|
|
A202785
|
|
Number of 3 X 3 0..n arrays with row and column sums equal.
|
|
2
|
|
|
14, 87, 340, 1001, 2442, 5215, 10088, 18081, 30502, 48983, 75516, 112489, 162722, 229503, 316624, 428417, 569790, 746263, 964004, 1229865, 1551418, 1936991, 2395704, 2937505, 3573206, 4314519, 5174092, 6165545, 7303506, 8603647
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
|
|
LINKS
|
|
|
FORMULA
|
Empirical: a(n) = (3/10)*n^5 + (3/2)*n^4 + (7/2)*n^3 + (9/2)*n^2 + (16/5)*n + 1.
G.f.: x*(7 - 2*x + x^2)*(2 + x + 4*x^2 - x^3) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6.
(End)
Empirical formula verified (see link): Robert Israel, May 02 2019
|
|
EXAMPLE
|
Some solutions for n=7:
..3..2..1....3..5..5....0..6..2....0..7..5....4..2..1....5..6..0....1..6..1
..2..0..4....5..6..2....2..1..5....6..1..5....3..2..2....0..4..7....5..2..1
..1..4..1....5..2..6....6..1..1....6..4..2....0..3..4....6..1..4....2..0..6
|
|
MAPLE
|
seq((3/10)*n^5 + (3/2)*n^4 + (7/2)*n^3 + (9/2)*n^2 + (16/5)*n + 1, n=1..30); # Robert Israel, May 02 2019
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|