[seqfan] Re: A117963

Paul D Hanna pauldhanna at juno.com
Wed Jul 25 05:19:08 CEST 2012

     The second and third formulae are equivalent; 
thus it suffices to show only that: 
  a(n) == Fibonacci(n+1) (mod 3). 
Given the g.f.  
  A(x) = A(x^3)*(1 - 4*x^3 - x^6)/(1 - x - x^2), 
suppose we define F(x) such that  
  F(x) = F(x^3)*(1 - x^3 - x^6)/(1 - x - x^2), 
then it is not hard to see that 
  A(x) == F(x) (mod 3).  
But now F(x) is simply 
  F(x) = 1/(1 - x - x^2)
which is the g.f. for the Fibonacci sequence (with offset). 
Therefore the formulae hold. 
Best wishes, 
---------- Original Message ----------
From: "Harvey P. Dale" <hpd1 at nyu.edu>
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] A117963
Date: Tue, 24 Jul 2012 15:20:44 -0400

           I think the 2nd and 3rd formulae provided by Paul Hanna may
be wrong.




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