A072507

[franktaw at netscape.net]

This sequence (Start of n consecutive integers with n divisors, or 0 if 
no such number exists.) raises some interesting questions.

The first is to complete the proof that the remainder of the sequence 
is all zeros.  The comments in the sequence show that it is zero for 
every n greater than 2 except for 12, 24, and 120.

The next question is, what is the length, for a given n, of the longest 
sequence of integers with exactly n divisors?  Clearly, the value is 1 
for any odd n.  a(4) = 3; we have shown a(4) < 4, and A006558 gives 33 
as the start of a sequence of 3 numbers with 4 divisors.  Similarly, 
a(6) is 4 or 5, since 242 is the start of 4 consecutive numbers with 6 
divisors.  a(8) = 7; 171893 being the exemplar.  We can that a(10) < 8, 
since 8 consecutive values must have a multiple of 4.  A lower bound 
for 10 is 2, starting at 2511.  For 12, we can get 5 consecutive at 
204323; 24 is an upper bound (the only multiples of 24 with exactly 12 
divisors are 72 and 96).  The case n=14 is similar to 10, with 29888 
starting a pair.  For 16, 7 is again an upper bound, and there are 5 
starting at 178086.

So we have:
1 1
2 2
3 1
4 3
5 1
6 4 or 5
7 1
8 7
9 1
10 2 to 7
11 1
12 5 to 24
13 1
14 2 to 7
15 1
16 5 to 7

If the case n=6 can be resolved, I will go ahead and submit the 
sequence.

The final question is, what is the largest m such that there are 
infinitely many sequences of m consecutive integers with exactly n 
divisors?  The upper bounds above still apply, of course.  a(2) = 1, 
since 2 and 3 are the only consecutive primes.  Beyond that, I have no 
idea how to proceed.


Franklin T. Adams-Watters