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%I #96 Feb 22 2024 09:05:23
%S 1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,
%T -1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,
%U 0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1,1,0,-1
%N Periodic sequence 1,0,-1,...; expansion of (1+x)/(1+x+x^2).
%C Partial sums of signed sequence is shifted unsigned one: |a(n+2)| = A011655(n+1).
%C With interpolated zeros, a(n) = sin(5*Pi*n/6 + Pi/3)/sqrt(3) + cos(Pi*n/6 + Pi/6)/sqrt(3); this gives the diagonal sums of the Riordan array (1-x^2, x(1-x^2)). - _Paul Barry_, Feb 02 2005
%C From _Tom Copeland_, Nov 02 2014: (Start)
%C With a shift and a sign change the o.g.f. of this array becomes the compositional inverse of the shifted Motzkin or Riordan numbers A005043,
%C (x - x^2) / (1 - x + x^2) = x*(1-x) / (1 - x*(1-x)) = x*(1-x) + [x*(1-x)]^2 + ... . Expanding each term of this series and arranging like powers of x in columns gives skewed rows of the Pascal triangle and reading along the columns gives (mod-signs and indexing) A011973, A169803, and A115139 (see also A091867, A092865, A098925, and A102426 for these term-by-term expansions and A030528). (End)
%H Winston de Greef, <a href="/A057078/b057078.txt">Table of n, a(n) for n = 0..10000</a>
%H Ralph E. Griswold, <a href="https://www2.cs.arizona.edu/patterns/sequences/">Shaft Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (-1,-1).
%F a(n) = S(n, -1) + S(n - 1, -1) = S(2*n, 1); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, -1) = A049347(n). S(n, 1) = A010892(n).
%F From Mario Catalani (mario.catalani(AT)unito.it), Jan 08 2003: (Start)
%F a(n) = (1/2)*((-1)^floor(2*n/3) + (-1)^floor((2*n+1)/3)).
%F a(n) = -a(n-1) - a(n-2).
%F a(n) = A061347(n) - A049347(n+2). (End)
%F a(n) = Sum_{k=0..n} binomial(n+k, 2k)*(-1)^(n-k) = Sum_{k=0..floor((n+1)/2)} binomial(n+1-k, k)*(-1)^(n-k). - Mario Catalani (mario.catalani(AT)unito.it), Aug 20 2003
%F Binomial transform is A010892. a(n) = 2*sqrt(3)*sin(2*Pi*n/3 + Pi/3)/3. - _Paul Barry_, Sep 13 2003
%F a(n) = cos(2*Pi*n/3) + sin(2*Pi*n/3)/sqrt(3). - _Paul Barry_, Oct 27 2004
%F a(n) = Sum_{k=0..n} (-1)^A010060(2n-2k)*(binomial(2n-k, k) mod 2). - _Paul Barry_, Dec 11 2004
%F a(n) = (4/3)*(|sin(Pi*(n-2)/3)| - |sin(Pi*n/3)|)*|sin(Pi*(n-1)/3)|. - _Hieronymus Fischer_, Jun 27 2007
%F a(n) = 1 - (n mod 3) = 1 + 3*floor(n/3)) - n. - _Hieronymus Fischer_, Jun 27 2007
%F a(n) = 1 - A010872(n) = 1 + 3*A002264(n) - n. - _Hieronymus Fischer_, Jun 27 2007
%F Euler transform of length 3 sequence [0, -1, 1]. - _Michael Somos_, Oct 15 2008
%F a(n) = a(n-1)^2 - a(n-2)^2 with a(0) = 1, a(1) = 0. - _Francesco Daddi_, Aug 02 2011
%F a(n) = A049347(n) + A049347(n-1). - _R. J. Mathar_, Jun 26 2013
%F E.g.f.: exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/3. - _Stefano Spezia_, May 16 2023
%F a(n) = -a(-1-n) for all n in Z. - _Michael Somos_, Feb 20 2024
%e G.f. = 1 - x^2 + x^3 - x^5 + x^6 - x^8 + x^9 - x^11 + x^12 - x^14 + x^15 + ...
%p A057078:=n->1-(n mod 3); seq(A057078(n), n=0..100); # _Wesley Ivan Hurt_, Dec 06 2013
%t a[n_] := {1, 0, -1}[[Mod[n, 3] + 1]] (* _Jean-François Alcover_, Jul 05 2013 *)
%t CoefficientList[Series[(1 + x) / (1 + x + x^2), {x, 0, 40}], x] (* _Vincenzo Librandi_, Nov 03 2014 *)
%t LinearRecurrence[{-1, -1},{1, 0},90] (* _Ray Chandler_, Sep 15 2015 *)
%o (PARI) {a(n) = [1, 0, -1][n%3 + 1]}; /* _Michael Somos_, Oct 15 2008 */
%o (Haskell)
%o a057078 = (1 -) . (`mod` 3) -- _Reinhard Zumkeller_, Mar 22 2013
%o (Sage)
%o def A057078():
%o x, y = -1, 0
%o while True:
%o yield -x
%o x, y = y, -x -y
%o a = A057078(); [next(a) for i in range(40)] # _Peter Luschny_, Jul 11 2013
%Y Cf. A049310, A010892, A011655.
%Y A049347(n) = a(-n).
%Y Cf. A005043, A011973, A169803, A115139, A091867, A092865, A098925, A102426, A030528.
%Y Cf. A002264, A010060, A010872, A049347, A061347.
%K easy,sign
%O 0,1
%A _Wolfdieter Lang_, Aug 04 2000
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