your opinion indeed

Christian G.Bower bowerc at usa.net
Fri Dec 18 06:43:32 CET 1998


This is a response to a math-fun posting.
I'm CC'ing to seqfan because it's sequence related.

Dave Wilson wrote:

> Once I understood what Wouter were doing, I could reproduce his
> sequence.  It is indeed a very curious sequence.

> Let S_n have the generating function PROD(k = 1 to n, 1+x^k).  As n
> increases, S_n closes in termwise on S_infinity = A000009, the
> partitions into distinct integers = partitions into odd integers.

> Now look at the sequence

>     T_n = STRIP(n, A000009 - S_n)

> where STRIP(n, S) removes the leading n zeroes from sequence S.  Then,
> as n approaches infinity, T_n approaches the sequence T_infinity =

> 0 1 2 3 5 7 10 14 19 25 33 43 55 70 88 110 137 169 207 253 307 371 447
> 536 640 762 904 1069 1261 1483 1739 2035 2375 2765 3213 3725 4310 4978
> 5738 6602 7584 8697 9957 11383 12993 14809 16857 19161 21751 24661

> T_infinity turns out to be the running sum of A000009!

> Is there an easy explanation?  Does an analogous thing happen if we
> replace Q with P, the partition numbers?

Yes to both questions.
S_n represents partitions into distinct integers from {1,2,...,n}
S_infinity - S_n represents partitions into distinct integers with
at least one integer >=n.

What do we get for S_infinity(55) - S_50(55)?
Each partition has exactly one part >=50. Divide those into that
part being 50, 51, 52, 53, 54, and 55.

For 50 we get for the remainder Q(5).
For 51 we get for the remainder Q(4).
...
For 55 we get for the remainder Q(0)=1.
So we get Q(0)+Q(1)+...+Q(5) the running sum.

Yes, an analogous thing happens if we replace Q with P.
We get the running sum of the partition numbers A000041 (which is in the
EIS as A000070).

Interesting things happen with similar sequences.

Take ordered partitions.
I.e. S_n has generating function 1/(1-x-x^2-...-x^n).
So S_infinity = 2^(n-1)
T_infinity = (n+2)*2^(n-1) A001792

Take cycle ordered partitions (i.e. ordered in a circle).
T_infinity = 2^n.
This is also true for asymmetric cycle ordered partitions.


Is anyone on the list familiar with combinatorial species
(aka Joyal theory)?

The pattern suggests the following formula:

Take any species A.

Let S_n be the unlabeled enumeration of A(E_{1,2,...,n}).
E_{1,2,...,n} is the species of sets whose cardinality is
in {1,2,...,n}.

Then T_infinity is the unlabeled enumeration of A'(E+)/(1-X)
E+ is the species of nonempty sets.
X is the species of singletons.

Christian


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