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$$ 2 ^ 3 = 8 = 2 + 1 \times 6 \equiv 2 \pmod 6 \text ; $$ $$ 2 ^ 5 = 32 = 2 + 1 \times 30 \equiv 2 \pmod {30 } \text ; $$ $$ 2 ^ { 13 } = 8192 = 2 + 39 \times 210 \equiv 2 \pmod { 210 } \text ; $$ $$ 2 ^ { 61 } \equiv 2 \pmod { 30030 } \text ; $$ $$ 2 ^ { 121 } \equiv 2 \pmod { P(17) } \text ; $$ $$ 2 ^ { 361 } \equiv 2 \pmod { P(19) } \text ; $$ $$ 2 ^ { 360 q + r } \equiv {2^r} \pmod { P(19) } \text . $$ $$ 2 ^ { 3961 } \equiv 2 \pmod { P(23) } \text ; $$ 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

Stéphane Jaouen
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  • Please use mathjax. – Martund May 01 '21 at 13:33
  • It's hard to guess what you are trying to generalize. Obviously $2^n-2=2\times (2^{n-1}-1)$ factors. Is that what you are asking? – lulu May 01 '21 at 13:34
  • The number $510510$ doesn't even appear in your post. It's really not at all clear what you are talking about. – lulu May 01 '21 at 13:51
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    Let $p_n#$ be the nth primorial. For example $p_7#=2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17=510510$. You are looking for a sort of multiplicative order of $2$ modulo $p_n#$. You want to find the smallest $k$ such that $p_n#$ divides $2^k-2$. – 2'5 9'2 May 01 '21 at 16:01
  • You have made a large number of small edits to this question. Please note that this kind of editing is rather disruptive---your edits bump the question to the top of the queue (thereby pushing other questions off of the front page), and have the potential to create a "moving target" for answerers. Please decide what it is that you want to ask, ask that question, and then leave things alone. If you are having trouble with the formatting, please feel free to use the sandbox for long and coimplex posts. – Xander Henderson May 08 '21 at 14:29

2 Answers2

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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 $2^n\equiv2$.

Gerry Myerson
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    OK, but do you get the idea? Do you see how to get the right answer for $29$ from the right answer for $23$ and the smallest solution to $2^n\equiv1\bmod{29}$? and from there how to get the answer for $31$, and then for $37$, and so on, however many you want? – Gerry Myerson May 11 '21 at 10:05
  • I found that : https://oeis.org/A002326. It is very interesting: $2^6=64=1+7\times9$ for example. – Stéphane Jaouen May 11 '21 at 17:03
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    And so on... However many I want... I'VE FINALLY UNDERSTOOD. Thank you so much, Mr Myerson : 27720q+r – Stéphane Jaouen May 11 '21 at 19:05
  • If you have found my answer helpful, Stéphane, you have the option of "accepting" it, by clicking in the check mark next to it. – Gerry Myerson May 11 '21 at 23:12
  • Alors, peut-être vous avez "upvoté" ma réponse, mais vous ne l'avez pas "accepté" (pardonnez ma Français, s'il vous plaît). – Gerry Myerson May 11 '21 at 23:22
  • OK, now I see that you have accepted. Sorry if you had already done that earlier, but I don't think it showed up until just now. – Gerry Myerson May 11 '21 at 23:29
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what you can do is to write $2^n(\mod 510510) \equiv 2$ as $2^n - 510510 \left \lfloor \frac {2^n-1}{255255} \right \rfloor = 2$. Try solving this equation and maybe you can find the value of $n$. I solved it and got $n = 1$. Maybe you can get some other solutions. Hope it helps

p_square
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