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I posted a version of this to stackexchange and got 12 up-votes and no answers in somewhat more than a day. Someone in a comment construed it as asking for a lot of novel research including figuring out how to make the statement precise. But the actual question was whether someone had already done those things.

I was looking at this set of prime factorizations: $$ \begin{align} 1100 & = 2\times2\times5\times5\times11 \\ 1101 & =3\times 367 \\ 1102 & =2\times19\times29 \\ 1103 & =1103 \\ 1104 & = 2\times2\times2\times2\times 3\times23 \\ 1105 & = 5\times13\times17 \\ 1106 & = 2\times7\times79 \end{align} $$ I noticed that all of the first 10 prime numbers, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, appear within the factorizations of only seven consecutive integers (and within only six of them). (The next prime number, 31, has its multiples as far from 1100 as it could hope to get them (1085 and 1116).) So no nearby number could hope to be divisible by 29 or 23, nor even by 7 for suitably adjusted values of "nearby". Consequently when you're factoring nearby numbers, you're deprived of those small primes as potential factors by which they might be divisible. So nearby numbers, for lack of small primes that could divide them, must be divisible by large primes. And accordingly, not far away, we find $1099=7\times157$ (157 doesn't show up so often---only once every 157 steps---that you'd usually expect to find it so close by) and likewise 1098 is divisible by 61, 1008 by 277, 1096 by 137, 1112 by 139, 1095 by 73, 1094 by 547, etc.; and 1097 and 1109 are themselves prime.

So if an unusually large number of small primes occur unusually close together as factors, then an unusually large number of large primes must also be in the neighborhood.

Are there known precise results quantifying this phenomenon?

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This looks interesting.(btw, what are the seven consecutive integers.) But Michael, maybe can you try to formulate more precisely what is the phenomenon you ask about? I suppose the firt question is:Given k and a much larger r, what is the smallest integer m such that the k consecutive integers from m contains as factors all first r primes. And the second is if it is always the case that integers nearby to m will have unusually large factors. I guess that integers nearby m will not have unusual larger factors given their number of digits and given the few small primes they must miss. –  Gil Kalai Oct 19 '11 at 23:52
    
Hint: Jacobsthal's function. Gerhard "Ask Me About Jacobsthal's Function" Paseman, 2011.10.19 –  Gerhard Paseman Oct 20 '11 at 0:22
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OK, I waited long enough to be asked. Many of the issues arising when you consider divisibility criteria among integers in a small interval arise when considering prime gaps or in looking at integers coprime to a given number. Trying to find good bounds on Jacobsthal's function involved me in situations similar to what Michael Hardy describes. Other than possibly references to literature on prime constellations and admissible sets, I am not sure quite what is being asked, so I am not yet giving an answer. Gerhard "The Questions Are So Close" Paseman, 2011.10.19 –  Gerhard Paseman Oct 20 '11 at 1:17
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Are you sure that the effect you are seeing isn't just caused by the small number 1100 you are looking at? 29^2 = 841 is already almost of the same size (more than half). There is only one other prime, 31, whose square is less than 1100. What are the numbers around $1100\dots 1106$ supposed to be divisible by? The only choices are "big" primes or powers of "really" small primes. Among the big primes you give are the examples 61 and 73, which are not impressively big. –  j.p. Oct 20 '11 at 9:56
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You didn't mention the numbers 1107 and 1110, which are divisible by 37 rsp. 41. In total you are looking at 12 numbers around $1100\dots 1106$: from 1094 to 1099 and from 1107 until 1112. Two of them are primes, another two have prime factors between 30 and 60 and the remaining 8 have prime factors bigger than 60. I'd say that a random number is divisible by a prime between 30 and 500 with probability 43.3% and by a prime between 60 and 500 by 33.0%. As 33.0% is 77.3% of 43.3%, which is quite close to the observed 8 out of 10, I'd say that nothing special is going on here. Sorry. –  j.p. Oct 20 '11 at 10:01

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