# [seqfan] Re: Orderly Numbers (and related sequences) submitted

Alonso Del Arte alonso.delarte at gmail.com
Sat Nov 7 22:13:52 CET 2009

```I think I'm understanding it better now. I remember some time ago having
worked out that only powers of primes have a prime number of divisors, and
that all odd primes (except 3) are congruent to +/- 1 mod 6 (these are
well-established facts of number theory, in any case), so that with an
understanding of the basic concept of orderly numbers, the rest of your
explanation follows logically.

As for the primality of orderly numbers with multiple values of k, at this
point I can only say that there is no composite below 10000 with more than
one value for k. So it vastly speeds up computer calculations in that range
to limit the search to primes (e.g., Prime[Range] instead of
Range).

Al

On Fri, Nov 6, 2009 at 8:03 PM, Andrew Weimholt
<andrew.weimholt at gmail.com>wrote:

> Only prime orderly numbers appear to have multiple values of k, so for
> the following remarks, let's assume that is the case. The following is
> a more generalized version of Tony's remarks about the non-existence
> of an orderly number with precisely 6 values of k....
>
> The only numbers with a prime number of divisors are powers of primes.
> If q is prime, then q^x has x+1 divisors. So to find an orderly number
> with x values of k, where x+1 is prime, we must look for a prime of
> the form q^x+2. All primes greater than 3 are congurent to +/- 1 mod
> 6, so if x is even (which it is if x+1 is an odd prime), then q>3 ==>
> q^x is conguent to 1 mod 6, which means q^x + 2 is divisible by 3 and
> therefore not prime.
> Therefore the only primes of the form q^x + 2, where x>0 is even, are
> those with q=3.
> So the only orderly number with precisely 2 values of k is 11, because
> 11 = 3^2 + 2
> The only orderly number with precisely 4 values of k is 83, because 83 =
> 3^4 + 2
> 3^6 + 2 is NOT prime, so there is no orderly number with precisely 6
> values of k.
> 3^10 + 2 is prime (59051), so it is the only orderly number with
> precisely 10 values of k
> 3^12 + 2 is NOT prime, so there is no orderly number with preciseuly
> 12 values of k.
>
> For precisely 13 values of k, since 13+1 is not prime, you can also
> look for primes of the form q*r^6+2 where q and r are prime.
> You can try
> 3*r^6 + 2
> q*3^6 + 2
> or
> q*r^6 + 2, where q == -1 mod 6
>
> Andrew
>
> On 11/6/09, Alonso Del Arte <alonso.delarte at gmail.com> wrote:
> > That makes sense. Something tells me I should be able to extend that same
> >  logic for 6 to the cases of 10, 12 and 13 with relative ease, but I
> can't,
> >  as if a cloud was blocking the obvious answer from my sight. (The stress
> >  of getting the word out about my Symphony on eBay is getting to me).
> Despite
> >  the cloud, a sequence of smallest orderly numbers with precisely n
> values of
> >  k is starting to emerge:
> >
> >  11, 17, 83, 47, [nonexistent], 107, 227, 569, ?, 317, ?, ?, 2027, 947,
> ...
> >
> >  The subsequence 107, 227, 569 yields a tantalizing result in A117877.
> >
> >  Also, are all these numbers primes? And is 11 the only orderly number
> with
> >  two values of k?
> >
> >
> >  On Fri, Nov 6, 2009 at 2:21 AM, T. D. Noe <noe at sspectra.com> wrote:
> >
> >  > >For what it's worth, I think A167408 ought to have the keyword
> "nice." It
> >  > is
> >  > >amazing that for all the years the OEIS has existed, no one had
> thought of
> >  > >sending this sequence in before. Also, the more I think about the
> concept,
> >  > >the more tantalizing questions that arise. For instance, what is the
> >  > >smallest orderly number with precisely six values of k? (My
> calculations
> >  > >suggest it's either a very large number or does not exist).
> >  >
> >  > I agree that it should be labelled "nice".
> >  >
> >  > The smallest orderly number with 6 values of k is a prime p such that
> p-2
> >  > has 6 divisors greater than 1.  This means p-2 is the q^6 for some
> prime q.
> >  > So just look for the first prime of the form q^6+2.  However, there is
> no
> >  > such prime because for prime q>3, q^6+2 is divisible by 3.
> >  >
> >  > BTW, I recently added a comment to the sequence showing that there are
> an
> >  > infinite number of orderly numbers that are powers of primes.
> >  >
> >  > Tony
> >  >
> >  >
> >  >
> >  > _______________________________________________
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> >
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