[seqfan] Re: Summing partitions
Maximilian Hasler
maximilian.hasler at gmail.com
Tue Jul 10 12:40:49 CEST 2012
I'm not sure whether I understood well, but it looks to me that this
should rather be termed as a ternary relation than as addition.
If one agrees upon a (sequential) numbering of partitions, then one can imagine
* the sequence a(P) = number of ("smaller"(?)) partitions than can be
"added" to P
* the sequence b(P) = number of distinct partitions than can be
created from P in that way
* the 2-dim tables
- T(P,Q) = number of distinct partitions that can be created from P and Q
- R(P,Q) = "smallest" partition that can be created from P and Q, 0(or
-1) if none.
- S(P,Q) = "largest" partition that can be created from P and Q, 0(or
-1) if none.
where smallest/largest can be defined in several meaningful ways,
as index in any of the standard enumerations.
Maximilian
On Mon, Jul 9, 2012 at 3:49 PM, <franktaw at netscape.net> wrote:
> The discussion last month about A002219 brought to mind something
> I was playing with a while ago, but didn't get far enough to start
> submitting sequences.
>
> We can define a non-deterministic sum of two partitions as a partition
> where each part is either one of the parts of one the two addends, or
> the sum of a part from each addend, with each addend part being used
> exactly once. Alternatively, one can view this as splitting a partition into
> two, with each part either split between the two new partitions, or put
> entirely into one of them. (This is again a non-deterministic operation.)
>
> Note that a partition P can be an addend to partition Q precisely when
> P <= Q in the standard ordering.
>
> A number of sequences suggest themselves:
> * A table where T(n,k) is the number of possible sums of the n-th and
> the k-th partition (in one of the canonical orderings).
> * A table where T(n,k) is the number of ways to generate the n-th
> partition as the sum of two partitions, one of size k.
> - There are two variants of this: one with k from 0 to sum(P(n)), and
> one with k from 0 to floor(sum(P(n))/2). These will differ for
> T(2*k,k),
> the first counts ordered pairs, the second counts unordered pairs.
> - One could also look at the sums in more detail, e.g. for a sum part
> of size 3, if there are parts of size 3, 2, and 1 in both addends, the
> part could be generated as 3+0, 2+1, 1+2, or 0+3.
> * The row sums of the above: total ways to split the n-th partition.
> * Row sums of these row sums: total ways to split all partitions of n.
>
> Franklin T. Adams-Watters
>
> Incidentally, someone else may have pointed this out, but:
>>
>> This leads to the observation
>> a000041(2n) - a002219(n) >= sum_{k=0..n-1} a000041(k)
>> (and the two sides of this inequality could themselves be added to the
>
> OEIS).
>
> The right hand side is A000070 (right-shifted).
>
>
>
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