0,1

Contribution from Matthew Lehman (matt.comicopia(AT)gmail.com), Nov 17 2008: (Start)

In the Fibonacci sequence, F(n) = F(n-1) + F(n-2),

for every ith number, F(n+i) = A(i)*F(n) + B(i)*F(n-i),

B(i) is given by this sequence,

where B(i) = (-1)^(i+1).

A(i) = F(2*i-1)/F(i-1).

For every Fibonacci number, F(n+1) = F(n) + F(n-1).

For every 2nd Fibonacci number, F(n+2) = 3*F(n) - F(n-2).

For every 3rd Fibonacci number, F(n+3) = 4*F(n) + F(n-3).

For every 4th Fibonacci number, F(n+4) = 7*F(n) - F(n-4).

For every 5th Fibonacci number, F(n+5) = 11*F(n) + F(n-5).

(End) (From Vasiliy Danilov (danilovv(AT)usa.net))

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)

Tanya Khovanova, Recursive Sequences

Eric Weisstein's World of Mathematics, Inverse Tangent

Eric Weisstein's World of Mathematics, Stirling Transform

Wikipedia, Grandi's series [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Nov 21 2009]

Index entries for sequences related to linear recurrences with constant coefficients

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 [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 20 2009]

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

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

(PARI) a(n)=1-2*(n%2) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 20 2009]

Sum_{0<=k<=n} a(k) = A059841(n) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Nov 21 2009]

sign,easy,new

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

Comment on Fibonacci square unit creation fallacy and Mathematica command added by Alonso Delarte (alonso.delarte(AT)gmail.com), Nov 30 2009