Since $j!$ is obviously divisible by $9$ if $j\ge 9$ , we have that $$\sum_{j=1}^n j!$$ is divisible by $9$ for $n\ge 8$ and is obviously greater than $9$, hence cannot be a semiprime
I do not know pfgw deep enough. There could be an option to search for factors. Primality tests cannot be parallelized , in contrary to factorizations.
the idea is simple : We check whether $b^{n-1}\equiv 1\mod n$ holds, only if it does, it makes sense to continue because otherwise, we know already that n must be composite. (Assuming that $b$ and $n$ are coprime)
This is what pfgw usually does. This is usually the start of a primality test.
And in this test, we use repeated squaring , but we have to calculate the values one after another.
Robert Frost began with the verification of a "super-Mersenne-number". I do not know whether the program still runs. He estimated to need 3 years. The number, if prime, would be a new record.
I searched for semiprimes of the form of $n^n+n!$, where $n \in \Bbb{N}$, for a range of $n \le 2 \times 10^4$ on PARI/GP and found semiprimes of the form $n^n+n!$ only for $n=2, 3, 7$.
We can write $n^n+n!$ as:
$$n^n+n!=n(n^{n-1}+(n-1)!)$$
Therefore we can also alternatively look for primes of ...
1933 digits, I do not know whether this can in general be done. I think, it is still within reach if many computers work together. With some luck, "small" factors are found and the factorization can be completed.
What do you think ? Which number of digits for RSA will be safe forever ?
I have never understood how this algorithm should work at all. Perhaps , it is based on assumed quantum phenomens that do not actually exist. And I do not mean the effects making a quantum computer possible.
I searched for semiprimes of the form of $n^n+n!$, where $n \in \Bbb{N}$, for a range of $n \le 2 \times 10^4$ on PARI/GP and found semiprimes of the form $n^n+n!$ only for $n=2, 3, 7$.
We can write $n^n+n!$ as:
$$n^n+n!=n(n^{n-1}+(n-1)!)$$
Therefore we can also alternatively look for primes of ...
