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Search: id:A033999
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| 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; listen)
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OFFSET
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0,1
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COMMENT
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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)
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LINKS
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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
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FORMULA
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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]
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MAPLE
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A033999 := n->(-1)^n;
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PROGRAM
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(PARI) a(n)=1-2*(n%2) [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 20 2009]
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CROSSREFS
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Adjacent sequences: A033996 A033997 A033998 this_sequence A034000 A034001 A034002
Sequence in context: A143622 A076479 A155040 this_sequence A057077 A157895 A063747
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KEYWORD
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sign,easy
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AUTHOR
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Vasiliy Danilov (danilovv(AT)usa.net) Jun 15 1998
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