A033999 a(n) = (-1)^n.

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

Author

Vasiliy Danilov (danilovv(AT)usa.net), Jun 15 1998

Keywords

• sign
• easy

Comments

(-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

Periodic with period length 2. - Ray Chandler, Apr 03 2017

From Bernard Schott, May 11 2022: (Start)

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).

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.

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.

Note that the converse of the theorem is true iff a(n) is monotonic (End).

Examples

			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 + ...
		

References

• J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10000
• Paul Barry, Characterizations of the Borel triangle and Borel polynomials, arXiv:2001.08799 [math.CO], 2020.
• F. Javier de Vega, An extension of Furstenberg's theorem of the infinitude of primes, arXiv:2003.13378 [math.NT], 2020.
• S. K. Ghosal and J. K. Mandal, Stirling Transform Based Color Image Authentication, Procedia Technology, 2013 Volume 10, 2013, Pages 95-104.
• Tanya Khovanova, Recursive Sequences
• Mathematics Stack Exchange, Convergence of series implies convergence of Cesàro mean, 2013.
• László Németh, The trinomial transform triangle, J. Int. Seqs., Vol. 21 (2018), Article 18.7.3. Also arXiv:1807.07109 [math.NT], 2018.
• ProofWiki, Cesàro mean.
• Michael Somos, Rational Function Multiplicative Coefficients
• Eric Weisstein's World of Mathematics, Inverse Tangent
• Eric Weisstein's World of Mathematics, Stirling Transform
• Wikipedia, Ernesto Cesàro.
• Wikipedia, Grandi's series
• Wikipedia, +/-1-sequence
• Wikipedia, Dirichlet eta function
• Wikipédia, Lemme de Cesàro (in French).
• Index entries for linear recurrences with constant coefficients, signature (-1).
• Index to sequences related to inverse of cyclotomic polynomials

Crossrefs

Cf. A056594, A059841, A063007, A122803.

About Cesàro mean theorem: A114112, A237420.

Cf. A072691 (abs. val. Dgf at s=2), A197070 (abs. val. Dgf at s=3), A267315 (abs. val. Dgf at s=4).

Programs

Haskell

a033999 = (1 -) . (* 2) . (`mod` 2)
a033999_list = cycle [1,-1] -- Reinhard Zumkeller, May 06 2012, Jan 02 2012

Magma

[(-1)^n : n in [0..100]]; // Wesley Ivan Hurt, Nov 19 2014

Maple

A033999 := n->(-1)^n: seq(A033999(n), n=0..100);

Mathematica

Table[(-1)^n, {n, 0, 88}] (* Alonso del Arte, Nov 30 2009 *)
PadRight[{}, 89, {1, -1}] (* Arkadiusz Wesolowski, Sep 16 2012 *)

PARI

a(n)=1-2*(n%2) /* Jaume Oliver Lafont, Mar 20 2009 */

Python

def A033999(n): return -1 if n % 2 else 1 # Chai Wah Wu, May 24 2022

Formula

G.f.: 1/(1+x).

E.g.f.: exp(-x).

Linear recurrence: a(0)=1, a(n)=-a(n-1) for n>0. - Jaume Oliver Lafont, Mar 20 2009

Sum_{k=0..n} a(k) = A059841(n). - Jaume Oliver Lafont, Nov 21 2009

Sum_{k>=0} a(k)/(k+1) = log(2). - Jaume Oliver Lafont, Mar 30 2010

Euler transform of length 2 sequence [ -1, 1]. - Michael Somos, Mar 21 2011

Moebius transform is length 2 sequence [ -1, 2]. - Michael Somos, Mar 21 2011

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

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

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

D.g.f.: (2^(1-s)-1)*zeta(s) = -eta(s) (the Dirichlet eta function). - Ralf Stephan, Mar 27 2015

From Ilya Gutkovskiy, Aug 17 2016: (Start)

a(n) = T_n(-1), where T_n(x) are the Chebyshev polynomials of the first kind.

Binomial transform of A122803. (End)

a(n) = exp(i*Pi*n) = exp(-i*Pi*n). - Carauleanu Marc, Sep 15 2016

a(n) = Sum_{k=0..n} (-1)^k*A063007(n, k), n >= 0. - Wolfdieter Lang, Sep 13 2016