[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
9,40,...}.
Neil
--
Neil Fernandez
More information about the SeqFan
mailing list