[seqfan] Re: partitions in A120452 and A344613
William Orrick
will.orrick at gmail.com
Tue Jun 8 15:24:10 CEST 2021
Dear Richard,
The two sequences are equal. A comment to A344613 says that "a(n) is the
number of odd-length partitions of 2n whose conjugate partition has exactly
two odd parts." Consider the Ferrers graph of the partition. It must have
two odd-length columns and, because the length of the partition is odd, one
of those must be the first column. There are two cases:
(1) the two odd-length columns are of equal length, and hence must be the
first two columns. In this case the smallest part is 2 and all rows except
the last come in equal-length pairs,
(2) the two odd-length columns are unequal, and hence the smallest part is
1. If the row containing the shorter odd column has length k, the following
row must have length k-1. (Otherwise there would be two columns with the
shorter odd length.) All rows except these two and the last, length-1 row
come in equal-length pairs.
We can now exhibit a bijection between these partitions and the
partitions of A120452: create an empty partition, P. Identify a pair of
same-length rows and add a single part of the same length of the form
oo...o to P. Then delete the pair. Continue until there are no same-length
pairs. What remains is either a single length-2 row, in which case we add
the part * to P, or three rows of lengths k, k-1, and 1, in which case we
add the part oo...o* to P, where o occurs k-1 times. The partition P will
never contain o* because that would require k=2 in the latter case, which
would mean two rows of length 1 (the length-(k-1) row and the length-1
row). This won't occur because pairs of equal-length rows have already been
elminated.
-Will Orrick-
On Fri, Jun 4, 2021 at 5:05 AM Richard J. Mathar <mathar at mpia-hd.mpg.de>
wrote:
> Are these two sequences duplicates (with A344613 an additional a(0).):
> http://oeis.org/?q=id:A344613|id:A120452
> There is a simple sum formula in A344613 which may equal one of the
> partitions formulas in A120452 (supposing one can decipher Mma).
>
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