Has this problem been raised ?

[Robert Israel]

More generally, given any n and any positive integers a and b,
if m > 1/((a+b+1)^(1/n) - (a+b)^(1/n))
there will be i,j,k with floor((i/m)^n) = a,
floor((j/m)^n) = b, floor((k/m)^n) = a + b.

Cheers,
Robert

On Sun, 2 Mar 2008, Max Alekseyev wrote:

> What is the problem?
> To find a solution (i,j,k,m,n) to this equation? In integers?
> Anyway, one needs just to take m large enough so that every floor in
> the equation turns into 0.
>
> Regards,
> Max
>
> On Sun, Mar 2, 2008 at 7:24 PM, Alexander Povolotsky <apovolot at gmail.com> wrote:
>> Hi,
>>
>>  floor((i/m)^n) + floor((k/m)^n) = floor((l/m)^n)
>>
>>  n > 2, m > 1, i > 0,  k > 0, l > 0, i != k
>>
>>  Has this problem been raised and resolved (one way or another) ?
>>
>>  Thanks,
>>  Best Regards,
>>  Alexander R. Povolotsky
>>
>



This seemed like a sequence which could change with time,

Neil

%S A000001 1, 2, 3, 4, 5, 6, 7, 8, 9, 11
Conjecture has been verified for nontrivial monotone graph properties.




Jonathan Post said:

I'm guessing that it's not worth submitting the emirp subsequence of
Lekraj Beedassy's A138044, which is itself a subsequence of A138043.


Base-dependent sequences - especially those
involving silly names like emirps - are usually
not so interesting.  





I had not realized that Jonathan Post wrote to the
whole mailing list about this!  [Surely it would
have been enough to write to me!]

I replied to him saying that the sequence
was rejected because it was time-dependent

> I'm wondering if this one got lost in the mail, or is still being
> edited? 
> %S A000001 1, 2, 3, 4, 5, 6, 7, 8, 9, 11
> %N A000001 The number of vertices for which the Evasiveness
... has not yet been verified





It is well known that one can construct a square-free string of infinite
length using the digits 0, 1, 2 (see A085794). I was interested to know
how much "slack" there was in the system, so I've been experimenting with
the same construction when specific substrings are disallowed.

Things start to get interesting already with substrings of length 3: every
permutation of 012 must appear in an infinite squarefree string, but if
a digit is repeated (eg 010) it seems possible to disallow the substring
and still to have an infinite string. It seems possible even to manage
without two such substrings, if they are chosen correctly.

Of particular interest is the pair (010, 212). I cannot be certain that
this yields an infinite string, since I am relying only on observation of
the strings yielded in a backtracking construction - I have seen it reach
to a length of 2000 digits, well past the point I usually take as an
indication of infiniteness, but it reaches it much more slowly than other
more obviously infinite selections, with much more extensive backtracking;
at this length my simplistic code already starts to become quite slow.

Here is the interesting bit: if I additionally disallow the n-digit prefix
of the apparently infinite string yielded by that pair *for any n*, it
appears to be enough to make an infinite squarefree string impossible.

That is to say, it appears that if (010, 212) does indeed allow an infinite
additional substring that appears in that string, of any length, is enough
to make the maximum length string finite.

Alternatively expressed: it appears that any infinite squarefree string that
avoids the substrings 010 and 212 must necessarily contain every other
allowable substring, of any length.

Since this is all reliant on observation only, I would be interested if
anybody is aware of this, or can suggest proof or disproof of any of
the facts I think I have observed.

Some numbers:

Here are the apparent first 400 digits of the string for (010, 212):

0120210121020120210201210120210121020121
0120210201202101210201202102012101202102
0120210121020121012021012102012021020121
0120210121020121012021020120210121020121
0120210121020120210201210120210201202101
2102012021020121012021012102012101202102
0120210121020120210201210120210201202101
2102012101202101210201202102012101202102
0120210121020120210201210120210121020121
0120210201202101210201210120210121020120

Here are the maximum lengths of a squarefree string when the n-digit prefix
of the above string is disallowed in addition to 010 and 212 (offset=1):
.. and I have verified that this is finite at least to n=160 (with max length
approx 699). Note the distinct patterns of a(2^n) = 2a(2^(n-1))+2, and
a(2^n-1) = a(2^(n-1)) + 2^n-1; if these patterns can be proved to hold, that
is itself enough to prove the infiniteness of the (010, 212) string.

Here are the number of squarefree strings for given n (offset=1):

Of course these are valid sequences only if the string for (010, 212) is
indeed infinite, and does indeed start as shown above.

Hugo