# Consecutive integers with no large prime factors

I need answer of following Question for my study of an irrational number. (The raw problem is slightly different.)

Let $k$ be an arbitrary large positive integer, and let $A$ is a positive integer satisfying $Ce^{2^k}\le A\le Ce^{2^k}$ and have no prime factor larger than $2^k$.(Conventionally, $C$'s are certain positive constants.) Let $y_0$ be a positive integer which suffices $y_0< A$. We now think about $2^k$ products $$P_s=(y_0+As+1)(y_0+As+2)\cdots(y_0+As+k)\qquad (0\le s< 2^k).$$ Question is "Can we find some $s$'s such that $P_s$ has no prime factor larger than $2^k$?".

It is helpful not only answer for this question, but also introducing relating paper or research.(I can't discover relating research.)

If this question is nonsense or ridicurous, sorry for asking this question.

Sorry, I got some help which asserts some mistakes in my previous question.So probably, this question contains some mistakes. If you discover some of mistakes, it's helpful asserting that.

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There seems to be a misprint in your first pair of inequalities involving $A$. Should the two $C$s be different constants? – Harald Hanche-Olsen Aug 3 '12 at 17:38
I do not understand the question. $P_0$ is just $k!$ so of course has no prime factor larger than $2^k$. Could please you clarify what you actually want to ask. Also, in case you do not know it the key-word 'smooth number' could be of help. – user9072 Aug 3 '12 at 17:50
for Olsen, conventionally, $C$ can be different values at the different place. for quid, sorry, you're right. I must modify the question, but whole my study is too large to write here, so please give me some hours. p.s. I researched about smooth number little, but I don't get similar problem. – Yuta Suzuki Aug 3 '12 at 17:58

Yuta is correct, Størmer's method, or preferably, D.H. Lehmer's 1963 refinement of that method, applies to a finite set of primes.

If the set contains $k$ primes, then you have $2^k-1$ Pell equations to solve (Størmer's original method involved solving $3^k$ Pell's.

I have had a particular interest in this subject for some years, and have provided most of the high-end values at OEIS, see http://oeis.org/A002072. I intend to raise some questions arising from this work in a new posting here at mathoverflow.

But to Yuta's original question, the only way I can think of to identify smooth numbers in a particular range is to use a smart sieving algorithm. But if the interval is really large, and the number of primes is also very large, this may not be practical.

By the way, in Lehmer's paper, "Størmer" appears as "Störmer", I have yet to determine which is correct! I suspect Lehmer was probably correct.

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Why would a Norwegian spell his name like a German? Anyway, he didn't. Here's a bio in Norwegian: snl.no/.nbl_biografi/Carl_St%C3%B8rmer/utdypning and another: web.archive.org/web/20080505005456/http://www.fys.uio.no/plasma/… Even so, Viggo Brun spelled the name with an umlaut when writing about him in English. – Gene Ward Smith Jan 11 '13 at 5:24
You are absolutely right. Indeed, within Lehmer's paper, it's only in the title that an umlaut is used, within the text proper it's always "Størmer". Apologies to Wiki and anybody else I might have inadvertently offended! :-) – Jim White Jan 12 '13 at 3:00

You are asking for consecutive runs of smooth numbers. I do not have the keyboard to spell Stormer with a stroke over the o, but http://en.wikipedia.org/wiki/St%C3%B8rmer's_theorem has information for you. Unfortunately, I do not know of bounds for the largest pair of consecutive smooth numbers, but perhaps you can find out and report back here. I will say that I suspect a sequence of k such numbers will not exist once you reach numbers the size of A.

read your link to wikipedia, and I think stormer's theorem can be applied to finite sets of primes P. But in my question, the number of primes not greater than 2k tends to $\infty$ as $k\to\infty$. So perhaps stormer's theorem can not be applied to my question. Is it right? For bounds, I'm researching now. – Yuta Suzuki Aug 4 '12 at 4:01