A131561 Period 3: repeat [1, 1, -1].
1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1, -1, 1, 1
Offset: 0
Examples
G.f. = 1 + x - x^2 + x^3 + x^4 - x^5 + x^6 + x^7 - x^8 + x^9 + x^10 + ...
References
- Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.
Links
- Index entries for linear recurrences with constant coefficients, signature (0,0,1).
Programs
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Magma
&cat [[1, 1, -1]^^30]; // Wesley Ivan Hurt, Jul 02 2016
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Maple
A131561 := proc(n) op((n mod 3)+1,[1,1,-1]) ; end: seq(A131561(n), n=0..120); # R. J. Mathar, Oct 18 2007
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Mathematica
Table[(-1)^Mod[n-1,3], {n, 0, 120}] (* Michael De Vlieger, Mar 07 2015 *) PadRight[{},120,{1,1,-1}] (* Harvey P. Dale, Mar 15 2021 *)
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PARI
a(n)=1-2*(n%3==2) /* Jaume Oliver Lafont, Mar 24 2009 */
Formula
a(n) = (4*cos((2*n - 1) * Pi/3) + 1) / 3. - Federico Acha Neckar (f0383864(AT)hotmail.com), Sep 02 2007
G.f.: (1+x-x^2)/((1-x)*(x^2+x+1)). - R. J. Mathar, Nov 14 2007
G.f.: (1+x-x^2)/(1-x^3). - Jaume Oliver Lafont, Mar 24 2009
a(n) = (-1)^((n-1) mod 3). - Christopher Richmond, Oct 07 2011
a(n) = a(n-1)^2 - a(n-1) - a(n-2), for a(0),a(1) = 1,1; or same repeating pattern with 1,-1 or -1,1 as initial values. - Richard R. Forberg, Jun 13 2013
a(n+1) = A257075(n) for all n in Z. - Michael Somos, May 13 2015
a(n) = a(n-3) for n>2. - Wesley Ivan Hurt, Jul 02 2016
Product_{n >= 1} (1 + a(n-1)*x^n) = 1 + x + x^2 + x^5 + x^7 + x^12 + x^15 + ... = Sum_{n >= 0} x^A001318(n), a companion identity to Euler's pentagonal number theorem. - Peter Bala, Aug 30 2017
E.g.f.: (exp(x) + 2*exp(-x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Oct 19 2024
Extensions
Edited by N. J. A. Sloane, Sep 15 2007
Comments