Consecutive numbers with less than $k$ prime factors?

Question

Let $k$ be an integer. Consider the consecutive numbers with less than $k$ distinct prime factors.

• Are there arbitary large differences between those numbers ?

  With other words :

• Are there arbitary many consecutive numbers having at least $k$ distinct prime factors for every natural number $k$ ?

  For $k=3$, the jumping champions are :

? n=1;maxi=0;while(maxi<100,m=n;gef=0;while(gef==0,m=m+1;w=factor(m);w=component (w,1);if(length(w)<3,gef=1));d=m-n;if(d>maxi,maxi=d;print(n," ",m," ",d));n= m)

1   2   1
29   31   2
229   232   3
643   647   4
1307   1312   5
2663   2669   6
6849   6856   7
9059   9067   8
17012   17021   9
28607   28617   10
48917   48929   12
104659   104672   13
249569   249583   14
414398   414413   15
427543   427561   18
512821   512843   22
1039429   1039453   24
7319799   7319825   26
14927063   14927093   30
40252217   40252249   32
40899439   40899472   33
41544214   41544253   39
118049629   118049669   40

Answer 1

Construction of the sequence :

Denote $p_n$=nth-prime number.

Search a number $u$ with

$u\equiv 1\ \ mod\ (\ p_1...p_k\ )$

$u\equiv 2\ \ mod\ (\ p_{k+1}...p_{2k})$

...

$u\equiv m\ \ mod\ (\ p_{(m-1)k+1}...p_{mk})$

This is possible because of the chinese remainder theorem.

Then $u-m,...,u-1$ have at least $k$ distinct prime factors.