[seqfan] Re: polynomial-to-product transform

Neil Fernandez primeness at borve.org
Fri Nov 7 11:28:28 CET 2008

In message <8CB0EBCD3E3B032-E48-2040 at webmail-md19.sysops.aol.com>,
franktaw at netscape.net writes

>Why do you have step 2?  It seems simpler to just start with
>C1 = A* =  1 + a(1)x + a(2)x^2 + a(3)x^3 +...

Good idea. I think the main reason was to avoid having b(1)=a(1) for all
sequences, but omitting step 2 does make it simpler.

>Then you're getting A* = (1+b(1)x)(1+b(2)x^2)...,
>which unlike your formula has just one term of each
>degree.  By first dividing by (1+x) you are really doing an
>extra transform first.

Omitting step 2 takes the primes to {2,3,-1,9,-4,0,-16,89,...}
and the Fibonacci numbers (beginning 1,1,) to {1,1,1,2,2,3,4,8,8,14,18,2

Neil Fernandez

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