[seqfan] Re: Questions on A230492
David Wilson
davidwwilson at comcast.net
Fri Dec 27 19:22:27 CET 2013
This means that if there are square elements of A230492 other than 209^2,
they are either of the form p^2*q^2 and extremely large, or of some other
prime signature. So I was wrong in my initial guess, it is possible that
209^2 could be the only prime element.
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Lars
> Blomberg
> Sent: Friday, December 27, 2013 2:00 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Re: Questions on A230492
>
> I have found no more prime pairs for n <= 21000.
> In fact, no prime occurs for p after n=990.
>
> /Lars Blomberg
>
> -----Ursprungligt meddelande-----
> From: David Wilson
> Sent: Thursday, December 26, 2013 4:56 PM
> To: 'Sequence Fanatics Discussion list'
> Subject: [seqfan] Re: Questions on A230492
>
> I think it's likely that there are other squares in this sequence.
>
> 209^2 = 11^2*19^2 has prime signature p^2*q^2, so perhaps we should look
> for other elements with this signature.
>
> p^2*q^2 will be in A230492 if it satisfies the critical equality
>
> [1] 1 + p + q + p^2 + p*q = q^2.
>
> Solving [1] empirically for integer (p, q), we get solutions
>
> (1, 3)
> (11, 19)
> (79, 129)
> (545, 883)
> (3739, 6051)
> (25631, 41473)
> (175681, 284259)
> (1204139, 1948339)
> (8253295, 13354113)
> (56568929, 91530451)
> ...
>
> The theory of quadratic Diophantine equations tells us that these
solutions
> will satisfy a set of linear recurrences. In this case we find that
>
> p(n) = 7p(n-1) - p(n-2) + 3.
> q(n) = 7q(n-1) - q(n-2) - 1.
>
> or, if you like pure recurrences
>
> p(n) = 8p(n-1) - 8p(n-2) + p(n-3).
> q(n) = 8q(n-1) - 8q(n-2) + q(n-3).
>
> With these recurrences, we can quickly generate (p, q) pairs satisfying
[1].
>
> With the tools at my disposal, I was only able to test q < 2^64, and in
that
> range (p, q) = (11, 19) was the only pair of primes I found. However, the
> divisor patterns do not seem to rule out a larger pair of primes (p, q)
for
> which p^2*q^2 would be in A230492. Perhaps a seqfan with power tools
> could find such a pair.
>
> > -----Original Message-----
> > From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of
> > Vladimir Shevelev
> > Sent: Wednesday, December 25, 2013 1:11 PM
> > To: Sequence Fanatics Discussion list
> > Subject: [seqfan] Re: Questions on A230492
> >
> > Dear Michel and SeqFans,
> >
> > I would like to pay attention to one more question.
> > Using b-file of A230492 containing 250 terms, I found among them only
> > one
> > square: a(19)=43681=209^2. Does exist the next one?
> > It is clear, that square of a prime is not in the sequence.
> > In case of a semiprime p*q, it is easy to see that p and q should be
> dstinct.
> > Let p<q. Then, it is easy to see that, a key equality is
> q^2-1=(p+1)*(p+q).
> > For example, for p=11, q=19, we have the equality 360=12*30.
> > Therefore, (11*19)^2 is in the sequence.
> >
> >
> > Regards,
> > Vladimir
> >
> > ________________________________________
> > From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of
> > michel.marcus at free.fr [michel.marcus at free.fr]
> > Sent: 19 December 2013 18:16
> > To: seqfan at list.seqfan.eu
> > Subject: [seqfan] Questions on A230492
> >
> > Hi SeqFans,
> >
> > I have recently extended A230492 to 250 terms. Unless mistaken, it
> > still
> verify
> > the property that even terms are perfect.
> > I have also checked that the 5 first perfect numbers (A000396) belong
> > to
> this
> > sequence (the 6th one is taking some time).
> >
> > Is it possible to prove that perfect numbers are in A230492, but no
> > other even numbers ?
> > And what could be said about the odd terms ?
> >
> > I have searched the OEIS and found only 2 other sequences with the
> > same
> > property:
> > A034897 Hyperperfect numbers.
> > A225417 Composite numbers which contain their sum of aliquot parts as
a
> > substring.
> > Are there others ?
> >
> > Thank you for your help.
> > Michel
> >
> >
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> >
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> >
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