[seqfan] Re: a partition sequence

[franktaw at netscape.net]

You're missing:

2 1
2

and

2 1
1
1

Franklin T. Adams-Watters

-----Original Message-----
From: Christopher Hunt gribble <cgribble263 at btinternet.com>

Dave,

For n = 5, I get the following subset of plane partitions when I apply 
my
understanding of your restrictions:

5

4 1

3 2

3
1
1

2
2
1

2
1
1
1

1
1
1
1
1

According to your calculations, there should be 2 more.  What am I 
missing?

Best regards,
Chris Gribble

-----Original Message-----
 From: seqfan-bounces at list.seqfan.eu 
[mailto:seqfan-bounces at list.seqfan.eu]
On Behalf Of David Scambler
Sent: 10 October 2011 6:40 AM
To: seqfan at list.seqfan.eu
Subject: [seqfan] Re: a partition sequence

After finding multiple bugs, I think this might be correct, up to 260 
chars,
n=44.

1, 1, 2, 3, 5, 9, 14, 21, 32, 47, 67, 98, 136, 199, 277, 387, 537, 745,
1012, 1390, 1860, 2509, 3354, 4480, 5912, 7830, 10267, 13471, 17553, 
22857,
29588, 38313, 49301, 63391, 81184, 103803, 132228, 168196, 213259, 
269898,
340752, 429284, 539581, 676960, 847417

Can anyone else compute a few terms to check it?

Dave


David Newman davidsnewman at gmail.com
Sun Oct 2 23:22:11 CEST 2011
I'd like someone to check my numbers before I propose this sequence.

Let a(n) be the number of planar partitions whose summands are distinct
going across rows and have distinct frequencies going down columns.

I get a(1)=1, a(2)=2, a(3)=3, a(4)=5, a(5)=9, a(6)=14, a(7)=21 for the
values that I've computed by hand.

For example the allowed partitions of 6 are:

...


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