curious eta-GFs and two news seqs

[Joerg Arndt]

Yes, number of partitions of 2n into even square-free parts is equal to 
number of partitions
of n into odd square-free parts.

Trivially, doubling every odd square-free part gives an even 
square-free number, and these
will sum to 2n.

Conversely, no even square-free part can be divisible by 4, so half of 
such a part must be
odd.  So halving each part in a partition of 2n into distinct even 
square-free parts will
give a partition of n into distinct odd square-free parts.

Franklin T. Adams-Watters

-----Original Message-----
From: Joerg Arndt <arndt at jjj.de>

addendum:

...

\\ not in OEIS:  partitions of 2*n into distinct even sqrfree parts
v=Vec( prod(n=1,N, 1+moebius(2*n)^2*x^(n) ) );
v2=Vec( prod(n=1,L2, 
(etaplus(x^(2*(2*n-1)^2))/etaplus(x^((2*n-1)^2)))^(-moebius(2*n-1)))
);
(v-v2)==0 \\ OK up to 3000 terms
\\ ***** NOTE: apparently SAME AS partitions into distinct odd sqrfree 
parts
v=Vec( prod(n=1,N, 1+moebius(2*n-1)^2*x^(2*n-1) ) );
(v-v2)==0 \\ OK up to 3000 terms




ja> From seqfan-owner at ext.jussieu.fr  Fri May 16 21:17:22 2008
ja> Date: Fri, 16 May 2008 21:05:42 +0200
ja> From: Joerg Arndt <arndt at jjj.de>
ja> To: seqfan at ext.jussieu.fr
ja> Subject: Re: curious eta-GFs and two news seqs
ja> ...
ja> 
ja> \\ not in OEIS:  partitions of 2*n into distinct even sqrfree parts
ja> v=Vec( prod(n=1,N, 1+moebius(2*n)^2*x^(n) ) );
ja> v2=Vec( prod(n=1,L2, (etaplus(x^(2*(2*n-1)^2))/etaplus(x^((2*n-1)^2)))^(-moebius(2*n-1))) );
ja> (v-v2)==0 \\ OK up to 3000 terms
ja> \\ ***** NOTE: apparently SAME AS partitions into distinct odd sqrfree parts
ja> v=Vec( prod(n=1,N, 1+moebius(2*n-1)^2*x^(2*n-1) ) );
ja> (v-v2)==0 \\ OK up to 3000 terms

I do not understand this. If 2*n is any even number, the count
of partitions into distinct, odd and squarefree elements
(distinct odd members of A005117, members of A056911) and the count of partitions
into distinct, even and squarefree elements (distinct even
member of A005117, distinct members of A039956) is not the same.

For n=3 the partitions of 6 are
6=6
6=1+5
(o.k., one partition each)

For n=4 the partitions of 8 are
8=2+6
8=3+5=1+7
(not o.k, because there is 1 partition in even, 2 partitions in odd parts)

For n=5 the partitions of 10 are
10=10
10=3+7
(o.k, one partition each)

For n=6 the partitions of 12 are
12=2+10
12=5+7=1+11
(not o.k, one partition in even squarefree, two partitions in odd squarefree parts)
...

For n=9 the partitions of 18 are
18=2+6+10
18=7+11=5+13=3+15=1+17
(not o.k, one partition in even squarefree, four partitions in odd squarefree parts)
...

Related existing sequences: A087188 and A073576.

Richard




Dear Richard,

Just a note to thank you for many excellent comments.

In an hour I will have got caught up
with /your/comments through last Saturday!

I am planning some major changes to the way edits
are handled by the associate editors,
that should make life easier for everyone (especially me).

But this is going to take a few weeks.

I will copy this to the seqfan list - Please note: 
there will be changes that will make things easier,
as suggested by several friends!

Best regards

Neil

PS Richard, about A019484: you pointed out that the 
present description was bogus.  You also said:
The sequence does not seem to follow any 3-term recurrence at all!

You were right on both counts, of course.  Superseeker found
the right g.f., which is

New version of OEIS in about an hour