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A033999 (-1)^n. 85
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

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

REFERENCES

S. K. Ghosal, J. K. Mandal, Stirling Transform Based Color Image Authentication, Procedia Technology, 2013 Volume 10, 2013, Pages 95-104.

LINKS

Table of n, a(n) for n=0..88.

Tanya Khovanova, Recursive Sequences

M. Somos, Rational Function Multiplicative Coefficients

Eric Weisstein's World of Mathematics, Inverse Tangent

Eric Weisstein's World of Mathematics, Stirling Transform

Wikipedia, Grandi's series

Wikipedia, +/-1-sequence

Index to sequences with linear recurrences with constant coefficients, signature (-1).

FORMULA

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

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

D.g.f.: (2^(1-s)-1)*zeta(s).

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

Sum_{0<=k<=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>1, b(p^e) = -1 if p>2 and e>1. - 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(2*cos((2*k+1)*Pi/(4*n)), k=0..2*n-1), 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

EXAMPLE

1 - x + x^2 - x^3 + x^4 - x^5 + x^6 - x^7 + x^8 - x^9 + x^10 - x^11 + x^12 + ...

MAPLE

A033999 := n->(-1)^n;

MATHEMATICA

Table[(-1)^n, {n, 0, 88}]

PadRight[{}, 89, {1, -1}] (* Arkadiusz Wesolowski, Sep 16 2012 *)

PROG

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

(Haskell)

a033999 = (1 -) . (* 2) . (`mod` 2)

a033999_list = cycle [1, -1]  -- Reinhard Zumkeller, May 06 2012, Jan 02 2012

CROSSREFS

Sequence in context: A143622 A076479 A155040 * A000012 A162511 A157895

Adjacent sequences:  A033996 A033997 A033998 * A034000 A034001 A034002

KEYWORD

sign,easy

AUTHOR

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

EXTENSIONS

Comment on Fibonacci square unit creation fallacy and Mathematica command added by Alonso del Arte, Nov 30 2009

STATUS

approved

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Last modified October 23 00:52 EDT 2014. Contains 248411 sequences.