A131534 Period 3: repeat [1, 2, 1].

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

Offset: 0

Author

Paul Curtz, Aug 26 2007

Keywords

• nonn
• easy

Comments

Partial sums of A106510. Inverse binomial transform of A024495 (without leading zeros). - Philippe Deléham, Nov 26 2008

a(n) = A130196(n) - A022003(n) = A080425(n) - A130196(n)+2 = A153727(n)/A130196(n). - Reinhard Zumkeller, Nov 12 2009

Continued fraction expansion of A177346, (1+sqrt(10))/3. - Klaus Brockhaus, May 07 2010

From Daniel Forgues, May 04 2016: (Start)

a(n) = GCD of terms of the sequence S_n = {F_i+F_{i+1}+F_{i+2}+...+F_{i+2n}, i >= 0}, where F_i denotes a Fibonacci number. See A210209.

a(n) = GCD of terms of the sequence S_n = {L_i+L_{i+1}+L_{i+2}+...+L_{i+2n}, i >= 0}, where L_i denotes a Lucas number. See A229339. (End)

Links

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

Crossrefs

Cf. A167808. - Reinhard Zumkeller, Nov 12 2009

Cf. A000045, A022003, A024495, A080425, A101825, A106510, A130196, A153727, A177346, A210209, A229339.

Programs

Magma

[(-n mod 3)^(n mod 3) : n in [0..100]]; // Wesley Ivan Hurt, Aug 29 2014

Maple

A131534:=n->(-n mod 3)^(n mod 3): seq(A131534(n), n=0..100); # Wesley Ivan Hurt, Aug 29 2014

Mathematica

PadRight[{},120,{1,2,1}] (* Harvey P. Dale, Aug 06 2013 *)

PARI

a(n)=1+(n%3==1) \\ Jaume Oliver Lafont, Mar 20 2009

Formula

G.f.: (x+1)^2/((1-x)*(x^2+x+1)). - R. J. Mathar, Nov 14 2007

a(n) = 4/3 + (2/3)*cos(2*Pi*(n+2)/3). - Jaume Oliver Lafont, May 09 2008

a(n) = A101825(n+1). - R. J. Mathar, Jun 13 2008

a(n) = gcd(F(n)^2+F(n+1)^2, F(n)+F(n+1)). - Gary Detlefs, Dec 29 2010

a(n) = 2 - ((n+2)^2 mod 3). - Gary Detlefs, Oct 13 2011

a(n) = ceiling(n*4/3) - ceiling((n-1)*4/3). - Tom Edgar, Jul 22 2014

a(n) = 2 - abs(3*floor(n/3)+1-n). - Mikael Aaltonen, Jan 02 2015

a(n) = 1+[3|(2n+1)], using Iverson bracket. - Daniel Forgues, May 04 2016

a(n) = a(n-3) for n>2. - Wesley Ivan Hurt, Jul 05 2016

E.g.f.: (4*exp(x) - exp(-x/2)*(cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Aug 04 2025