Timeline for answer to Maximal length of consecutive numbers that are multiples of a set of primes can be estimated on the size of the set of primes alone by GH from MO
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| when toggle format | what | by | license | comment | |
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| Jan 26 at 0:04 | comment | added | philipxy | @user1868607 New questions belong in new posts, not comments. | |
| Jan 25 at 18:46 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Jan 24 at 14:58 | comment | added | Eduardo Rodríguez Golvano | With this interpretation it is the same as the original question. For any arithmetic progression $a_i$, $0 \leq i \leq K-2$ such that every $a_i$ is divisible by at least one prime from $P$ there exists an integer $N$ such that $N+i$ is divisible by the same primes from P as $a_i$, $0 \leq i \leq K-2$, assuming d is coprime to all the primes in P. | |
| Jan 24 at 3:40 | comment | added | fedja | @GHfromMO "It is not clear to me what the OP really means by..." Erm... I see only one meaningful interpretation (after all simple reductions): Let $k$ be a positive integer. Estimate from above the smallest $K=K(k)$ such that for any positive integer $d$ and any set $S$ of $k$ primes not dividing $d$, every arithmetic progression of integers with difference $d$ and of length $K$ contains an element not divisible by any prime in $S$. Do you see any alternative? Anyway, if the OP disagrees with this interpretation, he/she is welcome to clarify :-) | |
| Jan 24 at 3:21 | comment | added | GH from MO | @fedja It is not clear to me what the OP really means by "if one considers sequences of numbers at distance d where no natural number of P dividing some number in the sequence divides d" A precisely formulated question (with set-theoretic notation and quantors) as in the original post would be helpful, and a new post would serve this the best (comments tend to be more condensed than main posts). Also, it is discouraged to use the comments section for lengthy discussions, follow-up questions, etc. At any rate, I will stop here. | |
| Jan 24 at 3:12 | comment | added | fedja | @GHfromMO Why? This thread is by no means overloaded yet, and it would be nice to have everything in one place instead of multiple cross-references. Do you have a full answer BTW? (The estimate should not depend on the progression difference $d$ either). What is clear to me now is that we care about divisor sets consisting of primes only, but getting a clean power bound (if it holds) by completely elementary means in this generality still seems somewhat challenging) | |
| Jan 24 at 0:22 | comment | added | GH from MO | @fedja Please encourage the OP to ask the arithmetic progression version in a separate post rather than discussing the problem here. | |
| Jan 23 at 23:43 | comment | added | fedja | Do you care about the exact power, or not too much? | |
| Jan 23 at 23:35 | comment | added | user1868607 | @fedja thanks for your feedback, a polynomial bound in $k$ is particularly attractive in my context | |
| Jan 23 at 22:27 | comment | added | fedja | The absolutely trivial bound for the initial problem is $N\le 2k\cdot 4^{2k}$. Indeed, for any $P$, the product of primes less than $P$ is $M\le 4^P$ (Chebushev), so considering only numbers congruent to $1$ modulo $M$, we get $N/M\le k+(N/M)\sum_p 1/p\le k+(N/M)(k/P)$ where $p$ runs over the first $k$ primes after $P$. Now just take $P=2k$. This bound trivially extends to the setting of arbitrary numbers $\ge 2$ (replace each number by some its prime divisor) and to the arithmetic progression setting when the divisor set consists of primes. Of course, one can do much better in that case too. | |
| Jan 23 at 20:41 | comment | added | GH from MO | @user1868607 I suggest that you post ask your new question in a separate post, with precise notation etc. as in the present post. | |
| Jan 23 at 19:28 | vote | accept | user1868607 | ||
| Jan 23 at 19:21 | vote | accept | user1868607 | ||
| Jan 23 at 19:21 | |||||
| Jan 23 at 19:12 | comment | added | GH from MO | Let $P$ in an arbitrary finite subset of $\mathbb{Z}_{\geq 2}$, and $P'$ is the set of prime divisors of the elements of $P$, then $\Phi(P)\subset\Phi(P')$. Hence $\Lambda(P)\leq\Lambda(P')$, and for $\Lambda(P')$ you can apply Iwaniec's bound. BTW, if you like my answer, please accept it officially (so that it turns green). Thanks in advance! | |
| Jan 23 at 19:00 | vote | accept | user1868607 | ||
| Jan 23 at 19:01 | |||||
| Jan 23 at 17:31 | history | answered | GH from MO | CC BY-SA 4.0 |