Timeline for Increasing integral sequence of intermediate growth which is periodic modulo almost all primes
Current License: CC BY-SA 3.0
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Mar 12 at 23:11 | history | edited | GH from MO |
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Jul 22, 2015 at 19:15 | comment | added | Greg Martin | Another construction, that I suspect you will want to exclude, is to take an exponentially growing pan-periodic sequence (like $2^n$) and replace the $n$th term by its least nonnegative residue modulo the product of all the primes up to $\sqrt n$, say. This sequence grows slower than $(e+\epsilon)^{\sqrt n}$ and, modulo any fixed prime, is congruent to the original sequence after finitely many terms. | |
Jul 22, 2015 at 19:11 | history | edited | Greg Martin | CC BY-SA 3.0 |
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Jul 22, 2015 at 14:35 | comment | added | Roland Bacher | joro, thanks for your remark. You are right, I consider this to be trivial. | |
Jul 22, 2015 at 12:00 | comment | added | joro | Interleaving periodic sequences is again periodic. I suppose this is trivial for you: $a(2n)=poly_1(n),a(2n+1)=poly_2(n)$? | |
Jul 22, 2015 at 9:16 | history | edited | Roland Bacher | CC BY-SA 3.0 |
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Jul 22, 2015 at 8:59 | history | asked | Roland Bacher | CC BY-SA 3.0 |