I will prove that the limit exists and is precisely the Golomb-Dickman constant $\lambda$. First, let me sketch how one can show that the expected value of $\frac{\log(P_1(n))}{\log(n)}$, where $P_1(n)$ denotes the largest prime factor of $n$, is precisely $\lambda\approx0.62433$ via probability theory.
We say that an integer $n\geq1$ is $y$-smooth if every prime divisor of $n$ is $\leq y$. We also say that $n\geq1$ is $y$-powersmooth if every prime power divisor of $n$ is $\leq y$. Let's denote by $S(x,y)$ and $PS(x,y)$ the set of $y$-smooth and $y$-powersmooth numbers $\leq x$ respectively. It is a theorem of Dickman that
$$F_S(\nu):=\lim_{x\to+\infty}\frac{1}{x}|S(x,x^{\nu})|=\rho(1/\nu)$$
where $\rho(t)$ is Dickman's function satisfying the differential equation $t\rho'(t)+\rho(t-1)=0$.
For large $x\gg0$, choose a random $tx\in[0,x]$ where $t\in[0,1]$ (that is, consider the random variable $xT\sim U(0,x)$ where $U(0,x)$ is the discrete uniform distribution that "looks" continuous for large $x$). What is the probability that $\frac{\log(P_1(tx))}{\log(tx)}\leq\nu$? Well
$$tx\in S(x,x^\nu)\iff P_1(tx)\leq x^\nu\iff \frac{\log(P_1(tx))}{\log(tx)}\leq\frac{\nu\log(x)}{\log(tx)}=\frac{\nu}{1+\frac{\log(t)}{\log(x)}}\sim\nu$$
so, for large $x\gg0$, we have $P\left(\frac{\log(P_1(xT))}{\log(xT)}\leq\nu\right)\sim P(xT\in S(x,x^\nu))=\frac{1}{x}|S(x,x^\nu)|\sim F_S(\nu)$. This means that $F_S(\nu)$ is the cumulative distribution function of $\frac{\log(P_1(n))}{\log(n)}$ so we can compute the expected value as
$$\lim_{x\to+\infty}\frac{1}{x}\sum_{n\leq x}\frac{\log(P_1(n))}{\log(n)}=\int_0^1\nu F_S'(\nu)d\nu=\int_0^1\frac{-\rho'(1/\nu)}{\nu}d\nu$$
$$=\int_0^1\rho\left(\frac{1}{\nu}-1\right)d\nu=\int_0^\infty\frac{\rho(t)}{(1+t)^2}dt=\lambda$$
Now we denote the greatest prime power divisor of $n\geq1$ as $Q_1(n)$. For example,
$$Q_1(24)=Q_1(2^3\times3)=2^3=8\text{ and }Q_1(72)=Q_1(2^3\times3^2)=3^2=9$$
We want to compute the expected value of $\frac{\log(Q_1(n))}{\log(n)}$ is a similar manner as before. The key idea is to note that, for large $x\gg0$ and $tx\in[0,x]$ chosen uniformly, we have
$$tx\in PS(x,x^\nu)\iff Q_1(tx)\leq x^\nu\iff\frac{\log(Q_1(tx))}{\log(tx)}\leq\frac{\nu\log(x)}{\log(tx)}\sim\nu$$
so, again, $P\left(\frac{\log(Q_1(xT))}{\log(xT)}\leq\nu\right)\sim\frac{1}{x}|PS(x,x^\nu)|$. Thus, if we had $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$ (which I'll prove in just a moment), it would immediately follow that
$$\lim_{x\to+\infty}\frac{1}{x}|PS(x,x^\nu)|=\lim_{x\to+\infty}\frac{1}{x}|S(x,x^\nu)|=F_s(\nu)=\rho(1/\nu)$$
$$\Rightarrow\lim_{x\to+\infty}\sum_{n\leq x}\frac{\log(Q_1(n))}{\log(n)}=\int_0^1\nu F_S'(\nu)d\nu=\int_0^\infty\frac{\rho(t)}{(1+t)^2}dt=\lambda$$
as desired. So it suffices to show that $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$.
