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Apr 27, 2017 at 10:17 vote accept CommunityBot
Apr 24, 2017 at 18:43 answer added Uwe Stroinski timeline score: 2
Apr 21, 2017 at 18:45 comment added Joe Silverman @GerhardPaseman As far as I know, there are no significant theorems saying that most primitive divisors appear with low multiplicity, although in practice that's surely the case. I think that in general, for questions of this sort, it's best to first consider the simpler case of the sequence $2^n-1$, or $A^n-1$. And just the fact we don't know that $2^p-1\not\equiv0\pmod{p^2}$ for infinitely many primes indicates our lack of understanding for these sorts of questions. The OP might have better luck with an average question, i.e., bound growth of $\sum_{n<X}P_n$.
Apr 21, 2017 at 16:50 comment added Gerhard Paseman Also, if n itself has a small prime factor k, then you can reduce the exponent (n-1) by subtracting something near n/k. This is significant for n which are not coprime to 6. (Maybe even replace the right hand side by $\alpha^{\phi(n)}$.) Gerhard "Use Primitive Facts For Optimization" Paseman, 2017.04.21.
Apr 21, 2017 at 16:42 comment added Gerhard Paseman @JoeSilverman, do we know that primitive divisors occur with low multiplicity? Otherwise a large prime power could be a factor. On a brighter note, Granville showed that for some sequences, primitive divisors occur to an odd power for n not too small. Gerhard "Can't Be Good Without Bad" Paseman, 2017.04.21.
Apr 21, 2017 at 13:15 comment added Joe Silverman You can get a tiny improvement by noting that if $p_1,\ldots,p_k$ are the primitive divisors in increasing order, then first (as you note) $p_1\ge n-1$, and then $p_2\ge p_1+2=n+1$, etc. So $(n-1)(n+1)\cdots(n+2k-3)\le\alpha^{n-1}$.
Apr 21, 2017 at 12:30 history asked user40023 CC BY-SA 3.0