[seqfan] Re: A117963

[Harvey P. Dale]

Thanks to Paul, Tony, and Alois.  I had misread the "==" to be "=" but
now I understand.
Best,
Harvey

-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of T. D.
Noe
Sent: Wednesday, July 25, 2012 12:04 PM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A117963

a(n) == Fibonacci(n+1) (mod 3) means

a(n) - Fibonacci(n+1) is a multiple of 3.

Best,

Tony


At 11:55 AM -0400 7/25/12, Harvey P. Dale wrote:
>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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