[seqfan] Re: A117963
Paul D Hanna
pauldhanna at juno.com
Wed Jul 25 05:19:08 CEST 2012
Harvey,
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,
Paul
---------- 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.
Best,
Harvey
_______________________________________________
Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan
mailing list