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The look-and-say sequence starting with $S_1=1$ is,

$$S_n = 1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211,\dots$$

If $L_n$ is the number of digits of the $n$th term then,

$$\lim_{n\to\infty} \frac{L_{n+1}}{L_n}=\lambda\tag{1}$$

where $\lambda = 1.303577\dots$ is an algebraic number of degree 71.

(A discussion of this sequence and the constant can be found in Nathaniel Johnston's blog.)

Questions:

  1. Is the limit (1) an artifact of base-10? If the terms of $S_n$ are translated into some other base-$n$ representation, say in binary, will the limit still be $\lambda$?
  2. I find it intriguing that $\lambda$ happens to be an algebraic number of such high degree. Without using obvious examples like $x^n=1$, or n-nacci constants $x^n(2-x)=1$, or contrived ones $\phi^{1/m}, \lambda^{1/n}$ etc, is there a constant that appears in a non-trivial number-theoretic context (like a limiting ratio, etc) that is algebraic and has a higher degree than $\lambda$?
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Regarding 1, have you tried generating a bunch of terms of the sequence in other bases to see what happens? –  Matthew Conroy Dec 29 '13 at 1:05

3 Answers 3

up vote 9 down vote accepted

First-of-all, the key to the analysis of the look-and-say-sequence is the transition matrix $T$ of the "elements of audio-active decay", as John H. Conway has called them. This matrix can be used to give a closed form for the number of digits and asymptotic results are found by considering the eigenvalues of $T$. That is: look-and-say is like Fibonacci, just with 92 instead of 2. So:

1) The look-and-say-sequence does not depend much on the chosen base, with one important restriction: Much of the regularity in the behavior relies on the fact that no other number than $1,2,3$ can appear in the sequence. However, this is still true for any base $\geq 4$.

So for any base $\geq 4$ the matrix $T$ and hence $\lambda$ will be the same. For bases $2$ and $3$, the recursion (if there is any!) will probably be much different, so a priori there is no reason why $\lambda_2$ and $\lambda_{10}$ should be related.

2) The constant $\lambda$ is algebraic because it's an eigenvalue of an integer matrix. So if you take a recurrence relation where $T$ has a minimal polynomial with degree higher than 71, you should be able to give a lambda with higher degree. But I guess that this construction is not what you meant as it isn't as natural as the "audioactive decay".


Edit: Regarding examples for high order algebraic numbers: The onset of a $7$-cycle of the logistic map has degree 114 as stated in http://mathworld.wolfram.com/AlgebraicNumber.html

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1. Hm, I think I didn't phrase my question as I intended. In general, given any integer sequence such that if we define $L_n$ as the number of digits of the nth term in base 10 and $$\lim_{n\to\infty} \frac{L_{n+1}}{L_n}=\alpha\neq1$$ and $\alpha$ is algebraic, will the limit also be algebraic if we convert the terms into binary (or other base-m representation)? 2. Ah, the n-cycles of the logistic map. For $n=3$, I in fact gave to Weisstein the expression in terms of the silver constant –  Tito Piezas III Dec 29 '13 at 17:10

There is no "base" involved in the look-and-say sequence. As defined, each term of the sequence is not an integer, but a finite sequence of integers. It is usually written as a single integer because in the most common cases of interest, only digits 1,2,3 are involved so there is no possibility of confusion. But Conway's original theorem allows for arbitrary integers in the initial sequence, and according to the cosmological theorem these larger integers ultimately persist in one of two forms, which he called "isotopes of Np and Pu", with limiting ratio 0.

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1) You can use a similar approach to Nathaniel Johnston's Blog for the binary look-and-say sequence: 1,11,101,111011,11110101,... You can do it differently I suppose, but I got those 10 subsequences: 111011,11110101,100110,11100,10110,1110,111100,1001100,11110,101100. You get a transformationmatrix with charactaristical polynomial $$ x^4(x+1)(x-1)^2(x^3-x^2-1) $$ Which has a non-trivial positive real root $$ x\approx 1.465571232 $$

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That's very interesting. I recognized the cubic as the minimal polynomial of $x = \frac{e^{\pi i/24}}{\sqrt{2}}\frac{\eta\,(\tau)}{\eta\,(2\tau)} = 1.4655\dots$ where $\tau=\frac{1+\sqrt{-d}}{2}$ and $d = 31$ which has class number $h(-d)=3$. The only other fundamental d with $h(-d)=3$ such that x is a cubic is $d = 23$ with $P(x)=x^3-x-1=0$ and the real root is the plastic constant. Do you know if there is a $n$-nary look-and-say sequence such that the characteristic polynomial has $P(x)$ as a factor? –  Tito Piezas III Dec 30 '13 at 17:06

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