Prev: Why does 2^n != 1 (mod n) is true for every n > 1?
Next: Check this nice fact out!
From: Jim Ferry on 6 Nov 2009 15:32
Consider the following sequence

1
11
12
2111
1321
11213111
1331112112
211221133211

where each line is formed by "look and say" from the
previous, and then reversed. (See
http://en.wikipedia.org/wiki/Look-and-say_sequence
for the standard look-and-say sequence.)

This sequence is A022481 in Sloane. (Equivalently,
and perhaps more naturally, the sequence of the
reverses of the above is A006711).

The lengths of these strings is A022476 in Sloane:

1, 2, 2, 4, 4, 8, 10, 12, 14, 20, 24, 30, 38, 54, 66, 92,
120, 160, 210, 284, 378, 490, 632, 852, 1134, ....

Whereas the asymptotic growth rate of the standard
look-and-say sequence is

1.303577269034296391257...

(an algebraic integer of degree 71), numerical
experimentation indicates the asymptotic growth
rate of the reverse sequence is approximately

1.327.

As in the standard case, this growth rate is
independent of the initial condition.

Is anyone aware of an analysis of this reverse
case? E.g., what are the "elements" and what
is the minimal polynomial for the asymptotic
growth rate?

I computed 56 terms before running out of
memory in Mathematica, which was not
enough to find a linear recurrence (which
would yield the minimal polynomial).
From: Jim Ferry on 7 Nov 2009 20:40
On Nov 6, 3:32 pm, Jim Ferry <corkleb...(a)hotmail.com> wrote:
> Consider the following sequence
>
> 1
> 11
> 12
> 2111
> 1321
> 11213111
> 1331112112
> 211221133211
>
> where each line is formed by "look and say" from the
> previous, and then reversed.  (Seehttp://en.wikipedia.org/wiki/Look-and-say_sequence
> for the standard look-and-say sequence.)
>
> This sequence is A022481 in Sloane.  (Equivalently,
> and perhaps more naturally, the sequence of the
> reverses of the above is A006711).
>
> The lengths of these strings is A022476 in Sloane:
>
> 1, 2, 2, 4, 4, 8, 10, 12, 14, 20, 24, 30, 38, 54, 66, 92,
> 120, 160, 210, 284, 378, 490, 632, 852, 1134, ....
>
> Whereas the asymptotic growth rate of the standard
> look-and-say sequence is
>
> 1.303577269034296391257...
>
> (an algebraic integer of degree 71), numerical
> experimentation indicates the asymptotic growth
> rate of the reverse sequence is approximately
>
> 1.327.
>
> As in the standard case, this growth rate is
> independent of the initial condition.
>
> Is anyone aware of an analysis of this reverse
> case?  E.g., what are the "elements" and what
> is the minimal polynomial for the asymptotic
> growth rate?
>
> I computed 56 terms before running out of
> memory in Mathematica, which was not
> enough to find a linear recurrence (which
> would yield the minimal polynomial).

I forgot to mention that a Maple user could
probably figure this out without much effort
using Ekhad's and Zeilberger's package HORTON:

http://www.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/horton.html

Unfortunately, I'm an adherent of another
denomination (the one with the infallible
supreme leader).

From: cbrown on 8 Nov 2009 01:04
On Nov 6, 12:32 pm, Jim Ferry <corkleb...(a)hotmail.com> wrote:
> Consider the following sequence
>
> 1
> 11
> 12
> 2111
> 1321
> 11213111
> 1331112112
> 211221133211
>
> where each line is formed by "look and say" from the
> previous, and then reversed.  (Seehttp://en.wikipedia.org/wiki/Look-and-say_sequence
> for the standard look-and-say sequence.)
>
> This sequence is A022481 in Sloane.  (Equivalently,
> and perhaps more naturally, the sequence of the
> reverses of the above is A006711).
>

I think it disappointing that the reverses do not have Sloan sequence
number 184220A ;). Or at least A101122031405060718.

Cheers - Chas
 | 
Pages: 1
Prev: Why does 2^n != 1 (mod n) is true for every n > 1?
Next: Check this nice fact out!