cp's OEIS Frontend

The (unproved) Curling Number Conjecture is that any starting sequence eventually leads to a "1". The starting sequence used here extends for a total of 142 steps before reaching 1. After than it continues as A090822.

Benjamin Chaffin has found that in a certain sense this is the best of all 2^45 starting sequences of at most 44 2's and 3's.

Note that a(362) = 4. The sequence is unbounded, but a(n) = 5 is not reached until about n = 10^(10^23) - see A090822.

The next three terms are conjectured to be 68, 76 and 77.

Here xy^k means the concatenation of the words x and k copies of y.

The name "Gijswijt's sequence" is due to N. J. A. Sloane, not the author!

Fix n and suppose a(n) = k. Let len_y(n) = length of shortest y for this k and let len_x = n-1 - k*len_y(n) = corresponding length of x. A091407 and A091408 give len_y and len_x. For the subsequence when len_x = 0 see A091410 and A091411.

The first 4 occurs at a(220) (see A091409).

The first 5 appears around term 10^(10^23).

We believe that for all N >= 6, the first time N appears is at about position 2^(2^(3^(4^(5^...^(N-1))))). - N. J. A. Sloane and Allan Wilks, Mar 14 2004

For a similar formula, see p. 6 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023

In the first 100000 terms the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005] respectively. - Allan Wilks, Mar 04 2004

When k=12 is reached, say, it is treated as the number 12, not as 1,2. This is not a base-dependent sequence.

Does this sequence have a finite average? Does anyone know the exact value? - Franklin T. Adams-Watters, Jan 23 2008

Answer: Given that "...the fraction of [1's, 2's, 3's, 4's] seems to converge, to about [.287, .530, .179, .005]..." that average should be the dot product of these vectors, i.e., about 1.904. - M. F. Hasler, Jan 24 2008

Second answer: The asymptotic densities of the numbers exist, and the average is the dot product. See pp. 56-59 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023

Which is the first step with two consecutive 4's? Or the shortest run found so far between two 4's? - Sergio Pimentel, Oct 10 2016

Answer: The first x such that the x-th and (x+1)-th element are 4, is 255895648634818208370064452304769558261700170817472823... ...398081655524438021806620809813295008281436789493636144. See p. 55 of Levi van de Pol article. - Levi van de Pol, Feb 06 2023

The "curling number" k = k(S) of a string of numbers S = s(1), ..., s(m) is defined as follows. Write S as XY^k for strings X and Y (where Y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of S.

The "tail length" t(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; t(S) is the number of terms that are appended to S before a 1 is reached.

If a 1 is never reached, set t(S)=oo (the Curling Number Conjecture says this will never happen).

A sequence S in {2,3}* is called "rotten" if either of t(2S) or t(3S) (or both) is strictly less than t(S).

Example: S = 32323 has curling number k=2, so we get 323232; now k=3, so we get 3232323; now k=3, so we get 32323233; now k=2, so we get 323232332; now k=1 so we stop. We added 4 terms before reaching 1, so t(S)=4.

On the other hand, 2S = 232323 only extends to 232323321..., so t(2S)=2 which means S is rotten.

Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n-1) starting strings). The rule for extending the string is this:

To get s(i+1), write the string s(1)s(2)...s(i) as xy^k for words x and y (where y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of the sequence so far. Then s(i+1) = k if k >=2, but if k=1 you must stop (without writing down the 1).

a(n) = sum of final length of string, summed over all 2^(n-1) starting strings.

Start with any initial string of n numbers s(1), ..., s(n), all = 2 or 3 (so there are 2^n starting strings). The rule for extending the string is this:

To get s(i+1), write the string s(1)s(2)...s(i) as xy^k for words x and y (where y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of the sequence so far (k is the curling number of s(1)s(2)...s(i)). Then s(i+1) = k if k >=2, but if k=1 you must stop (without writing down the 1).

a(n) = sum of final length of string, summed over all 2^n starting strings.

See A094004 for more terms. - N. J. A. Sloane, Dec 25 2012

See A217208 or A216730 for definition of tail.

Start with an initial string S of n numbers s(1), ..., s(n), all = 2 or 3. The rule for extending the string is this:

To get s(i+1), write the current string s(1)s(2)...s(i) as XY^k for words X and Y (where Y has positive length) and k is maximized, i.e. k = the maximal number of repeating blocks at the end of the sequence so far (k is the "curling number" of the string). Then set s(i+1) = k.

The "tail length" t(S) of S is defined as follows: start with S and repeatedly append the curling number (recomputing it at each step) until a 1 is reached; t(S) is the number of terms that are appended to S before a 1 is reached. If a 1 is never reached, set t(S)=oo .

The "Curling Number Conjecture" is that if one starts with any finite string and repeatedly extends it by appending the curling number k, then eventually one must reach a 1. This has not yet been proved.

The values shown for n >= 49 are only conjectures, because certain assumptions used to cut down the search have not yet been rigorously justified. However, we believe that ALL terms shown are correct.