# [seqfan] Re: A117963

Harvey P. Dale hpd1 at nyu.edu
Wed Jul 25 17:55:01 CEST 2012

```Paul:
I have tried to implement that formula in Mathematica but it doesn't
generate the terms of the sequence.  Any number, mod 3, cannot exceed 2
but there are many terms in the sequence which exceed that number.  What
don't I understand?
Best,
Harvey

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Paul D
Hanna
Sent: Tuesday, July 24, 2012 11:19 PM
To: seqfan at list.seqfan.eu
Subject: [seqfan] Re: A117963

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

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