The problem:
How to find this sum?
$$\sum_{a=0}^{\infty}\frac{1}{(\frac{(a(p-1)+b)!}{p^{q_a}} \mod p) \times p^q}$$
where:
$p \in Primes$
$b \in \mathbb{N}\quad$, $0 \leq b \leq p-2$, but not defined.
$q_a$ is the greatest power of $p$ who divides the term $(a(p-1)+b)!$
Details:
$b$ refers to the congruence modulus $p-1$, so $0 \leq b \leq p-2$.
$a(p-1)+b$ for differents $a$'s and $b$'s we can express all natural numbers,
so the summation looks like the sequence of $e$ but a bit modified.
$q_a$ can also be expressed as: $q_a=\lfloor{\frac{(a(p-1)+b)}{p}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^2}}\rfloor+\lfloor{\frac{(a(p-1)+b)}{p^3}}\rfloor+\ldots$
$\frac{(a(p-1)+b)!}{p^{q_a}} \bmod p$ is just the remainder of the division by $p$.
Expandind the sum we have:
$\frac{1}{(\frac{b!}{p^{q_0}} \mod p)p^{q_0}}+\frac{1}{(\frac{(p-1+b)!}{p^{q_1}} \mod p)p^{q_1}}+\frac{1}{(\frac{(2p-2+b)!}{p^{q_2}} \mod p)p^{q_2}}+\frac{1}{(\frac{(3p-3+b)!}{p^{q_3}} \mod p)p^{q_3}}+\ldots$
just to clarify, to $p=2$ and $b=0$ we have:
$\frac{1}{(0! \mod 2)2^0}+\frac{1}{(1! \mod 2)2^0}+\frac{1}{(\frac{2!}{2^1} \mod 2)2^1}+\frac{1}{(\frac{3!}{2^1} \mod 2)2^1}++\frac{1}{(\frac{4!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{5!}{2^3} \mod 2)2^3}+\frac{1}{(\frac{6!}{2^4} \mod 2)2^4}\ldots$
The motivation to find this sum is analyze some properties of the prime numbers using the expansion of $e$.
