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%I #167 Jan 28 2024 09:11:41
%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,
%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,-1,1,-1,1,-1,1
%N a(n) = (-1)^n.
%C (-1)^(n+1) = signed area of parallelogram with vertices (0,0), U=(F(n),F(n+1)), V=(F(n+1),F(n+2)), where F = A000045 (Fibonacci numbers). The area of every such parallelogram is 1. The signed area is -1 if and only if F(n+1)^2 > F(n)*F(n+2), or, equivalently, n is even, or, equivalently, the vector U is "above" V, indicating that U and V "cross" as n -> n+1. - _Clark Kimberling_, Sep 09 2013
%C Periodic with period length 2. - _Ray Chandler_, Apr 03 2017
%C From _Bernard Schott_, May 11 2022: (Start)
%C Cesàro mean theorem: When a(n) has a limit (finite or infinite) in the usual sense, then c(n) = (a(1)+...+a(n))/n has the same Cesàro limit, but the converse is false. This sequence is a counterexample in the case of a finite Cesàro limit (see A237420 for counterexample with an infinite Cesàro limit).
%C This sequence is not convergent in the usual sense because a(2n) = 1 while a(2n+1) = -1; the successive arithmetic means c(n) of the first n terms of the sequence are 1/1, 0/2, 1/3, 0/4, 1/5, 0/6, ... so c(2n) = 1/(2n+1) and c(2n+1) = 0, hence the Cesàro limit is 0 because c(n) -> 0 when n -> oo.
%C In fact, when sequence a(n) is "Period k: [a1, a2, ..., ak]", then the Cesàro limit c of this sequence is (a1+a2+...+ak)/k.
%C Note that the converse of the theorem is true iff a(n) is monotonic (End).
%D J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.
%H Paul Barry, <a href="https://arxiv.org/abs/2001.08799">Characterizations of the Borel triangle and Borel polynomials</a>, arXiv:2001.08799 [math.CO], 2020.
%H F. Javier de Vega, <a href="https://arxiv.org/abs/2003.13378">An extension of Furstenberg's theorem of the infinitude of primes</a>, arXiv:2003.13378 [math.NT], 2020.
%H S. K. Ghosal and J. K. Mandal, <a href="http://dx.doi.org/10.1016/j.protcy.2013.12.341">Stirling Transform Based Color Image Authentication</a>, Procedia Technology, 2013 Volume 10, 2013, Pages 95-104.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/514802/convergence-of-series-implies-convergence-of-cesaro-mean">Convergence of series implies convergence of Cesàro mean</a>, 2013.
%H László Németh, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL21/Nemeth/nemeth6.html">The trinomial transform triangle</a>, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also <a href="https://arxiv.org/abs/1807.07109">arXiv:1807.07109</a> [math.NT], 2018.
%H ProofWiki, <a href="https://proofwiki.org/wiki/Cesàro_Mean">Cesàro mean</a>.
%H Michael Somos, <a href="http://cis.csuohio.edu/~somos/rfmc.txt">Rational Function Multiplicative Coefficients</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/InverseTangent.html">Inverse Tangent</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StirlingTransform.html">Stirling Transform</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Ernesto_Cesàro">Ernesto Cesàro</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Grandi%27s_series">Grandi's series</a>
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/%C2%B11-sequence">+/-1-sequence</a>
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Dirichlet_eta_function">Dirichlet eta function</a>
%H Wikipédia, <a href="https://fr.wikipedia.org/wiki/Lemme_de_Cesàro">Lemme de Cesàro </a> (in French).
%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (-1).
%H <a href="/index/Pol#poly_cyclo_inv">Index to sequences related to inverse of cyclotomic polynomials</a>
%F G.f.: 1/(1+x).
%F E.g.f.: exp(-x).
%F Linear recurrence: a(0)=1, a(n)=-a(n-1) for n>0. - _Jaume Oliver Lafont_, Mar 20 2009
%F Sum_{k=0..n} a(k) = A059841(n). - _Jaume Oliver Lafont_, Nov 21 2009
%F Sum_{k>=0} a(k)/(k+1) = log(2). - _Jaume Oliver Lafont_, Mar 30 2010
%F Euler transform of length 2 sequence [ -1, 1]. - _Michael Somos_, Mar 21 2011
%F Moebius transform is length 2 sequence [ -1, 2]. - _Michael Somos_, Mar 21 2011
%F a(n) = -b(n) where b(n) = multiplicative with b(2^e) = -1 if e>0, b(p^e) = 1 if p>2. - _Michael Somos_, Mar 21 2011
%F a(n) = a(-n) = a(n + 2) = cos(n * Pi). a(n) = c_2(n) if n>1 where c_k(n) is Ramanujan's sum. - _Michael Somos_, Mar 21 2011
%F a(n) = (1/2)*Product_{k=0..2*n-1} 2*cos((2*k+1)*Pi/(4*n)), n >= 1. See the product given in the Oct 21 2013 formula comment in A056594, and replace there n -> 2*n. - _Wolfdieter Lang_, Oct 23 2013
%F D.g.f.: (2^(1-s)-1)*zeta(s) = -eta(s) (the Dirichlet eta function). - _Ralf Stephan_, Mar 27 2015
%F From _Ilya Gutkovskiy_, Aug 17 2016: (Start)
%F a(n) = T_n(-1), where T_n(x) are the Chebyshev polynomials of the first kind.
%F Binomial transform of A122803. (End)
%F a(n) = exp(i*Pi*n) = exp(-i*Pi*n). - _Carauleanu Marc_, Sep 15 2016
%F a(n) = Sum_{k=0..n} (-1)^k*A063007(n, k), n >= 0. - _Wolfdieter Lang_, Sep 13 2016
%e G.f. = 1 - x + x^2 - x^3 + x^4 - x^5 + x^6 - x^7 + x^8 - x^9 + x^10 - x^11 + x^12 + ...
%p A033999 := n->(-1)^n: seq(A033999(n), n=0..100);
%t Table[(-1)^n, {n, 0, 88}] (* _Alonso del Arte_, Nov 30 2009 *)
%t PadRight[{}, 89, {1, -1}] (* _Arkadiusz Wesolowski_, Sep 16 2012 *)
%o (PARI) a(n)=1-2*(n%2) /* _Jaume Oliver Lafont_, Mar 20 2009 */
%o (Haskell)
%o a033999 = (1 -) . (* 2) . (`mod` 2)
%o a033999_list = cycle [1,-1] -- _Reinhard Zumkeller_, May 06 2012, Jan 02 2012
%o (Magma) [(-1)^n : n in [0..100]]; // _Wesley Ivan Hurt_, Nov 19 2014
%o (Python)
%o def A033999(n): return -1 if n % 2 else 1 # _Chai Wah Wu_, May 24 2022
%Y Cf. A056594, A059841, A063007, A122803.
%Y About Cesàro mean theorem: A114112, A237420.
%Y Cf. A072691 (abs. val. Dgf at s=2), A197070 (abs. val. Dgf at s=3), A267315 (abs. val. Dgf at s=4).
%K sign,easy
%O 0,1
%A Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998
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