I'm not familiar with the formalism of plethysm, so I need some help in unpacking the plethystic substitution $h_n[n\mathbf{z}]$ found in eqns. 5.6 and 5.9 of "Combinatorics of labelled parallelogram polyominoes" by J-C. Aval, F. Bergeron, and A. Garsia. I suspect it unpacks as the partition polynomials of OEIS A035206.
For example, equating expressions in the paper,
$Frob(\mathbb{L}_{4,4}) = \frac{1}{n} \binom{2(n-1)}{n=1}h_n[n\mathbf{z}] |_{n=4}$
$ = 5 h_{1111} \binom{6}{6} +10 h_{211} \binom{6}{5} + 4 \binom{6}{4} h_{31} + 2 \binom{6}{4} h_{22} + 1 \binom{6}{3} h_{4} $
$ = 5 \cdot 1 h_{1111} +10 \cdot 6 h_{211} + 4 \cdot 15 h_{31} +2 \cdot 15 h_{22} + 1 \cdot 20 h_{4} $
$ = 5 h_{1111} + 60 h_{211} + 60 h_{31} +30 h_{22} + 20 h_{4} $,
so I conclude
$h_4[4\mathbf{z}] = Frob(\mathbb{L}_{4,4}) / 5 = 1 h_{1111} + 12 h_{211} + 12 h_{31} +6 h_{22} + 4 h_{4} $
where $\frac{1}{n\ }\ \ \frac{\left(2n-2\right)!}{\left(n-1\right)!\left(n-1\right)!}|_{n=4} = 5 $, a Catalan number of OEIS A000108.
The numerical coefficients (1, 12, 12, 6, 4,) comprise a row of A035206.
(The distinct binomial factors (1, 6, 15, 20), a row of A094527, and the factors in front of the binomials, (5,10,4,2,1), are present in unsigned A350499.)
Example 2 with the next Catalan number $\frac{1}{n\ }\ \ \frac{\left(2n-2\right)!}{\left(n-1\right)!\left(n-1\right)!}|_{n=5} = 14$ and suppressing the indexing for the partitions of $n$:
$Frob(\mathbb{L}_{5,5}) = \frac{1}{n} \binom{2(n-1)}{n}h_n[n\mathbf{z}] |_{n=5}$
$= 14 \binom{8}{8} h.. + 35\binom{8}{7} h.. +15 \binom{8}{6} h.. + 15 \binom{8}{6} h.. + 5 \binom{8}{5} h.. + 5 \binom{8}{5} h.. + 1 \binom{8}{4} h..$
$ = 14 \cdot 1 h.. + 35 \cdot 8 h.. +15 \cdot 28 h.. + 15 \cdot 28 h.. + 5 \cdot 56 h.. + 5 \cdot 56 h.. + 1 \cdot 70 h..$,
and
$h_5[5\mathbf{z}] = Frob(\mathbb{L}_{5,5}) / 14 = 1 h.. + 20 h.. + 30 h.. + 30 h.. + 20 h.. + 20 h.. + 5 h..$
(5, 20, 20, 30, 30, 20, 1) is a row of A035206.
(The distinct binomial factors (1, 8, 28, 56, 70) are a row of A094527, and the factors in front of the binomials, (14, 35, 15, 15, 5, 5, 1), are unsigned coefficients of A350499.)
So, the circumstantial evidence is promising, but I need some moral support in stating conclusively from an explicit definition / operational demonstration of the plethystic substitution that the coefficients of $h_n[n\mathbf{z}]$ are indeed given by the rows of A035206 and flagged by the partitions of $n$.
Restating this last paragraph in deference to Sam Hopkins as an explicit question:
From an explicit definition / operational demonstration of the plethystic substitution, can someone show that the coefficients of $h_n[n\mathbf{z}]$ are indeed given by the rows of A035206 and flagged by the partitions of $n$?
(I suspect that if I could master the content of "Alphabet splitting" or "Symmetric Functions & Combinatorial Operators on Polynomials" by Lascoux, I could answer the question for myself--this would serve as a good test of my understanding--but I'm impatient for confirmation or invalidation of this specific conjecture since it is related to some interesting combinatoric and algebraic structures.)
