# [seqfan] Re: Proposed solution, differs from A005132 after 23 terms (was Re: seqs whose |differences| are 1, 2, 3, 4, ...

Robert Munafo mrob27 at gmail.com
Thu Mar 11 23:41:23 CET 2010

```Franklin,

Thanks for the "one-to-one" clarification. I was only using the term because
one-to-one sequence a(n) satisfying |a(n+1)-a(n)|=n'?" and I didn't see the
meaning in your other messages describing the algorithms (I found them a bit
confusing).

In my language "onto" is a preposition so I don't like to use it as a noun
or adjective. But "bijection" is great so I'll use that.

On Thu, Mar 11, 2010 at 14:56, <franktaw at netscape.net> wrote:

> Yes, "indefinite postponement of hole-filling" is a reasonable way to
> describe why there is no minimum permutation.
>
> You are misusing "one-to-one".  This means only that no value is
> repeated.  The property that every number is present is called being
> "onto".  A sequence that is both one-to-one and onto is a "permutation".
>
> Alternatively, these can be described as injections, surjections, and
> bijections, respectively.
>
>
> -----Original Message-----
> From: Robert Munafo <mrob27 at gmail.com>
>
> I am now thinking that the original notion (a unique
> lexicographically-first one-to-one sequence) does not exist, because:
>
> * For certain values of N, any solution (lexicographically first
> subsequence) of length greater than N leaves all the holes that were
> present in the length-N solution, and
>
> * For each such N, there is another higher N with the same defect, and
>
> * If the holes are postponed indefinitely for all N, all possible
> one-to-one sequences are eliminated.
>
> This seems to be supported by the pattern that emerges from the
> algorithm I posted in the previous message. There is a rather simple
> pattern of
> consecutive "clumps" of holes; the holes are:
>
> 4,5, 14-19, 44-61, 134-187, 404-565, 1214-1699, 3644-5101, 10934-15307,
> ...
>
> with the initial hole in each clump given by i[0]=4; i[n]=3i[n-1]+2 and
> the number of holes in each clump given by n[n]=2*3^n.
>
> --
Robert Munafo  --  mrob.com

```