# [seqfan] Re: More about lex earliest cubefree 0,1 sequence

M. F. Hasler seqfan at hasler.fr
Tue May 30 13:03:34 CEST 2017

```Ed,
that is interesting.
At a first glance, your argument implies immediately the stronger statement
that the sequence cannot have any pattern of the form X,X,Z, where Z is
equal to X up to the last 3 digits of X,
i.e., Z[1 .. L-3] = X[1 ... L-3] where L = length(X).
For example, other excluded pattern are:
b,b,a ; b,b,c ;  c,c,a ; c,c,b ;  d,d,(any of a,...,e) ;
e,e,(any of a,...,k) ;  f,f,g ; f,f,h ; f,f,i ;  h,h,g ;
i,i,(any of f,...,j) ;  j,j,(any of f,...,j) ;  k,k,(any of a,...,l) ;
l,l,(anything) ;  m,m,n ;  n,n,m ; n,n,o ;
p,... oops, this one is not in lex order: should come before m!!

and so on, until: ...;  y,y,(any of m,...,z) ; z,z,(anything).

- Maximilian

On Tue, May 30, 2017 at 9:14 AM, L. Edson Jeffery <lejeffery2 at gmail.com>
wrote:

> (...) we can partition A282317 into blocks of the form (1).

It turns out that there are precisely (and coincidentally)
> twenty-six such possible distinct blocks(...):

a = {0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> b = {0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1}
> c = {0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1}
> d = {0, 0, 1, 0, 0, 1, 0, 1, 1}
> e = {0, 0, 1, 0, 0, 1, 1}
> f = {0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> g = {0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> h = {0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1}
> i = {0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1}
> j = {0, 0, 1, 0, 1, 0, 0, 1, 1}
> k = {0, 0, 1, 0, 1, 1}
> l = {0, 0, 1, 1}
> m = {0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> n = {0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1}
> o = {0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1}
> p = {0, 1, 0, 0, 1, 0, 0, 1, 1}
> q = {0, 1, 0, 0, 1, 0, 1, 1}
> r = {0, 1, 0, 0, 1, 1}
> s = {0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> t = {0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1}
> u = {0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 0, 1, 1}
> v = {0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1}
> w = {0, 1, 0, 1, 0, 0, 1, 0, 1, 1}
> x = {0, 1, 0, 1, 0, 0, 1, 1}
> y = {0, 1, 0, 1, 1}
> z = {0, 1, 1}
>
> So, A282317 can be represented as
> {a, a, b, a, a, c, a, a, d, a, a, c, a, a, d, a, a, c, a, b, ...}.
>
> (...)assume that X = {a,a,b} does appear later in A282317.

Then A282317 = {a, a, b, ..., Y, a, a, b, ...}, say, where Y is
> one of the above blocks.

However, for each possible Y, {Y,a,a,b} does contain a cube
> (because each block ends with {0,1,1})

```