Asymptotics of "ugly" function elucidate Goldbach's conjecture? Question
We now define the following "ugly" function:
$$ A_c(s,r,n,m)  =
  \begin{cases} 
       1  & \text{ if only $sr+nm=2c$ } \\ 0 & \text{otherwise}
  \end{cases}
$$
How does the "ugly" function asymptotically behave?
$$ \sum_{n=1}^c \sum_{m=1}^c \sum_{r=1}^c \sum_{s=1}^c A_c(s,r,n,m) \sim (?)$$
Does this connection between $A_c$ and $S(x)^2$ enlighten us further on Goldbach's conjecture?
$$ \sum_{c=2}^\infty \sum_{n=1}^c \sum_{m=1}^c \sum_{r=1}^c \sum_{s=1}^c A_c(s,r,n,m) (\mu(s) \omega(r) + \mu(m) \omega(n))x^{2c} = (\sum_{r=2}^\infty x^{p_r})^2 + x^4  $$
Relevance
$$ S(x)=\sum_{r=1}^\infty x^{p_r} $$
where $p_r$ is the $r$'th prime:
$$ \sum_{r=1}^\infty S(x^r) = \sum_{r=1}^\infty \frac{x^{p_r}}{(1-x^{p_r})}  $$
However, we also notice that:
$$ \sum_{r=1}^\infty S(x^r) = \sum_{r=1}^\infty \frac{x^{p_r}}{(1-x^{p_r})} = \sum_{r=1}^\infty \omega(r) x^r  $$
Where $\omega(r)$ is the number of distinct prime factors of $r$ and assigning $\omega(1)=0$ : http://mathworld.wolfram.com/DistinctPrimeFactors.html
(More about it here) 
Using the mobious inversion formula:
$$ S(x)=\sum_{r=1}^\infty x^{p_r} = \sum_{r=1}^\infty \sum_{s=1}^\infty \mu(s) \omega(r) x^{rs}$$
Hence, we notice the following relationship:
$$ S(x)=\sum_{r=1}^\infty x^{p_r} = \sum_{r=1}^\infty \sum_{s=1}^\infty \mu(s) \omega(r) x^{rs}$$
where $p_r$ is the $r$'th prime, $\mu$ is the mobius function and $\omega(r)$ is the number of distinct primes in $r$. For example $\omega(2^2 \times 3) = 2 $ also we assign $\omega(1)=0$ 
$$ S(x)^2 = (\sum_{r=1}^\infty x^{p_r})^2=\sum_{r=1}^\infty \sum_{j=1}^\infty x^{p_r+p_j} = \sum_{r=1}^\infty \sum_{s=1}^\infty (\mu(s) \omega(r) + \mu(m) \omega(n)) x^{rs+mn}$$
comparing the even powers and assuming Golbach's conjecture to be true:
$$ \sum_{sr+mn =2c}\mu(s) \omega(r) + \mu(m) \omega(n) = g(2c)  > 0$$
Where $g(2c)$ is the number of ways the primes can sum the number $2c$. For example $g(14=7+7=11+3=3+11)=3$
We note define the following:
$$ \underbrace{\mu(s) \omega(r)}_{a_{s,r}} + \underbrace{\mu(m) \omega(n)}_{a_{m,n}} = a_{s,r} + a_{m,n}$$
$$ \implies \sum_{sr+mn =2c} a_{s,r} + a_{m,n} = (?) $$
Taking a dummy case of $sr+mn =4$
$$ \sum_{sr+mn =4} a_{2,1} + a_{1,2} + a_{1,2} +a_{2,1}$$
$$ \implies \sum_{sr+mn =4} \underbrace{A_2(2,1,1,2)}_{1} (a_{2,1} + a_{1,2}) + \underbrace{A_2(1,2,2,1)}_{1}(a_{1,2} +a_{2,1})$$
Taking another dummy case of  $sr+mn =8$
$$ \sum_{sr+mn =8} a_{4,1} + a_{1,4} + a_{1,4} +a_{4,1} + a_{2,2} + a_{2,2}$$
$$\implies \sum_{sr+mn =8} \underbrace{A_4(4,1,1,4)}_1(a_{4,1} + a_{1,4}) + \underbrace{A_4(1,4,4,1)}_1 (a_{1,4} +a_{4,1}) +\underbrace{A_{4} (2,2,2,2)}_1 (a_{2,2}+a_{2,2}) $$
Making use of the $A_c$ in the general case:
$$ \sum_{n=1}^c \sum_{m=1}^c \sum_{r=1}^c \sum_{s=1}^c A_c(s,r,n,m) (\mu(s) \omega(r) + \mu(m) \omega(n)) = \sum_{sr+mn =2c} \mu(s) \omega(r) + \mu(m) \omega(n)  $$
Hence,
$$ \sum_{c=2}^\infty \sum_{n=1}^c \sum_{m=1}^c \sum_{r=1}^c \sum_{s=1}^c A_c(s,r,n,m) (\mu(s) \omega(r) + \mu(m) \omega(n))x^{2c} = (\sum_{r=2}^\infty x^{p_r})^2 + x^4  $$
P.S: I do not claim to have solved the conjecture. I am only curious on how viable this approach is .... Also, if you think I've skipped too many steps feel free to comment 
 A: It is somewhat straightforward to come up with a reasonable upper bound, and in fact to likely accurately approximate the leading order asymptotic.
You are looking to count the number of ways of writing $2c = sr + nm$. This can be written as $2c = X + Y$, where we are now interested in the number of ways of writing $X$ and $Y$ as products of two integers. That can be done in $d(X)d(Y)$ ways, where $d(n)$ is the number of divisors of $N$. Writing $Y = 2c - X$, we see this can be done in $d(X)d(2c - X)$ ways.
The only remaining consideration is which values $X$ can take? Recall that $X = sr$ and $1 \leq s,r \leq c$. It's clear that $X$ takes every value from $1$ to $c$, and every composite value from $c+1$ to $2c$. But $X$ doesn't take prime values from $c+1$ to $2c$.
Thus the count you're looking for takes the shape
$$ \sum_{X = 1}^{2c} d(X)d(2c - X) - \sum_{c < p < 2c} d(p)d(2X-p).$$
The divisor function has individual bound $d(n) \ll n^\epsilon$ for any $\epsilon > 0$, and on average $d(n) \ll \log n$. So we bound the first sum by
$$ \sum_{X = 1}^{2c} d(X) d(2c - X) \ll (2c)^{1 + \epsilon},$$
and where we should expect the bound to be barely larger than $(2c) (\log 2c)^2$, based on the correlation of $d(X)$ and $d(2c - X)$.
For the second sum, we note that $d(p) = 1$. This (probably) removes the problem of correlation, and we use the average bound $d(n) \ll \log n$ and the prime number theorem to bound the second by
$$ \sum_{c < p < 2c} d(2X - p) \ll \frac{c}{\log 2c} \cdot \log 2c = c.$$
[I should note that even if there is correlation, this term is no larger than the first, leading to the same initial upper bound].
So your leading order is or size $(2c)^{1 + \epsilon}$.
