Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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
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
@Gil: The phenomenon is that if lots of small primes are in a very short interval, then lots of unusually large primes, by comparison to the average sizes of prime factors of numbers of that size, are not far away. (And the seven consecutive integers are 1100 through 1106. One of those is prime and not divisible by any of the first 10 primes, but I didn't just say six integers because I wanted to be clear that they are close to each other: within an interval consisting of just seven consecutive integers. –  Michael Hardy Oct 20 '11 at 1:50
.....when I say lots of large primes are not far away, I mean of course (I hope my posting made this clear) that some of their multiples are not far away; they are not far away among the expressions on the rights sides of "$=$" above. –  Michael Hardy Oct 20 '11 at 2:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.