0,2

Also the number of subsets of {1,..,4n} containing exactly 2n elements with total sum n*(4n+1) (see also A060468 for a related sequence). This is of course the same as the number of partitions of n*(4n+1) having 2n distinct parts of length at most 4n. (cont.)

(cont.). This number is the coefficient of t^0 q^0 in the product('(t*q^k+1/(t*q^k)','k'=1..4*n). - Roland Bacher (Roland.Bacher(AT)ujf-grenoble.fr), May 02 2002

A bijection with a dissection as above of the 2n X 2n checkerboard is given by subtracting 1,2,3,..,2n of the smallest, second-smallest, etc. element in the subset.

Example (n=2 as above): {1,2,7,8} (yields the checkerboard partition {1-1,2-2,7-3,8-4}={0,0,4,4}), {1,3,6,8} (yields {1-1,3-2,6-3,8-4}={0,1,3,4}), {1,4,5,8} (yields {0,2,2,4}), {1,4,6,7} (yields {0,2,3,3}), {3,4,5,6} (yields {2,2,2,2}), {2,4,5,7} (yields {1,2,2,3}), {2,3,6,7} (yields {1,1,3,3}), {2,3,5,8} (yields {1,1,2,4}).

Appears to be the number of random walks of length 4n, moves +/-1, starting and ending at 0 and with signed area 0 under the path. It would be nice to have a lower bound of the form a(n) > c*2^{4n}/n^d - David_Mumford(AT)brown.edu, Jun 25 2003

a(n) = A029895(2n) = A067059(2n, 2n) = A107110(2n, 2n^2). a(n) seems to be close to (sqrt(75)/pi)*16^n/(20n^2+6n+1). - Henry Bottomley (se16(AT)btinternet.com), May 12 2005

For a 4 X 4 board (n=2) the 8 partitions are (4, 4, 0, 0), (4, 3, 1, 0), (4, 2, 1, 1), (4, 2, 2, 0), (3, 3, 2, 0), (3, 3, 1, 1), (3, 2, 2, 1), (2, 2, 2, 2).

Table[ Length@Select[ Partitions[ 2n^2 ], Length[ # ] <= 2n && First[ # ] <= 2n& ], {n, 1, 5} ] or faster: Table[ T[ 2n^2, 2n, 2n ], {n, 0, 24} ] with T[ m, a, b ] as defined in A047993.

Cf. A047993, A063075.

Sequence in context: A153525 A153553 A007347 this_sequence A005804 A162067 A086907

Adjacent sequences: A063071 A063072 A063073 this_sequence A063075 A063076 A063077

nonn

Wouter Meeussen (wouter.meeussen(AT)pandora.be), Aug 03 2001