Binomial transform, an introduction

By operations, we use Pascal's triangle as an infinite lower triangular matrix: A007318 at OEIS,
http://www.tinyurl.com/4zq4q:
1
1, 1
1, 2, 1
1, 3, 3, 1
1, 4, 6, 4, 1;
...
= P, then multiply P * [any sequence as a vector....]. Say the sequence = (1, 2, 4, 8,...)
Then A007318 * [1,2,4,8,....] = [1, 3, 9, 27,....]; i.e. powers of 3.
The "reason" behind this is is that taking 1,3,3,...items with distinct labels, we create a binomial frequency of items such that given 2 items, (say, 1 and 2); we have one of the 1 and one of the 2 (as in row 1 of Pascal's triangle = (1, 1). Next row = (1, 2, 1), so with 3 items: (1, 2, 4), we have one 1, two 2's, and one 4, = (1, 2, 1) dot (1, 2, 4) = (1 + 4 + 4) = 9, a power of 3.
To create strings with these binomial frequencies of terms, we access the "Lengths of Infinite Farey Tree Continued fractions", in A088696, as follows:
1
1, 2,
1, 2, 3, 2
1, 2, 3, 2, 3, 4, 3, 2
...; where these terms are based on a Gray code reflection principle: To get the next row, we bring down current row as the left half. ; say (1, 2, 3, 2), Now the "reflected" version of this =
(2, 3, 2, 1); but we add "1" to each term since each successive row introduces a new term. Then we append to (1, 2, 3, 2). That is, append (3, 4, 3, 2) to the right of (1, 2, 3, 2); getting:
(1, 2, 3, 2, 3, 4, 3, 2); representing a binomial frequency of terms labeled (1, 2, 3,...) since referring to Pascal's triangle, row 3 applies to 2^3 = 8 terms with a frequency of 4 distinct terms (1, 2, 3, 4); thus one 1, three 2's, three 3's, and one 4.
That's the "meaning" of the Binomial transform since the sums of these terms (1, 2, 3, 2, 3, 4, 3, 2) = 20, where the Binomial transform of (1, 3, 3...) = (1, 3, 8, 20,...).
However, our terms are (1, 2, 4, 8,...), not (1, 2, 3,...) so we index the (1, 2, 4, 8,..) terms with (1, 2, 3, 4,....), replacing the latter with the corresponding terms in (1, 2, 4, 8,...) getting:
1
1, 2
1, 2, 4, 2
1, 2, 4, 2, 4, 8, 4, 2
...where in the latter row we have one 1, three 2's, three 4's, and one 8, a binomial frequency., with sums of the rows = (1, 3, 9, 27,...); in other words, the binomial transform of
(1, 2, 4, 8,...) = (1, 3, 9, 27,...).