[seqfan] Re: definition of A002848

franktaw at netscape.net franktaw at netscape.net
Tue Feb 9 18:09:32 CET 2010

Possibly (I haven't really checked, but the pattern is right) A002849 
is the number of partitions of a subset of 1..n into triples X+Y=Z, 
with the maximum possible number of such triples.  A002848 would then 
be the number of such partitions that include n in one of the triples.

If this is correct, I would argue that A002849(1) and A002849(2) should 
both be 1, representing the empty partition.

Franklin T. Adams-Watters

-----Original Message-----
From: Andrew Weimholt <andrew.weimholt at gmail.com>

A002849 has the same definition, but different terms. One or both
definitions are wrong.

If you google "unity of combinatorics", the third result is a google 
books page.
R. K. Guy briefly discusses the X+Y=Z problem and the X+Y=2Z problem.

Given the numbers 1 to 3n, the goal is to partition them into triples 
such that
each triple is a solution to X+Y=Z. For example...


is one solution for n=4

here are two more for n=4



Unfortunately, this still doesn't shed much light on A002848 and
A002849, as the terms
do not seem to match.

As Guy notes, the X+Y=Z problem only has solutions for n == 0 or 1 mod 
whereas A002848 and A002849 only contain zeros for a(1) and a(2).

Something else is still missing.


On Mon, Feb 8, 2010 at 11:03 PM,  <franktaw at netscape.net> wrote:
> Nothing so straightforward is going to work, because the sequence 
> as they exist are not monotonic.  The first reference, "R. K. Guy,
> ``Sedlacek's Conjecture on Disjoint Solutions of x+y= z,'' in Proc.
> Conf. Number Theory. Pullman, WA, 1971, pp. 221-223.", looks to me 
> the place to start; unfortunately, I don't have access to this.
> Franklin T. Adams-Watters
> -----Original Message-----
> From: Rainer Rosenthal <r.rosenthal at web.de>
> Max Alekseyev wrote:
>> Does anybody understand the definition of A002848 and how it produces
>> the listed terms?
> Just a guess, to be verified:  "number of solutions of x+y+z=n"
> a(3)=1 solutions: 1+1+1=3
> a(4)=1 solutions: 1+1+2=4
> a(5)=2 solutions: 1+1+3=5 and 1+2+2=5
> a(6)=2 solutions: 1+1+4=6 and 1+2+3=6
> oops ... there is 2+2+2=6 as well.
> What a pity :-(
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