[seqfan] Re: The mysterious Layman sequences

[Neil Sloane]

And a second message from Remy said that he has checked the formula out to
n = 85000

That's fairly strong evidence, so can we prove it?   Let's define a
sequence S by the recurrence, with the appropriate initial values.  Then we
have explicit Binet-type formulas for S(2i) and S(2i+1).
The zeros of x^4-10*x^2+1 are 3^(1/2)-2^(1/2), -3^(1/2)+2^(1/2),
3^(1/2)+2^(1/2), -3^(1/2)-2^(1/2),
or numerically 0.317837246, -0.317837246, 3.146264370, -3.146264370.
Presumably it then follows that S(j) <= c*S(j+1), which should be easy to
check, and will prove one half of what we need.
The other half?  I'm not sure!



On Thu, May 13, 2021 at 3:48 PM Rémy Etc <remyetc9 at gmail.com> wrote:

> hi Neil,
>
> > the quite simple recurrence a(n) = 10*a(n-2) - a(n-4) ?  So I want a
> > numerical verification, as far out as is convenient, before I accept it!
> this formula is true at least up to n=5000.
>
> best regards,
>
> Rémy
>
> Le jeu. 13 mai 2021 à 20:03, Neil Sloane <njasloane at gmail.com> a écrit :
> >
> > Dear Seq Fans,
> > Rick Mabry has made some interesting comments about a family of sequences
> > called "Ratio-Determined Insertion Sequences" contributed many years ago
> by
> > the late John Layman. I propose to do some major editing of them, but
> first
> > I would like to know the answer to a specific question. It needs some
> > computing help, and the answer will determine how I edit
> >  these sequences.
> >
> > So here is the question, based on the test case A085376.
> > This involves a certain fraction, which is c := 31*37/(2^5*5^10) =
> 0.36704
> > exactly.
> >
> > Given c, we construct a triangle of numbers, as follows. The first row is
> > (1,1).
> > Given row k, we get the next row by repeating row k, except that between
> > every 2 adjacent terms x and y in that row, we insert their sum x+y iff y
> > <= c*x.
> >
> > The rows converge to a sequence, which is A085376.
> >
> > There is a conjecture there that this sequence satisfies a(n) =
> 10*a(n-2) -
> > a(n-4) for n >=5, with initial terms 1, 3, 11, 30
> >
> > Assuming that no one can prove this conjecture, I propose to replace the
> > existing definition with the recurrence, and state it as a conjecture
> that
> > it agrees with the sequence produced by the insertion rule. (I won't take
> > the space here to explain why I want to do this.)
> >
> > But first I would like to be sure that the conjecture is true.  So could
> > someone please generate a lot of terms using the present definition (the
> > insertion rule), and check that the recurrence is satisfied?
> >
> > What worries me is that c = 31*37/(2^5*5^10) = 0.36704 exactly is a
> rather
> > strange constant, and why should  the resulting sequence be explained by
> > the quite simple recurrence a(n) = 10*a(n-2) - a(n-4) ?  So I want a
> > numerical verification, as far out as is convenient, before I accept it!
> >
> > There are two text files written by Layman attached to the sequence. I
> have
> > not studied them carefully, maybe they contain the answer to the
> question.
> >
> > Best regards
> > Neil
> >
> > Neil J. A. Sloane, President, OEIS Foundation.
> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> > Phone: 732 828 6098; home page: http://NeilSloane.com
> > Email: njasloane at gmail.com
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
>