There was a question on MathOverflow which has since disappeared, that was on sums of at most $M$ B-smooth numbers. It asked several questions related to how many B-smooth numbers could be found in the interval $[N-d,N]$, where $N$ was given and $d$ was some parameter of size to be determined, but a starting guess for $d$ was $O( \log N)$. I can't refer to it, but I can refer to another question on factoring some number in an interval, namely this question.

These questions and others, and my thinking about the Jacobsthal function, lead me to present the following.
Recall *a positive integer $N$ is B-smooth if every prime factor
which divides $N$ is less than $B$,* i.e. $p \mid N
\implies p\le B$. Let me call $M$

**B-factorable**(there may be standard terminology but I do not recall it)

*if $M=PN$, where $N$ is B-smooth and $P$ is either $1$ or a prime larger than $B$*.

I suggested that for $B$ not large there were likely to be occurrences of B-factorable numbers in an interval of the form $[N-d,N]$ as above, and that if a probabilistic algorithm to find one was wanted, a simple one could be devised using trial factorization, Fermat tests, and other handy ingredients. Also, if $d$ were $O(\log N)$, the chances of success at finding one (and then later verifying the complete factorization using some computationally expensive test) were very high.

I am now convincing myself that the probability of success is in fact $1$, and this would jar (but not necessarily contradict) with the intuition I am developing about gaps between primes. So I ask for help with some questions to create some clarity.

In the following, $B=2$, and I will ask the reader to create her or his own variation with larger $B$. Also let $d = \lceil \log_2 N\rceil$ until we change it (so $d$ is essentially the number of bits used in writing $N$ in binary). First a (hopefully true) result.

1)For $N > 2$ there are $\pi(N)$ B-factorable numbers in the interval $(N/2,N]$ .

This result suggests (but does not imply) there are at least twice as many B-factorable numbers as primes in $[N-d,N]$.

2)True or false: for $N > 1$, there is at least one B-factorable number in the interval $[N -d, N]$ .

This should be true be considering how thin the complement of
B-factorable numbers is initially; even though they grow to an
eventual density of $1$, I imagine (but do not know) that they
do so slowly enough for **2)** to be true.

3)How small are the gaps between B-factorable numbers? In other words, how much can $d$ (as a function of $N$) be adjusted so as to make2)true?

Even though there are gaps between primes which are several times
larger than the average gap d, it is suspected such gaps get no
larger than $O(d^2)$, and observed gaps are much smaller. I am
working with the idea that such gaps get no larger than something
like $cd\log^3 d $ for some small constant $c < 4$. For B-factorable
numbers, a large gap between them is like asking for
$\log_2N$ many large gaps to appear in the primes at just the right places, and so that one can expect or even prove results much better than **2)**
above. Also, one can ask for $d$ as a function of $B$ and $N$, but I
prefer the simple version with $B=2$ for now.

**EDIT 2011.08.10** A computer run shows that there are no
2-factorable numbers in the interval of the $14$ numbers $[2312,2325]$. This is the smallest
of many such examples saying that **2)** is false for the defined $d$; my computer also says
$35647$ is the last of $30 $
consecutive numbers which are not 2-factorable. It may be that Greg Martin
is right and that some proofs in the literature can be expanded to
include this case. However, there is still $B>2$ to be considered.
**END EDIT 2011.08.10**

Gerhard "Ask Me About System Design" Paseman, 2011.08.10