This has been puzzling me for a few days now. Here is the relevant excerpt from the book mentioned in the title:
What four primes is he referring to, right at the end of the excerpt?
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2I think you must refer to the primes between $101$ and $120$. If we cross out the evens, three multiples, multiples of 5, and 7, we are left with.... $103, 107, 109$ and $113$. What I find difficult to understand is why he uses the word "actually". We were expecting to find more primes though we didn't know how many. ... And that's exactly what happened. We found four more. – fleablood Mar 17 '24 at 04:10
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I guess the "surprising" aspect is that we never needed to cross out any of the multiples of $11$ or higher. But that's because all multiples of $11$ less than $11^2$ (and not $11$ itself) would have to multiples of small numbers we did cross them out while doing the smaller numbers. – fleablood Mar 17 '24 at 04:14
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1@fleablood You might be on to something, but we should include 101, so that would be 5 more primes, not 4. – Daniel L Mar 17 '24 at 04:44
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Um, let's see. ... one, two, three.... hey! Okay... I have to admit. ... I don't really understand what is trying to be said. – fleablood Mar 17 '24 at 05:59
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1The point of that cryptic remark is that sieving out primes below $p_n$ works non only to find all primes below $,p_n^2,$ but actually - further - all of those below $p_n p_{n+1}$, as proved in the linked dupe. Here this yields four more primes - those between $11^2$ and $11\cdot 13,$ viz. $127,,131,, 137,, 139.\ \ $ – Bill Dubuque Mar 17 '24 at 07:10