Update 2/12/2024:
I'd like to ask for a reference that could provide an accessible bridge, for a pedestrian like me, between plethystic maneuvers and the umbral maneuvers I describe below involving specifically $h_n[n\mathbf{z}]$.
Ira Gessel in his comment gives the workable unpacking
$$h_k[n\mathbf{z}] = \frac{D_{t=0}^k}{k!} (H(t))^n = \frac{D_{t=0}^k}{k!} (1 + h_1t + h_2 t^2 +\cdots)^n.$$
One can get the same result in umbral maneuvers, e.g., for $n=4$ (with results illustrated above),
$$(a.' + a.'' + a.''' + a.'''')^4 / 4!$$
$$ = 12 x_2 x_1^2 + x_1^4 + 12 x_3 x_1 + 6 x_2^2 + 4 x_4$$
by multiplying out the umbral expression and aggegating factors (e.g., $a'.a.''a.'' =(a.')^1(a.'')^2$), then lowering the powers to subscripts (e.g., $(a.')^1(a.'')^2 = a_1'a_2''$), and then by erasing the primes and setting $a_j= j! x_j$ (e.g., $ a_1'a_2'' = a_1 a_2 = x_1 2! x_2$). Finally, identify $h_{1111} = x_1^4$, $h_{112} = x_1^2 x_2$, and so on. (The Appell Sheffer hybrid umbral op calculus applies then.)
For example, in excruciatig detail for $n=3$, changing notation for Wolfram Alpha to handle easily,
$(a'+a''+a''')^3 = (a + b +c)^3$
$ = a^3 + 3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + b^3 + 3 b^2 c + 3 b c^2 + c^3.$
Dropping the powers (the umbral maneuver) and letting $a=b=c=d$ (erasing primes) gives
$d_3 + 3 d_2 d_1 + 3 d_2 d_1 + 3 d_1 d_2 + 6 d_1^3 + 3 d_1 d_2 + d_3 + 3 d_2 d_1 + 3 d_1 d_2 + d_3$
$= 6 d_1^3 + 18 d_2 d_1 + 3 d_3$.
Letting $d_j = j! x_j$ gives
$ 6 d_1^3 + 18 d_2 d_1 + 3 d_3 = 6 x_1^3 + 36 x_2 x_1 + 18 x_3$
and normalizing with the factorial and identifying monomials with partitions of $n$ gives
$(6 x_1^3 + 36 x_2 x_1 + 18 x_3)/3! = x_1^3 + 6 x_2 x_1 + 3 x_3$
$=h_{111} + 6 h_{12} + 3 h_3$,
corresponding to row 3 of https://oeis.org/A035206.
The monomial summands in these expansions for fixed $n$ have the form $x_1^{e_1}x_2^{e_2} \cdots x_n^{e_n}$ with the associated numerical coefficient $\frac{n!}{\left(n-Se\right)!e_{1}!e_{2}!e_{3}!\cdots e_{n}!}$ with $Se = e_1+e_2+\cdots+e_n$, i.e., the sum of the exponents. Of course, for any of the monomials in the expansion, we have a partition of $n$, i.e., $n = 1\cdot e_1 + 2 \cdot e_2 + \cdots + n \cdot e_n$.
This is related to the Appell Sheffer formalism with
$H(t)^n = (1+h_1t+h_2t^2+\cdots)^n = (1+\bar{h}_1t+\bar{h}_2t^2/2!+\cdots)^n$
$ = (e^{\bar{h}.t})^n = e^{a't}e^{a''t}\cdots e^{(n \; primes)t}= e^{(a'+a'' + \cdots+a^{(n \; primes)})t}$
with $\bar{h}_k = k! h_k$ and with the umbral maneuvers / evaluations described earlier.
The Appell polynomials $A_k(x;\bar{h}_1,...,\bar{h}_k)$ with $A_k(0;\bar{h}_1,...,\bar{h}_k) = D_{t=0}^k (H(t))^n$ have the e.g.f. $(H(t))^n e^{xt}$, the lowering operator $D_x$, and an associated raising op from which recursions and other identities follow.
