Problems & Puzzles: Problems

Problem 20.- Divisors (II) K consecutive numbers with the same number of divisors.

Exist consecutive numbers with the same number of divisors. By example: d(243)=d(244)=6.

Moreover, it's conjectured (Erdös) that it should exist at least one set of K consecutive numbers, all with the same number of divisors, d, for any K=>2.

Now, let's talk about the least set for every K. In the following table are shown the known results, according to the R. K. Guy's "Unsolved Problems in Number Theory", 2nd edition (and my own very little search)

 K Least set d Source 2 2à3 (and the only set) 2 Anonymous † 3 33 à 35 4 R. K. Guy 4 242 à 245 6 R. K. Guy 5 11605 à 11609 (least) 40311 à 40315 (not least) 8 8 C. Rivera R. K. Guy 6 28374 à 28379 8 C. Rivera 7 171893 à 171899 8 S. Vandemergel, 1987 8 See below one value by Ivo Düntsch, not known if the smallest 48 Düntsch & Eggleton, 1990 9 17796126877482329126044à 17796126877482329126052 “presumably not the smallest of this kind", says Guy 48 Düntsch & Eggleton, 1990

Note: R. K. Guy says that "In 1990, Ivo Düntsch & Roger Eggleton discovered several such sequences of 7 numbers, two of 8 and one of 9, each with 48 divisors…"

Would you like to find:
a)  Three more cases for K = 7
b)  The smallest cases for K = 8 & 9?

(Ref.  2, B18, p. 73)

c) ADDENDUM (20/07/99):Sequences of consecutive numbers all with distinct quantity of divisors. I have found a sequence of 12 numbers starting at 16023339 and a sequence of 13 numbers starting at 16475964. Can you get larger sequences of this type?

Solution

Jud McCranie, informs (19/07/99) that "Sequence A006558 gives the smallest cases for k <= 7.  It gives the value you have for k=9 and it lists a value for k=8.  It is not known if these two are the smallest ones.  I think I found some more of length 7, I'll have to look it up.  I pointed out some time ago that 11605 was the correct value for k=5."

In another e-mail sent the 20/07/99, Jud sent the solution to a):
K = 7 for (180965 to 180971), (647381 to 647387) and (1039493 to 1039499). In all ranges it happened that d = 8. The first range was calculated also by Enoch Haga. Both believe that some hidden arithmetic properties of this kind of sequences must exist and need to be discovered to guide/execute an efficient search of larger ones.

In particular, Jud says:

"...If a number [of the sequence of K consecutive numbers] has a factor of 2^2, the number of divisors has a factor of 3.That means that for the numbers near it, it is necessary for their number of divisors must contain a factor of 3 in order to be the same.  A factor of 3 is obtained when a prime factor is raised to the power such as 2, 5, or 8, etc.  So for there to be a sequence of more than 7 terms with the same number of divisors, each of them must contain a prime factor to the power of 2, 5, 8, ...etc; since a sequence of 8 numbers must contain one with the factor 2^2.  Also, there are limits on the number of factors imposed. Suppose the number that has 2^2 as a factor has no other primes to powers higher than 1.  Then each of its neighbors must be of a special form - they must have 1 and only 1 prime factor to the power of 2, and the same number of prime factors; or 1 prime factor to the power of 5, and one fewer distinct primes; etc.  This is a pretty rare condition for 8 numbers in a row to meet."

***

The August 30, 2002, Ivo Düntsch, wrote by email this:

For K = 8, a starting number is

6,213,958,594,795,772,370,845.

I think we had others, but would have to check my old files. I have put our 1989 manuscript on the web - look at

On my request for telling if this number is the smallest for K=8, he replied:

I don't know.

***

On Set. 2005, Jud Mc Cranie wrote:

In 2002, I found the smallest value for k=8.  It is 1043710445721 and there are 24 divisors.

***

Bilgin Ali and Bruno Mishutka (bruno_mishutka@hotmail.com) wrote on Dec. 30, 2006:

Asunto: new information for your Problem 20, Divisors II

Mensaje: hi,

we have announced results for K=10,11 and 12 on the NMBRTHRY mailing list:

http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0612&L=nmbrthry&T=0&X=7EA06379063761515D&Y=bruno_mishutka%40hotmail.com&P=2853

Here is the text of our announcement:

Date: Fri, 29 Dec 2006 15:49:46 -0500
Sender: Number Theory List <NMBRTHRY@LISTSERV.NODAK.EDU>
From: Bruno Mishutka <bruno_mishutka@hotmail.com>
Subject: extending sequence in Guy's problem B18

hi everyone,

Bilgin Ali and I would like to announce that we have extended the work of
Duntsch and Eggleton as reported in Problem B18 in Guy's "Unsolved Problems
in Number Theory".

In particular, we have found the following.

(1) A sequence of 10 consecutive integers starting at 14366256627859031643,
each with 24 divisors.

(2) A sequence of 11 consecutive integers starting at
193729158984658237901148, each with 48 divisors.

(3) A sequence of 12 consecutive integers starting at
1284696910355238430481207644, each with 24 divisors.

We would also like to point out that Duntsch and Eggleton proved a
conjecture of Erdos (cited in Problem B18 of Guy's book) that there exist
arbitrarily long sequences of consecutive integers with the same number of
divisors, subject to Schinzel's Hypothesis H (see
http://www.cosc.brocku.ca/~duentsch/archive/equidiv.pdf ).

We would also like to point out one small error in their manuscript.

The entry for D(6,5) in Table 2 of their manuscript should read
10,093,613,546,512,321 not 10,093,613,546,512,121.

We have submitted our findings above to sequence A006558 of Sloane's Online
Encyclopedia of Integer Sequences (
http://www.research.att.com/~njas/sequences/Seis.html).

==================

Note that these are not known to be the least numbers with these properties. And in all likelihood, there are smaller examples.

Also note that Jud McCranie has found the smallest values for 8 and 9; they are 1043710445721 and 2197379769820, respectively. See http://www.research.att.com/~njas/sequences/A006558 as a reference.

***

Days after J. K. Andersen wrote:

Problem 20 only asks for up to 9 numbers, and the below doesn't say whether it's minimal, but maybe it's of interest anyway. I have checked the results.

Bruno Mishutka and Bilgin Ali found:
K = 10: A start is 14366256627859031643 with 24 divisors.
K = 11: A start is 193729158984658237901148 with 48 divisors.
K = 12: A start is 1284696910355238430481207644 with 24 divisors.

***

On June 12, 2015, Letsko Vladimir wrote:

I've found some numbers connected with your problem 20

A run of 10 consecutive positive integers each having exactly 12 divisors starts at 1545780028345667311380575449

A run of 13 consecutive positive integers each having exactly 24 divisors starts at 58032555961853414629544105797569

***

On Feb 28, 2017, Letsko Vladimir wrote again:

Some new results:

A run of 11 consecutive positive integers each having exactly 12 divisors starts at
25194373506124406718179246045;

A run of 14 consecutive positive integers each having exactly 24 divisors starts at 25335305376270095455498383578391968;

A run of 15 consecutive positive integers each having exactly 24 divisors starts at
176395563698184262497866317860996946591.

***

On March 17, 2017, Letsko wrote:

Hello Carlos!

New records again:

A run of 12 consecutive positive integers each having exactly 12 divisors starts at
3239053063539139069494921945;

A run of 16 consecutive positive integers each having exactly 24 divisors starts at 37981337212463143311694743672867136611416.

***

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