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%I #42 Jan 19 2023 02:02:56
%S 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,
%T -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,
%U -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
%N a(n) = (-1)^binomial(n,4): Periodic sequence 1,1,1,1,-1,-1,-1,-1,... .
%C Periodic sequence with period 8. More generally, it appears that (-1)^binomial(n,2^r) gives a periodic sequence of period 2^(r+1), the period consisting of a block of 2^r plus ones followed by a block of 2^r minus ones. See A033999 (r = 0), A057077 (r = 1) and A143622 (r = 3).
%C Nonsimple continued fraction expansion of A188943 = 1.767591879243... - _R. J. Mathar_, Mar 08 2012
%H Maciej Gawron, and Maciej Ulas, <a href="http://dx.doi.org/10.1016/j.disc.2015.12.016">On formal inverse of the Prouhet-Thue-Morse sequence</a>, Discrete Mathematics 339.5 (2016): 1459-1470. Also <a href="http://arxiv.org/abs/1601.04840">arXiv preprint</a>arXiv:1601.04840 [math.CO], 2016. The sequence appears on page 1464, prefixed by three 0's.
%F a(n) = (-1)^binomial(n,4) = (-1)^floor(n/4), since Sum_{k = 1..n-3} k*(k+1)(k+2)/3! = binomial(n,4) == floor(n/4) (mod 2) for n = 0,1,...,7 by calculation and both sides increase by an even number if we substitute n+8 for n.
%F a(n) = (1/4)*((n+4) mod 8 - n mod 8).
%F O.g.f.: (1+x+x^2+x^3)/(1+x^4) = (1+x)*(1+x^2)/(1+x^4) = (1-x^4)/((1-x)*(1+x^4)).
%F Define E(k) = Sum_{n>=0} a(n)*n^k/n! for k = 0,1,2,... . Then E(k) is an integral linear combination of E(0), E(1), E(2) and E(3) (a Dobinski-type relation).
%F a(n) = (-1)^A180969(2,n), where the first index in A180969(.,.) is the row index. - _Adriano Caroli_, Nov 18 2010
%F Euler transform of length 8 sequence [ 1, 0, 0, -2, 0, 0, 0, 1]. - _Michael Somos_, Sep 30 2011
%F G.f.: (1 - x^4)^2 / ((1 - x) * (1 - x^8)). a(n) = -a(-1 - n) for all n in Z. - _Michael Somos_, Sep 30 2011
%F E.g.f.: sin(x/sqrt(2))*sinh(x/sqrt(2)) + (sqrt(2)*sin(x/sqrt(2)) + cos(x/sqrt(2)))*cosh(x/sqrt(2)). - _Ilya Gutkovskiy_, Apr 15 2016
%e G.f. = 1 + x + x^2 + x^3 - x^4 - x^5 - x^6 - x^7 + x^8 + x^9 + x^10 + ...
%p with(combinat):
%p a := n -> (-1)^binomial(n,4):
%p seq(a(n),n = 0..103);
%t Table[(-1)^Binomial[n, 4], {n, 0, 100}] (* _Wesley Ivan Hurt_, May 20 2014 *)
%t a[ n_] := (-1)^Quotient[n, 4]; (* _Michael Somos_, May 05 2015 *)
%o (PARI) {a(n) = (-1)^(n \ 4)}; /* _Michael Somos_, Sep 30 2011 */
%o (PARI) x='x+O('x^99); Vec((1-x^4)^2/((1-x)*(1-x^8))) \\ _Altug Alkan_, Apr 15 2016
%o (Python)
%o def A143621(n): return -1 if n&4 else 1 # _Chai Wah Wu_, Jan 18 2023
%Y Cf. A033999, A057077, A130151, A143622.
%K easy,sign
%O 0,1
%A _Peter Bala_, Aug 30 2008
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