# 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

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 $N$.
• So even the idea to connect $S(x)^2$ to $D(2c)$ via mobius inversion and distinct prime counting factors $\omega(r)$ also exists? Could I get a reference for that too? Feb 3, 2016 at 14:45
• I don't have time to help you, but maybe the relation $g(2c)>N_{2}(c)$ where the quantity is defined in my question entitled About Goldbach's conjecture as well as in my blog ideasfornumbertheory.com could be useful Feb 3, 2016 at 17:30