%I A033999
%S A033999 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 A033999 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,
%U A033999 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
%V A033999 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,
%W A033999 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,
%X A033999 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 A033999 (-1)^n.
%C A033999 Contribution from Matthew Lehman (matt.comicopia(AT)gmail.com), Nov 17 
               2008: (Start)
%C A033999 In the Fibonacci sequence, F(n) = F(n-1) + F(n-2),
%C A033999 for every ith number, F(n+i) = A(i)*F(n) + B(i)*F(n-i),
%C A033999 B(i) is given by this sequence,
%C A033999 where B(i) = (-1)^(i+1).
%C A033999 A(i) = F(2*i-1)/F(i-1).
%C A033999 For every Fibonacci number, F(n+1) = F(n) + F(n-1).
%C A033999 For every 2nd Fibonacci number, F(n+2) = 3*F(n) - F(n-2).
%C A033999 For every 3rd Fibonacci number, F(n+3) = 4*F(n) + F(n-3).
%C A033999 For every 4th Fibonacci number, F(n+4) = 7*F(n) - F(n-4).
%C A033999 For every 5th Fibonacci number, F(n+5) = 11*F(n) + F(n-5).
%C A033999 (End) (From Vasiliy Danilov (danilovv(AT)usa.net))
%C A033999 Furthermore requiring F(0) = F(1) = 1, we have a(n) = (-1)^n = F(n)^2 
               - F(n - 1)F(n + 1), meaning that in trying to convert a square F(n)^2 
               square units in area into a rectangle F(n - 1) by F(n + 1) square 
               units in area, the square will be deficient by one square unit when 
               n is odd, and have a surplus of one square unit when n is even. (From 
               Alonso Delarte (alonso.delarte(AT)gmail.com), Nov 30 2009)
%H A033999 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A033999 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               InverseTangent.html">Inverse Tangent</a>
%H A033999 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               StirlingTransform.html">Stirling Transform</a>
%H A033999 Wikipedia, <a href="http://en.wikipedia.org/wiki/Grandi%27s_series">Grandi's 
               series</a> [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), 
               Nov 21 2009]
%H A033999 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A033999 G.f.: 1/(1+x). E.g.f.: exp(-x). D.g.f.: (2^(1-s)-1)*zeta(s).
%F A033999 Linear recurrence: a(0)=1, a(n)=-a(n-1) for n>0 [From Jaume Oliver Lafont 
               (joliverlafont(AT)gmail.com), Mar 20 2009]
%p A033999 A033999 := n->(-1)^n;
%t A033999 Table[(-1)^n, {n, 0, 88}]
%o A033999 (PARI) a(n)=1-2*(n%2) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), 
               Mar 20 2009]
%Y A033999 Sum_{0<=k<=n} a(k) = A059841(n) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), 
               Nov 21 2009]
%K A033999 sign,easy,new
%O A033999 0,1
%A A033999 Vasiliy Danilov (danilovv(AT)usa.net) Jun 15 1998
%E A033999 Comment on Fibonacci square unit creation fallacy and Mathematica command 
               added by Alonso Delarte (alonso.delarte(AT)gmail.com), Nov 30 2009