A132429 Period 4: repeat [3, 1, -1, -3].
3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1, -1, -3, 3, 1
Offset: 0
Links
- Index entries for linear recurrences with constant coefficients, signature (-1,-1,-1).
Programs
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Haskell
a132429 = (3 -) . (* 2) . (`mod` 4) a132429_list = cycle [3, 1, -1, -3] -- Reinhard Zumkeller, Aug 15 2015
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Magma
&cat [[3, 1, -1, -3]^^30]; // Wesley Ivan Hurt, Jul 10 2016
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Maple
A132429:=n->3 - 2 * (n mod 4); seq(A132429(n), n=0..100); # Wesley Ivan Hurt, Apr 18 2014
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Mathematica
PadRight[{}, 104, {3,1,-1,-3}] (* Harvey P. Dale, Nov 12 2011 *)
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PARI
a(n)=3-2*(n%4) \\ Jaume Oliver Lafont, Aug 28 2009
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Python
def A132429(n): return 3 - 2*(n & 3) # Chai Wah Wu, May 25 2022
Formula
G.f.: (3 + 4*x + 3*x^2)/((1+x)*(1+x^2)). - Jaume Oliver Lafont, Aug 30 2009
a(n) = (-1)^n + 2(-1)^((2n + (-1)^n - 1)/4). - Brad Clardy, Mar 10 2013
a(n) = 3 - 2*(n mod 4). - Joerg Arndt, Mar 10 2013
a(n) = (-1)^n + 2(-1)^floor(n/2). - Wesley Ivan Hurt, Apr 17 2014
From Wesley Ivan Hurt, Jul 10 2016: (Start)
a(n) + a(n-1) + a(n-2) + a(n-3) = 0 for n>2, a(n) = a(n-4) for n>3.
a(n) = 2*cos(n*Pi/2) + cos(n*Pi) + 2*sin(n*Pi/2). (End)
Comments