First observe that every $y$-smooth number is $y$-powersmooth. Thus we have that
$$0\leq|S(x,y)\setminus PS(x,y)|=|S(x,y)|-|PS(x,y)|$$
For every prime $p\leq y$, denote $M_p(x,y)=\{p^{1+\lfloor\log_p y\rfloor}k\leq x\}$ the set of multiples of $p^{1+\lfloor\log_p y\rfloor}$ which are $\leq x$ where $p^{1+\lfloor\log_p y\rfloor}$ is the least power of $p$ that exceeds $y$.
Now observe that, if $n\in S(x,y)\setminus PS(x,y)$, then $n$ is a product of powers of primes $\leq y$ as $n=\prod_{p_i\leq y}p_i^{\epsilon_i}$ where $\epsilon_i$ can be zero and, at the same time, $\exists p_i\leq y$ such that $p_i^{\epsilon}>y\Rightarrow \epsilon_i\geq1+\lfloor\log_{p_i}y\rfloor$. But this means that $n\in S(x,y)\setminus PS(x,y)$ implies that $\exists p\leq y$ prime such that $n\in M_p(x,y)$ so $S(x,y)\setminus PS(x,y)\subseteq\bigcup_{p\leq y}M_p(x,y)$. Thus, we can bound $|S(x,y)|-|PS(x,y)|$ by
$$|S(x,y)|-|PS(x,y)|=|S(x,y)\setminus PS(x,y)|\leq\left|\bigcup_{p\leq y}M_p(x,y)\right|\leq\sum_{p\leq y}|M_p(x,y)|$$
$$=\sum_{p\leq y}\left\lfloor\frac{x}{p^{1+\lfloor\log_p y\rfloor}}\right\rfloor\leq\sum_{p\leq y}\frac{x}{p^{1+\lfloor\log_p y\rfloor}}\leq\sum_{p\leq y}\frac{x}{p^{\log_p(y)}}=\frac{x}{y}\sum_{p\leq y}1=\frac{\pi(y)}{y}x$$
$$\Rightarrow\boxed{\therefore 0\leq\frac{|S(x,y)|-|PS(x,y)|}{x}\leq\frac{\pi(y)}{y}}$$
Finally, note that $\frac{\pi(y)}{y}\sim\frac{1}{\log(y)}\to0$ by the prime number theorem which, in particular, implies $\lim\limits_{x\to+\infty}\frac{|S(x,x^\nu)|-|PS(x,x^\nu)|}{x}=0$. In fact, we only need $\frac{\pi(y)}{y}\to 0$ (i.e. the zero natural density of prime numbers) which is not that hard to prove.$\ \square$
To me they are equally hard, since I can’t prove either. But consider this: Someone doesn’t proof Goldbach directly, but proves there are at least x solutions with not necessarily distinct primes, and then proves x > 1.7. This would prove Goldbach, but not your conjecture.
On the other hand, someone might prove an upper bound for the number of Goldbach counterexamples, which isn’t good enough for a proof of Goldbach. But turns out to be good enough for your conjecture, because you only examine n/log n numbers.
Best Answer
If $a^2+b^2=c^2$ where $a,b,c$ are relatively prime positive integers, with $b$ even, then there are integers $x,y>0$ such that $a=x^2-y^2$, $b=2xy$, and $c=x^2+y^2$. In your case, $a$ and $c$ must be primes $>19$. In particular, for $a=(x-y)(x+y)$ to be prime, we must have $x-y=1$ so we have $a=2y+1$, $b=2y(y+1)$, and $c=2y^2+2y+1$. The claim is then that if $a$ and $c$ are primes greater than $19$ then $b$ has at least $4$ distinct prime factors.
Simple mod $3$ and mod $5$ considerations show that $b$ must be divisible by $3$ and $5$. If $b$ has at most $3$ distinct prime factors, then, both $y$ and $y+1$ have no prime factors greater than $5$. It can be shown that there are only ten positive integers $y$ such that both $y$ and $y+1$ have no prime factors greater than $5$, namely $y=1,2,3,4,5,8,9,15,24,80$. Testing all of these values of $y$, none yield values of $a$ and $c$ that are both primes greater than $19$, so there are no examples where $b$ has only three distinct prime factors.