The definition given for A078756 is that it's the root of the g.f. of the prime indicator function. Is your $H$ the g.f. of the prime indicator function or the prime counting function?

$[z^{n-1}](\Pi(z) C(z)) = [z^{n-3}](z^{-2} \Pi(z) C(z))$. The introduction of the integral might be easier to see backwards: $\int \sum a_i z^i \textrm{d}z = \sum \int a_i z^i \textrm{d}z = \sum \frac{a_i z^{i+1}}{i+1}$

@StevenClark's coefficients are $$a_i = \begin{cases}\textrm{arbitrary} & \textrm{if } i = 1 \\ 2 & \textrm{if } i = 2 \\ 0 & \textrm{if } i+1 \textrm{ composite} \\ \frac{1}{\pi(i)} & \textrm{otherwise} \end{cases}$$ It's a direct encoding of the primes.

$$\pi(n+2) = \sum_{\vec{x}} \frac{c_{x_1-1}}{x_1} \cdot \frac{c_{x_2-1}}{x_1 + x_2} \cdot \frac{c_{x_3-1}}{x_1+x_2+x_3} \cdots \frac{c_{x_k-1}}{n}$$ where the sum is over compositions of $n$. Note quite the same as Bell polynomials, but possibly something in this form has been studied before.

Let $C_n$ be the multiset of non-zero values of $f(n, j)$. Then $v_r(n)$ is the $r$th power sum symmetric polynomial applied to $C_n$. Since the power sum symmetric polynomials generate the ring of symmetric polynomials, for any symmetric polynomial $P$ the g.f. $V_P(x) = \sum_{n \ge 0} P(C_n) x^n$ is a rational function. Perhaps other families of symmetric polynomials could shed some light.

Actually, no. $V_{16}, V_{20}, V_{22}$ probably each need a (further) quartic factor so that the evens consistently increase in degree by 2. The odds have degrees $1, 3, 5, 7, 7, 9, 11, 11, 13, 15, 15, 17, \ldots$ so that might be a consistent pattern without the need for further factors.

I get to $V_{23}$ but without proofs of correctness. Results here. The extrapolation of the second differences being Fibonacci numbers continues to hold for the quadratic term with an error of 1 at $V_{10}$, $V_{14}$, $V_{18}$, $V_{22}$, so perhaps the only artificial factor needed is $1-x^2$ for $r = 10 + 4n$.

The second differences of $2,4,7,11,17,26,40$ are $1,1,2,3,5$ so $-62x^2$ does make sense for $V_9$. (The slight flaw in this observation is that it would predict $-97x^2$ for $V_{10}$, but it does predict $-153x^2$ for $V_{11}$. Could there be a typo in your value of $V_{10}$?)

There must be several typos which need correcting for this to make sense. The double-subscripted $y_{i,j}$ aren't used outside of the first definition of $T_j$; $x$ is unbound (should it be bound by the existential quantifier?); and "find all $T_j$ such that ... there exists $T_j$" only makes sense if there's aliasing and the two references to $T_j$ are to different objects, but in that case neither of them is used anywhere.

Fair enough. In that case, here are the first 7999 values, computed exactly with Python.

@SamHopkins, if $x$ and $x+1$ are not adjacent, they can be swapped to get another alternating permutation. $x$ can't be adjacent to both $x-1$ and $x+1$ because then you have two consecutive ascents or two consecutive descents. There's a bit of work to be done to show that double-counting doesn't occur, but intuitively this feels like the explanation.

How far have you tested it numerically? My Sage code may be buggy, but it only seems to hold for $\ell \in \{1, 2\}$, $m > 0$.