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)

Index entries for sequences related to linear recurrences with constant coefficients

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]

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;

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

Sequence in context: A143622 A076479 A155040 this_sequence A057077 A162511 A157895

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

Adjacent sequences: A033996 A033997 A033998 this_sequence A034000 A034001 A034002

sign,easy,new

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