## 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