$$ 2 ^ 3 = 8 = 2 + 1 \times 6 \equiv 2 \pmod 6 \text ; $$ $$ 2 ^ 5 = 32 = 2 + 1 \times 30 \equiv 2 \pmod {30 } \text ; $$ $$ 2 ^ { 13 } = 8192 = 2 + 39 \times 210 \equiv 2 \pmod { 210 } \text ; $$ $$ 2 ^ { 61 } \equiv 2 \pmod { 30030 } \text ; $$ $$ 2 ^ { 121 } \equiv 2 \pmod { P(17) } \text ; $$ $$ 2 ^ { 361 } \equiv 2 \pmod { P(19) } \text ; $$ $$ 2 ^ { 360 q + r } \equiv {2^r} \pmod { P(19) } \text . $$ $$ 2 ^ { 3961 } \equiv 2 \pmod { P(23) } \text ; $$ Here are the results I'm trying to generalize. For example, I'm asking for an integer $ n $ such as $ 2 ^ n \equiv 2 \pmod {P(29)} $. I'm interested in primorial number system (primoradic, see stub OEIS). When you write $ 2 ^ n $ in primorial number system, periodicities appear, which can be proven using petit théorème de Fermat for example.
Using OEIS notations, $$2=(0:0:0:0:0:1:0)$$$$4=(0:0:0:0:0:2:0)$$$$8=(0:0:0:0:1:1:0)$$$$16=(0:0:0:0:2:2:0)$$$$32=(0:0:0:1:0:1:0)$$ You already see the periodicity 2 for the second coefficient : $$1-2-1-2-1-2-... $$For the third coefficient, the periodicity is 4$$(0-0-1-2-0-0-1-2-)$$ For the fourth coefficient, the periodicity is 12$$(0-0-0-0-1-2-4-3-6-5-3-...)$$
There are other results, for example: $10^{145}\equiv{10}\pmod{19\#}$ so that in primoradic $$10^{145}=(......:16:8:0:0:0:0:0:1:2:0)$$ So is $$10^{1585}\equiv{10}\pmod{23\#}$$ 1584=144x11.Every integer written base 2 can be written in primoradic.
It is at least interesting to minimize work on the computer(I use a spreadsheet) when you want to write in primoradic $ 2 ^ n $ when $ n $ is big. [with my spreadsheet, I have explored until $2^{16384}$, which is interesting because the 14-th Fermat's number is $2^{16384}+1$.See my question given in link about Fermat's numbers.] Thanks for your comments and your critics... sorry for my very bad English (French). Cordialement, Stéphane Jaouen.
P.S. : $$64=(0:0:0:2:0:2:0)$$ $$128=(0:0:0:4:1:1:0)$$ $$256=(0:0:1:1:2:2:0)$$ $$512=(0:0:2:3:0:1:0)$$ $$1024=(0:0:4:6:0:2:0)$$ $$2048=(0:0:9:5:1:1:0)$$ $$4096=(0:1:8:3:2:2:0)$$ $$8192=(0:3:6:0:0:1:0)$$ $$2^{31}=(9:14:7:9:3:6:4:1:1:0)$$ [using notations stub OEIS concerning Primorial number system] https://oeis.org/wiki/Primorial_numeral_system
As @alex points out in a comment, you want $n$ such that $2^n\equiv2\bmod p$ for all small primes $p$. You have $2^{3961}\equiv2\bmod p$ for all primes $p\le29$, so $2^{3960q+1}\equiv2\bmod p$ for those primes. $2^5\equiv1\bmod{31}$, so $2^{3960}\equiv1\bmod{31}$, so $2^{3961}\equiv2\bmod p$ for all primes $p\le31$.
I think it's easier to work with $2^n\equiv1\bmod p$. For each $p$ up to however high you want to go, the smallest $n$ satisfying this congruence is tabulated at the OEIS, and then you can combine those numbers to find the smallest value for primorial moduli. Then just add one to the answer to get $n$ with $2^n\equiv2$.