A210209 -id:A210209 - cp's OEIS frontend

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

Paul Curtz, Aug 26 2007

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)

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

Cf. A167808. - Reinhard Zumkeller, Nov 12 2009

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

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

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

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

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

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

A229339 GCD of all sums of n consecutive Lucas numbers.

Original entry on oeis.org

1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29, 2, 105, 1, 76, 1, 275, 2, 199, 1, 720, 1, 521, 2, 1885, 1, 1364, 1, 4935, 2, 3571, 1, 12920, 1, 9349, 2, 33825, 1, 24476, 1, 88555, 2, 64079, 1, 231840, 1, 167761, 2, 606965, 1, 439204, 1, 1589055, 2, 1149851, 1, 4160200, 1, 3010349, 2

Offset: 1

Alonso del Arte, Sep 23 2013

The sum of two consecutive Lucas number is the sum of four consecutive Fibonacci numbers, which is verified easily enough with the identity L(n) = F(n - 1) + F(n + 1). Therefore a(1) = a(2) = A210209(4).

			a(3) = 2 because any sum of three consecutive Lucas numbers is an even number.
a(4) = 5 because all sums of four consecutive Lucas numbers are divisible by 5.
a(5) = 1 because some sums of five consecutive Lucas numbers are coprime.

Colin Barker, Table of n, a(n) for n = 1..1000
Dan Guyer and aBa Mbirika, GCD of sums of k consecutive Fibonacci, Lucas, and generalized Fibonacci numbers, Journal of Integer Sequences, 24 No.9, Article 21.9.8 (2021), 25pp; arXiv preprint, arXiv:2104.12262 [math.NT], 2021.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,1,0,-1,0,-3,0,0,0,1).

Cf. A210209, A022112, A022088, A022098, A106291 (Pisano periods of the Lucas sequence).

a[n_] := a[n] = If[n <= 14, {1, 1, 2, 5, 1, 4, 1, 15, 2, 11, 1, 40, 1, 29}[[n]], 3*a[n - 4] + a[n - 6] - a[n - 8] - 3*a[n - 10] + a[n - 14]]; Array[a, 64] (* Giovanni Resta, Oct 04 2013 *)
CoefficientList[Series[(x^12 - x^11 + 2 x^10 - 5 x^9 - 2 x^8 - x^7 - 6 x^6 + x^5 - 2 x^4 + 5 x^3 + 2 x^2 + x + 1) / (-x^14 + 3 x^10 + x^8 - x^6 - 3 x^4 + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 09 2014 *)
LinearRecurrence[{0,0,0,3,0,1,0,-1,0,-3,0,0,0,1},{1,1,2,5,1,4,1,15,2,11,1,40,1,29},70] (* Harvey P. Dale, Jul 21 2021 *)
Table[GCD[LucasL[n + 1] - 2, LucasL[n] + 1], {n, 0, 50}] (* Horst H. Manninger, Dec 25 2021 *)

Vec(x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1) + O(x^100)) \\ Colin Barker, Nov 09 2014

a(n) = 3*a(n-4) + a(n-6) - a(n-8) - 3*a(n-10) + a(n-14) for n > 14. - Giovanni Resta, Oct 04 2013
G.f.: x*(x^12 -x^11 +2*x^10 -5*x^9 -2*x^8 -x^7 -6*x^6 +x^5 -2*x^4 +5*x^3 +2*x^2 +x +1) / (-x^14 +3*x^10 +x^8 -x^6 -3*x^4 +1). - Colin Barker, Nov 09 2014
From Aba Mbirika, Jan 04 2022: (Start)
a(n) = gcd(L(n+1)-1, L(n+2)-3).
a(n) = Lcm_{A106291(m) divides n} m.
Proofs of these formulas are given in Theorems 15 and 25 of the Guyer-Mbirika paper. (End)

cp's OEIS Frontend

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

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