Brockhaus, Apr 02 2010
Continued fraction expansion of (2 + sqrt(6))/4. - Klaus Brockhaus, Apr 02 2010
Digital root of the powers of any number congruent to 8 mod 9. - Alonso del Arte, Jan 26 2014

			0.18181818181818181818181818181818181818181...

Cecil Balmond, Number 9: The Search for the Sigma Code. Munich, New York: Prestel (1998): 203.

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

Cf. A000012 (all 1's sequence), A010731 (all 8's sequence), A174925 (decimal expansion of (2 + sqrt(6))/4). [Klaus Brockhaus, Apr 02 2010]
Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 4, A100402; c = 5, A070366; c = 7, A070403.
Cf. sequences listed in Comments section of A283393.
Cf. A010888.

&cat[ [1, 8]: n in [0..52] ]; // Klaus Brockhaus, Apr 02 2010

&cat [[1,8]^^60]; // Bruno Berselli, Mar 10 2017

Table[Mod[8^n, 9], {n, 0, 99}] (* Alonso del Arte, Jan 26 2014 *)
PadRight[{},120,{1,8}] (* Harvey P. Dale, Jun 03 2015 *)

A010689(n):=if evenp(n) then 1 else 8$
makelist(A010689(n),n,0,30); /* Martin Ettl, Nov 09 2012 */

a(n)=1; if(n%2==1, 8, 1) \\ Felix Fröhlich, Aug 11 2014

[power_mod(8,n,9)for n in range(0,105)] # Zerinvary Lajos, Nov 27 2009

From Paul Barry, Sep 16 2004: (Start)
G.f.: (1 + 8*x)/((1 - x)*(1 + x)).
a(n) = (9 - 7*(-1)^n)/2.
a(n) = 8^(ceiling(n/2) - floor(n/2)).
a(n) = gcd((n-1)^3, (n+1)^3). (End)
E.g.f.: cosh(x) + 8*sinh(x). - Stefano Spezia, Feb 09 2025
a(n) = A010888(8*a(n-1)). - Stefano Spezia, Mar 20 2025

Definition edited and keywords cons, cofr added by Klaus Brockhaus, Apr 02 2010

A129203 a(n) = numerator(3/(n+1)^3)*(3/2 + (-1)^n/2).

Original entry on oeis.org

6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1, 6, 3, 2, 3, 6, 1

Offset: 0

Paul Barry, Apr 03 2007

(1/(2*Pi))*Integral_{t=0..2*Pi} exp(i*(n+1)*t)*((t-Pi)/i)^3 = (A129202(n)*Pi^2 - a(n))/A129196(n), i=sqrt(-1).

Periodic with period 6. - Alois P. Heinz, Oct 03 2012

Muniru A Asiru, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,1).

Cf. sequences listed in Comments section of A283393.

List(List([0..10],n->3/(n+1)^3*(3/2+(-1)^n/2)),NumeratorRat); # Muniru A Asiru, Jul 01 2018

[Numerator(3/(n+1)^3)*(3/2 + (-1)^n/2): n in [1..100]]; // Vincenzo Librandi, Jul 01 2018

&cat [[6, 3, 2, 3, 6, 1]^^20]; // Vincenzo Librandi, Jul 01 2018

a:= n-> [6, 3, 2, 3, 6, 1][irem(n, 6)+1]:
seq(a(n), n=0..119);  # Alois P. Heinz, Oct 03 2012

Array[Numerator[3/(# + 1)^3] (3/2 + (-1)^#/2) &, 105, 0] (* or *)
PadRight[{}, 105, {6, 3, 2, 3, 6, 1}] (* Michael De Vlieger, Jun 30 2018 *)
CoefficientList[ Series[-(x^5 + 6x^4 + 3x^3 + 2x^2 + 3x + 6)/(x^6 - 1), {x, 0,
   105}], x] (* or *)
LinearRecurrence[{0, 0, 0, 0, 0, 1}, {6, 3, 2, 3, 6, 1}, 105] (* Robert G. Wilson v, Jul 28 2018 *)

a(n)=[6, 3, 2, 3, 6, 1][n%6+1] \\ Charles R Greathouse IV, Oct 28 2014

def Gauss_factorial(N, n): return mul(j for j in (1..N) if gcd(j, n) == 1)
def A129203(n): return Gauss_factorial(3, n+1)
[A129203(j) for j in (0..71)] # Peter Luschny, Oct 01 2012

G.f.: (6 + 3*x + 2*x^2 + 3*x^3 + 6*x^4 + x^5)/(1 - x^6).
a(n) = cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3) + cos(Pi*n/3)/3 - sqrt(3)*sin(Pi*n/3)/3 + 7*cos(Pi*n)/6 + 7/2.
a(n) = numerator(6/(n+1)^2). - Paul Barry, Apr 03 2007
a(n) = denominator of coefficient of x^6 in the Maclaurin expansion of -exp(-(n+1)*x^2). - Francesco Daddi, Aug 04 2011
a(n) = 3_(n+1)! = Gauss_factorial(3, n+1) = Product_{1<=j<=3, gcd(j,n+1)=1} j. - Peter Luschny, Oct 01 2012
a(n) = denominator((n+1)/6). - Jon Hearn, Nov 10 2013
a(n) = denominator of 2*(1/(12*n)+1)*n^n; related to gamma function approximation for positive integers less the factor sqrt(Pi/2)/(exp(n)*sqrt(n)). - Thomas Blankenhorn, Jun 21 2018
a(n) = 6/gcd(n+1,6). - Ridouane Oudra, Jul 29 2022

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A000034 Period 2: repeat [1, 2]; a(n) = 1 + (n mod 2).

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A010685 Period 2: repeat (1,4).

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A010689 Periodic sequence: Repeat 1, 8.

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A129203 a(n) = numerator(3/(n+1)^3)*(3/2 + (-1)^n/2).

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A129197 a(n) = numerator( 3*(3+(-1)^n)/(n+1)^3 ).

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