Keith's final example, the numbers 2^X * 3^Y, is relatively easy to generate
in sorted order, using a finite amount of (arbitrarily large) numbers.
Looking at the sequence, and the ratios:
1,2,3,4,6,8,9,12,16,18,24,27,32,36,48,54,64,...
2, 3/2, 4/3, 3/2, 4/3, 9/8, 4/3, 4/3, 9/8, 4/3, 9/8, 32/27, 9/8, 4/3,  
9/8, 32/27, ...
At any given term of the sequence, there are only a few possible ratios
for computing the next term.  After 6, the ratios 2 and 3/2 are retired.
It looks like the number of active or potentially active ratios is some
small number like 4 or 6, and the new-hire and retirement rules should be
easy to figure out.  This isn't exactly a trivial sequence, but given
the bounded number of memory-numbers required, and the brevity of the
implied Next-Term program, it would be fair to call it near-trivial.

The 2,3 ratio series also has a kind of "self-similarity":  If you imagine
a read-head moving over the series, writing new ratios some distance ahead,
a ratio is either copied forward from the read-head, or split into two new
terms, whose product is the read-head term.

The situation for three primes looks interesting.  Is the set-size of
required ratios bounded?  Can they perhaps be computed "on-the-fly"
for each term, with a bounded number of arithmetic operations?

In the 2,3 case, imagine the first quadrant of the lattice of (X,Y)
integer points, and a broom of slope -log3/log2 that starts sweeping
at (0,0).  We want to work out the order of the lattice points being
swept up; and when we need to widen our broom.
The 3D problem with 2,3,5 looks harder.  Our broom morphs from a
line segment into a triangle, whose edges are determined by the ratios
of the two-prime series.

If we're willing to keep a record of the series in memory, maybe we could
have two read heads located further back, say at terms T/2 and T/3, that
offer candidates for the next term after T?  Then the series is the union
of the double of the T/2 read-head, and the triple of the T/3 read-head.
Does this work for 2,3,5?

Rich

-----
Quoting "Keith F. Lynch" <kfl@KeithLynch.net>:
> Dan Asimov <dasimov@earthlink.net> wrote:
>> I made a table of the Loeschian numbers Lo(K,L) for K >= L.
>> and then tried to *connect each one with the next largest one*
>> with a line segment on paper.
>> But it was hard to see any pattern in how they connect (below).
>> So the question boils down to:
>> Is there a concise formula for the "next largest number" in a set
>> like Lo = {Lo(K,L) | K, L in Z} ???  I.e., f(0) = 1, f(1) = 3, f(3)
>> = 4, ....
> I wouldn't think so.  There are lots of examples of two-dimensional
> integer sequences, aka tables, where there's a simple pattern to
> every row and to every column, but no pattern to all the numbers
> sorted in order.
> Examples include:
> * The multiplication table (excluding the 1s row and column).
>   The entries are the composite numbers (A002808), which are of
>   course exactly as patternless as the primes.
> * Power numbers (A001597), i.e. the union of the squares, cubes, etc.
>   Only in 2002 was it even proven that 8 and 9 are the only adjacent
>   members of the set.
> * Numbers in Pascal's triangle (other than the first two rows and
>   columns), aka the binomial coefficients (A006987).
> * Nontrivial polygonal or figurate numbers, i.e. the union of the
>   triangular numbers (other than 1 and 3), the squares (other than 1
>   and 4), the pentagonal numbers (other than 1 and 5), etc. (A090466).
> * Numbers that are a power of 2 plus a power of 3 (A004050).
>   Similarly with any other two primes.
> * Numbers that are a power of 2 times a power of 3 (A003586).
>   Similarly with any other two primes.
> Can anyone think of a nontrivial example of such a table sequence that
> *does* have a simple formula for finding the next one?  I can't.  (By
> "trivial," I think I just mean tables that contain all numbers or that
> contain only a small number of distinct numbers.)