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 



Asymptotics of "ugly" function elucidate Goldbach's conjecture?
Your sum can be rewritten as
$$ D(N)=\sum\_{n=1}^{N-1}d(n)d(N-n),$$
where $N=2c$, and $d(m)$ is the number of divisors of $m$. This is a so-called "binary additive divisor sum", and it has been studied widely. In particular, Ingham (1927) showed that
$$ D(N) = (1+o(1))\frac{6}{\pi^2}\sigma\_1(N)(\log N)^2,$$
where $\sigma\_1(N)$ is the sum of divisors of $N$. 


You can read more about the history of this problem, including several more precise results (with lower order terms) in Motohashi's [classical article](https://eudml.org/doc/82369) on the subject.


