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Let $n>1$ be an integer and $F_n:=2^{2^n}+1$.

Result: $n$ even: $F_n\equiv17\pmod{210}$

Result: $n$ odd: $F_n\equiv47\pmod{210}$

Proof: Suppose $n$ even, $F_n - 1\equiv16\pmod{210}$. Then $2^{2^{n+2}}=({2^{2^n}})^4\equiv16^4\pmod{210}\equiv16\pmod{210}$ and $F_{n+2}\equiv17\pmod{210}$. The proof is the same if $n$ is odd.

I found that result using primoradic (see stub OEIS: https://oeis.org/wiki/Primorial_numeral_system).

That's the way I found that result see here, which I did't know before and I wonder if there are other results with $2310$, $30030,\ldots$ (primorials)

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Stéphane Jaouen
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    For $2310$, the sequence is eventually periodic of order $4$. Specifically, it goes ${5, \overline {17,257,857,457}}$. This is just a simnple matter of modular arithmetic. I assume the other moduli work similarly. – lulu May 10 '21 at 18:08
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    Update, $30030$ also yields something periodic of order $4$. Specifically ${5,\overline {17,257,5477,16637}}$. Again, just a routine modular computation. The period is $5$ for $510510$ and it is $12$ for $9699690$. I don't really see any pattern here, but it's just a few examples. – lulu May 10 '21 at 18:10
  • What interests me is not so much this result as the fact that the use of primoradic led me to it – Stéphane Jaouen May 10 '21 at 18:16
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    I don't see any connection at all. Obviously the sequence is periodic. We can write $F_n=(F_{n-1}-1)^2+1$, so if you ever repeat a value (which, obviously, you must if you are working $\pmod m$) then the sequence is periodic after that. – lulu May 10 '21 at 18:18
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    As lulu mentions, the sequence is obviously periodic. Let me merely add that the periods being small should not be surprising - if $n$ is a product of primes, then the period mod $n$ is the lcm of periods mod primes dividing $n$. Primorials are products of few small primes, so the periods couldn't be large either. – Wojowu May 10 '21 at 18:57
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    Via the method in my answer to your prior question: $,F_{n+2} = (F_n-1)^4+1,$ and $\color{#0a0}{17}$ and $\color{#c00}{47}$ are fixed points of $,f(x) = (x-1)^4+1\bmod {210},$ the complete list being $1,2,16,\color{#0a0}{17},22,26,31,37,\color{#c00}{47},52,61,71,82,86,92,101,106,107,121,122,127,131,136,142,152,157,166,176,187,191,197,206,,$ which are the solutions of $,x\equiv 1,2\pmod{!2,3,5,7}$ along with $,x\equiv 3,5\pmod{!7}.\ $ With $,z=x-1,$ the fixed points are roots of $,z^4-z,$ so $0$ and cube roots of $,1\pmod{p}\ \ $ – Bill Dubuque May 10 '21 at 19:01
  • @lulu : for 2310, ok. 17=0x210+17; 257=1x210+47; 857 = 4x210+17 ; 467 = 2x210+47. Thanks – Stéphane Jaouen May 15 '21 at 06:42

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