What about it?

m=128208201+j=2m(j1)!+1j(j1)!j1/820= ? \Huge { \sum_{m=1}^{2^{820}} } \large \left \lfloor \left \lfloor \frac{820}{1 + \displaystyle \sum_{j=2}^m \left \lfloor\frac{(j-1)!+1}{j} -\left \lfloor\frac{(j-1)!}{j}\right \rfloor \right \rfloor }\right \rfloor ^{1/820} \right \rfloor = \ ?

You may use this List of Primes as a reference.

Note: By definition, a=b+1b1=0 \displaystyle \sum_{a=b+1}^b 1 = 0 .

This summation is rather infamous.
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