a question about A134865

[David W. Wilson]

I have a sneaking suspicion that A134865 might in fact be finite.

Let's look for the smallest element of A134865 with divisor 7.

Suppose 7 | a(n). Its divisors are 1,7 with 1,2 divisors.
The smallest numbers with 1,2 divisors are 1,2, and so
lcm(7,1,2) = 14 | a(n).

The divisors of 14 are 1,2,7,14 with 1,2,4 divisors.
The smallest numbers with 1,2,4 divisors are 1,2,6,
and so lcm(14,1,2,6) = 42 | a(n).

Repeating this argument, we find that 168, 840, and finally 2520
divide a(n). So if 7 | a(n), we must have 2520 | a(n).  We are
lucky at this point, 2520 is actually an element of the sequence.

Following this same process with small primes gives

2
3->6
5->10->30->120
7->14->42->168->840->2520
11->22->66->264->1320->27720->332640
13->26->78->312->1560->32760->4324320
17->34->102->408->2040->42840->5654880->73513440->1470268800
   ->27935107200->83805321600->1927522396800->13492656777600

The unexpected thing here is that our final two values have prime
divisor 23. In other words, 17 | a(n) => 23 | a(n). The process,
(luckily for it) ends here, since 13492656777600 is an element of
the sequence.

Starting with 23, we wind up at the same place by a different path:

23->46->138->552->2760->57960->7650720->99459360->33816182400
   ->1927522396800->13492656777600

Starting with 29, we get

29->58->174->696->3480->73080->9646560->125405280->42637795200
   ->2430354326400->391287046550400->11738611396512000
   ->4002866486210592000->?

I don't know what follows, but at this point 31 has been drawn
in. My gut feeling is that sufficiently large starting prime
p will always pull in a larger prime, meaning that no element
of a is divisible by p. Indeed, I suspect that any sufficiently
large starting value for this process will pull in arbitrarily
large primes, which would imply that the sequence is finite.

> -----Original Message-----
> From: Don Reble [mailto:djr at nk.ca]
> Sent: Saturday, May 17, 2008 2:00 PM
> To: njas at research.att.com; jhbubby at mindspring.com
> Cc: seqfan at ext.jussieu.fr
> Subject: Re: a question about A134865
> 
> > There are 2 numbers, 332640 and 352800, that I want to check and see
> > if they qualify for A134865.
> 
> %S A134865
> 1,2,4,6,12,24,36,48,120,240,360,720,2520,5040,7560,10080,15120,20160,
> %T A134865
> 45360,50400,100800,332640,352800,665280,705600,4324320,8648640,
> %U A134865 17297280,21621600,43243200
> 
> --
> Don Reble  djr at nk.ca
> 
> 
> --
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