[seqfan] Re: polynomial-to-product transform
franktaw at netscape.net
franktaw at netscape.net
Fri Nov 7 07:54:38 CET 2008
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 +...
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.
-------
The rule for "sign" and "nonn" is that the "nonn" keyword
should be used if all the terms in the database are non-
negative, even if there are later terms that are negative.
Use "sign" only if you are entering negative terms.
Franklin T. Adams-Watters
-----Original Message-----
From: Neil Fernandez <primeness at borve.org>
An integer sequence can be transformed as follows:
1) from the sequence A = {a(0), a(1),...},
construct the series A* = 1 + a(1)x + a(2)x^2 + a(3)x^3 +...
2) divide A* by (1+x) to get C1 = (1 + b(1)x +...)
3) divide C1 by (1+b(1)x) to get C2 = (1 + b(2)x^2 +...)
4) divide C2 by (1+b(2)x^2) to get C3 = (1 + b(3)x^3 +...)
3) divide C3 by (1+b(3)x^3) to get C4 = (1 + b(4)x^4 +...)
...
giving A* = the product (1+x)(1+b(1)x)(1+b(2)x^2)...
from which we get the sequence B = {b(1),b(2),...}
If A is the prime sequence then B = {1,2,1,3,2,-4,2,5,4,-6,4,4,10,-36,..
.}
If A is the Fibonacci sequence, beginning {1,2,...}, then B = {0,2,1,4,2
,1,4,18,8,8,18,17,40,50,88,396,210,...}
Neither of these were in the OEIS. When submitting the second, I got an
error message for leaving both "nonn" and "sign" unchecked, so I checked
"nonn" but I'm not sure that the sequence doesn't contain any negative
terms.
If we call B the Polynomial-to-Product transform of A, written PTP(A),
then questions arising include:
* for what sequences A in the OEIS is PTP(A) also in the OEIS?
* for what sequences A in the OEIS is the inverse transform PTP^(-1)(A)
also in the OEIS? (If A is the prime sequence, then the inverse
transform is {1,3,5,14,28,57,...}
* for what A does PTP(A)=A?
Neil
--
Neil Fernandez BA PhD
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