[seqfan] Re: definition of A002848
Richard Guy
rkg at cpsc.ucalgary.ca
Tue Feb 9 21:12:33 CET 2010
Indeed, but I'm as confused, perhaps even more so,
than the rest of you. Neil must have constructed
this (these?) from my papers. For considerably
more detail, see Richard Nowakowski's 1975 Calgary
MSc thesis on the Langford-Skolem problem. This
sh'd've been pursued further and publicized ...
I don't understand A002848 or A002849. No
example is given in either case. My guess is
that it shd read Number of sol'ns of x+y=z
from {1,2,...,3m} instead of ...,n}. But
that still doesn't make clear what is going on.
Another good place to start is my Unity of
Combinatorics lecture:
Richard K.~Guy, The unity of combinatorics,
{\it Proc.\ 25th Iran.\ Math.\ Conf., Tehran}
(1994), {\it Math.\ Appl.}, {\bf329} 129--159,
Kluwer Acad.\ Publ., Dordrecht, 1995;
{\it MR} {\bf96k}:05001.
I'm not clear what Neil had in mind, but
perhaps he himself will remember. The problem
was to partition the numbers from 1 to 3m
into m triples which each satisfy the
equation x + y = z. There are numerous
connexions with other combinatorial objects.
As someone has pointed out, there are
solutions only if m == 0 or 1 mod 4.
For m = 1, the unique solution is 1 + 2 = 3.
For m = 4, there are 8 solutions:
1 5 6 1 5 6 2 5 7 1 6 7
2 8 10 3 7 10 3 6 9 4 5 9
4 7 11 2 9 11 1 10 11 3 8 11
3 9 12 4 8 12 4 8 12 2 10 12
2 4 6 2 6 8 3 4 7 3 5 8
1 9 10 4 5 9 1 8 9 2 7 9
3 8 11 3 7 10 5 6 11 4 6 10
5 7 12 1 11 12 2 10 12 1 11 12
I don't think that we found all solutions
for any larger values of m. Computers
and computation have advanced quite a bit
in the last 36 years, and some of you can
probably push this quite a way. Nowakowski
gives (some) solutions for m = 4k and
4k+1 for all k.
But I don't think this answers your questions.
Come in, Neil! R.
On Tue, 9 Feb 2010, David Newman wrote:
> Pardon my ignorance, but is there an option of writing to Guy and asking
> him?
>
> On Tue, Feb 9, 2010 at 12:09 PM, <franktaw at netscape.net> wrote:
>
>> 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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