Why does count of Z -Transform of sequence change? I was looking at the reference [1] below and noted the author defined the Z-transform for [1, 2, 3] as $$[6, \frac{11}{4}, 2]$$
I worked it out as follows:
$$X[z]=\sum_{n=0}^2x[n]z^{-n}$$
$$=x[0]z^0+\frac{x[1]}{z}+\frac{x[2]}{z^2}$$
$$=1+\frac{1}{z}+\frac{2}{z^2}$$
so from there the following results:
$$X[1]=1+\frac{2}{1}+\frac{3}{1}=6$$
$$X[2]=1+\frac{2}{2}+\frac{3}{2^2}=\frac{11}{4}$$
$$X[3]=1+\frac{2}{3}+\frac{3}{3^2}=2$$
Hence
$$X[z]=[6, \frac{11}{4}, 2]$$
However, in the 2nd part when plugging in for z it implies X[z] counts from 1 to 3 i.e. X[0] does not exist only X[1], X[2], and X[3]. Before the transform, the count of x[n] was from 0 to 2 so x[0], x[1], and x[2] exist while x[3] does not. Can anyone explain why the count changes? I've tried using n=1 to 3 to carry out the transform but a different result comes out. I've also changed z to z=0 to 2 but doesn't work either.
References
[1] http://algorithmicassertions.com/quantum/2014/04/27/The-Not-Quantum-Laplace-Transform.html

When I wrote that post I used 0-based indexing for the input because that's just what programmers do by default, and I used 1-based indexing for the output because of the division-by-zero issue. I don't know if it's more common for the indices to start at 1 for the input.
Basically, don't worry too much about it. It's just a convention, and it's not hard to convert between the various possibilities.
