OEIS A122888 is: $1, 1, 1, 1, 2, 2, 1, 1, 3, 6, 9, 10, 8, 4, 1, 1, 4, 12, 30, 64, 118, 188, 258, 302, 298, 244, 162, 84, 32, 8, 1, 1, 5, 20, 70, 220, 630, 1656,\dots$
I'm having trouble understanding the definition:
$T^{n}_{k}$ is the number of strings of length $k-1$ on the alphabet $ (1,2,\dots,n) $ such that between every two occurrences of a letter i there is an occurrence of a letter strictly larger than i. For example, for $n = 3, k = 4$ we have the strings $121, 131, 232$ and the six permutations of $123$.
$T=\left[ \begin{array}{cccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0& \dots \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0& \dots \\ 1 & 2 & 2 & 1 & 0 & 0 & 0 & 0& \dots \\ 1 & 3 & 6 & 9 & 10 & 8 & 4 & 1& \dots \\ 1 & 4 & 12 & 30 & 64 & 118 & 188 & 258& \dots \\ 1 & 5 & 20 & 70 & 220 & 630 & 1656 & 4014& \dots \\ 1 & 6 & 30 & 135 & 560 & 2170 & 7916 & 27326& \dots \\ 1 & 7 & 42 & 231 & 1190 & 5810 & 27076 & 121023& \dots \\ 1 & 8 & 56 & 364 & 2240 & 13188 & 74760 & 409836& \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots& \ddots \\ \end{array} \right]$
In addition I wonder what is the meaning behind multiplying the columns of $A122888$ by $(1−x)^k$ . I understand that it gives us $A122890$ ($H$) and that in a certain sense it is more "predictable" but what is the reasoning behind the choice of $(1−x)^k$ ?
$H=\left[ \begin{array}{ccccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &\dots\\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 &\dots\\ 0 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 0 &\dots\\ 0 & 0 & 1 & 5 & 0 & 0 & 0 & 0 & 0 &\dots\\ 0 & 0 & 0 & 10 & 14 & 0 & 0 & 0 & 0&\dots \\ 0 & 0 & 0 & 8 & 70 & 42 & 0 & 0 & 0&\dots \\ 0 & 0 & 0 & 4 & 160 & 424 & 132 & 0 & 0&\dots \\ 0 & 0 & 0 & 1 & 250 & 1978 & 2382 & 429 & 0&\dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots& \vdots& \ddots \\ \end{array} \right]$
For $n=3$ and $k=4$, the strings on the alphabet $(1,2,3)$ length $k-1=3$ are $27$: $$111,112,113,121,122,\dots,331,332,333.$$ We consider the strings in the list such that:
0) the letters are all different: $123,132,213,231,312,321$
1) between every two occurrences of a letter $1$ there is an occurrence of a letter $2$ or $3$: $121,131$.
2) between every two occurrences of a letter $2$ there is an occurrence of a letter $3$: $232$.
Therefore $T^{3}_{4}=6+2+1=9$.
As regards your second question, if we multiply a g.f. $f(x)=\sum_j a_jx^j$ by $(1-x)^k$ we obtain a new g.f. $\sum_n b_nx^n$ where the sequence of coefficients $(b_n)_n$ is a sort of binomial transform of the sequence $(a_j)_j$: $$b_n=\sum_{j=0}^n \binom{k}{j}(-1)^ja_{n-j}.$$