invert transform, an introduction

Refer to Aug. 24-th 2010 blog; "INVERT transform"

INVERT transform

The basic links at OEIS, http://www.tinyurl.com/4zq4q showing the
arxiv article listed - by Bernstein and Sloane.
The operations are presented in this blog, by way of example: find the INVERT
transform of (1, 2, 3,..).
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1. Given an offset of 1 for S(x) = (x + 2x^2 + 3x^3 + 4x^4 +...); we plug about 6 of such terms into Wolfram alpha with S(x) in the following:
INVERT transform = (1 / (1 - S(x)) - 1.
So we plug into the field: 1/(1 - x - 2x^2 - 3x^3 - 4x^4 - 5x^5), enter.
Then subtract 1 getting (x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...) where the coefficients = the even indexed Fibonacci numbers (1, 3, 8, 21,...). Therefore, the INVERT transform of (1, 2, 3,...) = (1, 3, 8, 21, 55,...).
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The reverse operation = the INVERTi transform and can be defined as
(-1 / (1 + S(x)) + 1; and in an operational mode, we enter into WolframAlpha:
-1 / (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5), then enter.
Subtract 1, getting (x + 2x^2 + 3x^3 + 4x^4 + ...). Therefore, the INVERTi
transform of (1, 3, 8, 21,...) = (1, 2, 3,...).
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A property of INVERT transforms (say, B(x) is the INVERT transform of A(x);
then we preface B(x) with a 1 = (1 + B(x)) . Then A(x) * (1 + B(x)) = B(x).
That is, the INVERT transform shifts to the left when prefaced with a 1 and "convolved" with A(x). Example using A(x) = (1, 2, 3,...) and B(x) = (1, 3, 8,..).
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Again, using WolframAlpha, we plug in:
(x + 2x^2 + 3x^3 + 4x^4 + 5x^5) * (1 + x + 3x^2 + 8x^3 + 21x^4 + 55x^5 + ...).
getting (x + 3x^2 + 8x^3 + 21x^4).
Thus, (1, 1, 3, 8, 21, 55,....) has shifted to the left to (1, 3, 8, 21,...) when
convolved with (1, 2, 3,..).