[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...
1+11=12,
2+6=8,
3+7=10,
4+5=9
is one solution for n=4
here are two more for n=4
1+11=12
2+7=9
3+5=8
4+6=10
1+5=6
2+8=10
3+9=12
4+7=11
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
4,
whereas A002848 and A002849 only contain zeros for a(1) and a(2).
Something else is still missing.
Andrew
On Mon, Feb 8, 2010 at 11:03 PM, <franktaw at netscape.net> wrote:
> Nothing so straightforward is going to work, because the sequence
terms
> 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
like
> 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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