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When I look at the count of distinct least prime factors for a range of consecutive integers, I am seeing the same minimum number appear again and again. I am wondering if this number represents the true minimum.

Consider a range of consecutive integers defined by $R(x+1,x+c) = x+1, x+2, x+3, \dots, x+c$ with $C(x+1,x+c)$ = the count of distinct least prime factors for $R(x+1,x+c)$

Let $p_k$ be the the $k$th prime which is the greatest prime less than or equal to $\left\lfloor\frac{c}{2}\right\rfloor$

I am finding that for $x \ge 1$, $c \ge 4$, the mimimum $C(x+1,x+c)$ that I can find is $k+2$.

Consider $c=61$, with $\left\lfloor\frac{61}{2}\right\rfloor = 30$ and the greatest prime less than $30$ is $p_{10}=29$. Using the Chinese Remainder Theorem, I can find $x$ where the count of distinct least prime factors is $10+2=12$. In this case, $x=210504408624479$

Here is another example. consider $c=225$, with $\left\lfloor\frac{225}{2}\right\rfloor = 112$ and the greatest prime less than $112$ is $p_{29}=109$. In this case, the minimum that I find is $29+2=31$ with $x=2082073176015230647073633038337038148767197882834487$.

I have a simple java application that allows me to find $x$ for a given $c$ with $k+2$ distinct least prime factors.

Is there a counter example that shows an $x,c$ where $C(x+1,x+c) < k+2$?

Edit: Added $x \ge 1$ and $c \ge 4$ to clarify bounds.

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I think that the statement "I am finding that for any $x$, any $c$, the mimimum $C(x+1,x+c)$" should be reformulated, clarifying that if we set the value of $x$, then $c$ is free to run on its domain and vice versa, by specifying also which is the aforementioned domain of the pair $(x,c)$ (since the closed interval of positive integers $C$ depends on both $x$ and $c$), otherwise we could simply take $c:=2$ and conclude that it does not exist any pair of consecutive integers containing $k+2$ prime numbers, since $\min(k) : p_k \in \mathbb{P} = 1$ and then $\min_k(p_k)+2 \geq 3$ if $k \in \mathbb{Z}^+$ by hypothesis.

Basically, I have constructed my counterexample here by simply taking $\overline{c} := 2$ so that $\left\lfloor\frac{\overline{c}}{2}\right\rfloor = \left\lfloor\frac{2}{2}\right\rfloor = 1$ and it follows that $[x+1, x+2]$ cannot contain more primes than the cardinality of the set {x+1, x+2} itself (i.e., two), but it is trivial to point out that $\nexists p_k \in \mathbb{P} : p_k \leq \left\lfloor\frac{\overline{c}}{2}\right\rfloor$, while for $c:=4$ we can find $k+2=3$ primes if $x=1$ (i.e., by considering the peculiar set {2, 3, 4, 5} such that $p_k=p_1=2$ is (least or) equal to $\left\lfloor\frac{4}{2}\right\rfloor$).

Anyway, I am starting to think about Rosser's Theorem (or its further improvements as Dusart's bound), maybe it would just be enough to give you a proper answer (since $p_k > k \cdot k(\log(k))$ holds for any $k$ as above).

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  • $\begingroup$ Yeah, my only two concerns arose from the proper setting of the $c$ domain (which was unspecified and $c \in \mathbb{Z}^+$ doesn't work if we assume that $p_k$ has to exist by hypothesis... on the other hand, as I wrote, $c \geq 4$ is perfectly fine, IMHO, since $2$ is a very special prime), whereas my second concern was about the opportunity to clarify better in the text (I know, it is a trivial thing) that we cannot set both $c$ and $x$ at the same time. $\endgroup$ Commented May 1, 2023 at 1:38
  • $\begingroup$ I have updated the bounds. For $𝑐=4, 𝑝_1=2=\left\lfloor\frac{4}{2}\right\rfloor$ so that $𝑘=1$ and the minimum expected is $1+2=3$ distinct least prime factors which matches $\{2,3,5\}$ in your example. Am I misunderstanding your point? $\endgroup$ Commented May 1, 2023 at 1:38
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    $\begingroup$ Thank for explaining. I agree. I should have added the bounds in the original question. :-) $\endgroup$ Commented May 1, 2023 at 1:39
  • $\begingroup$ You are welcome :-) $\endgroup$ Commented May 1, 2023 at 1:40

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