Effect of $L_n[1/4,c]$ integer factorization on RSA-2048 Using the L-notation, integer factorization of an integer $n$ has the best known complexity of $L_n[1/3,c]$ using general number field sieve. Would discovery of an algorithm with complexity $L_n[1/4,c]$ be of any consequential significance to security of RSA-2048? If my computation is correct, for $c \approx 2$, it shows the complexity is reduced from $10^{36}$ to $10^{23}$, which seems doable with today's hardware. But I'm not confident in my computation or my assumptions about the capabilities of today's hardware.

Edit: here are my computations and assumptions: $\log(2^{2048})\approx1420$. $L_n[1/3,2]\approx \exp [2 (1420)^{1/3} (\log 1420)^{^{1 - 1/3}}]\approx 10^{36}$. $L_n[1/4,2]\approx \exp [2 (1420)^{1/4} (\log 1420)^{^{1 - 1/4}}]\approx 10^{23}$. Assuming $10^9$ possibilities can be tested every second and there are $10^6$ such processing units, it takes 3 years to crack RSA-2048: $10^{23} / (10^{9}\cdot60\cdot60\cdot24\cdot365)\approx 3 \times 10^6$.
Unfortunately, deriving actual numbers from asymptotic bounds makes little sense (after all, $10^{36}$ and $10^{23}$ are both $O(1)$). Getting an idea of the actual cost of running an algorithm requires a fine analysis of the algorithm itself; asymptotic results are not sufficient. An $L_n[1/4]$ algorithm for factoring would be a great theoretical result, but it wouldn't necessarily perform better than the current algorithms on, say, RSA-2048.

Fear the hidden constant factors.