I'm interested in the smallest solution to a family of "excluded congruences." To be precise, let $p_1 < \ldots< p_k$ be a sequence of primes and consider the constraints $$ x \not\equiv a_1 \pmod {p_1}\\ \vdots\\ x \not\equiv a_k \pmod {p_k}\,, $$ where $0 \leq a_i < p_i$ for each $i$. I'd like a reasonably general bound on the smallest (non-negative, say) $x$ that solves this. (It would be nice to have a bound on x as a function of $p_1$ and $p_k$, the smallest and largest primes in the sequence.) Of course $x$ need be no larger than $\prod p_i$ by Chinese remaindering, and $x$ can clearly be taken to be no more than $k$ if $k < p_1$.

It seems clear that something follows from Brun's sieve-like machinery: Along these lines, I'd be grateful for a reference to a version of the sieve that yields something of this form.

Added 12/14/2012: A 1924 theorem of Rademacher provides a nontrivial bound on the question. The statement below is from D. Charles's Sieve Methods book, pg. 45. Let $p_1, \ldots, p_r$ be primes, and let $a_i < p_i$, $b_i < p_i$ be non-negative integers for $1 \leq i \leq r$. Let $D > 1$ be an integer with $\gcd(D, p_i) = 1$ for each $i$, $1 \leq i \leq r$, and $Λ$ is an integer, $0 < Λ < D$, such that $\gcd(Λ,D) = 1$.

Let $P(D; x; p_1, a_1, b_1; p_2, a_2, b_2; \ldots; p_r, a_r, b_r)$ denote the cardinality of the set $$\{ n \leq x \mid n \equiv Λ \;(\bmod\; D); n \not\equiv a_i\; (\bmod\; {p_i}) ; n \not\equiv b_i\; (\bmod \;{p_i})\}\,.$$ If $p_1 < p_2 < \cdots < p_r$ and $p_i > 2$, then $$P(D;x; p_1,a_1,b_1;\ldots ; p_r;a_r;b_r) > \frac{Cx}{D \ln^2 p_r} - C'p_r^{7.9}$$ where $C$ and $C'$ are positive constants.

So, the presence of this theorem somewhat muddies my question. Can things be significantly simplified in my case (where I have a single excluded congruence class modulo each prime, and no other explicit constraints)?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.