We have a sequence which generalize A329369: $$a(2n+1, p, q) = a(n, p,q), a(2n, p , q) = pa(n, p,q) + qa(n - 2^{f(n)}, p,q) + a(2n - 2^{f(n)}, p,q), a(0, p, q) = 1$$ where $f(n)$ is A007814, exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$. Then $$a((4^n-1)/3, 2, q) = a((4^n-1)/3, q, 2) = \sum\limits_{k=0}^{n}(q-1)^{n-k}T(n,k)$$ where $T(n,k)$ is A087736, $n \geqslant 0$, $q\in\mathbb{R}$. Also $$T(n,k)=[0, 1, 3, 6, 10, 15, 21,\cdots] \operatorname {DELTA} [1, 3, 6, 10, 15, 21, 28,\cdots]$$ Here $\operatorname {DELTA}$ defined as follows:

The operator $\operatorname {DELTA}$ takes two sequences $r = (r_0, r_1,\cdots)$, $s = (s_0, s_1,\cdots)$ and produces a triangle $T(n, k)$, $0 \leqslant k \leqslant n$, as follows: Let $q(k) = x\cdot r_k + y\cdot s_k$ for $k \geqslant 0$; let $P(n, k)$ ($n \geqslant 0, k \geqslant -1$) be defined recursively by $P(0, k) = 1$ for $k \geqslant 0$; $P(n, -1) = 0$ for $n \geqslant 1$; $P(n, k) = P(n, k-1) + q(k)P(n-1, k+1)$ for $n \geqslant 1, k \geqslant 0$. Then $P(n, k)$ is a homogeneous polynomial in $x$ and $y$ of degree $n$ and $T(n, k)$ = coefficient of $x^{n-k}y^k$ in $P(n, 0)$.

Is there a way to prove it? Are similar sums with a triangle possible for another $p$? If so, what are these triangles? Are they set similarly using the operator $\operatorname {DELTA}$?