|
Search: id:A122670
|
|
|
| A122670 |
|
If n mod 4 = 2 or n mod 4 = 3 then a(n) = 0 else let m=floor(n/4), then a(n) = (2*m)!/m!. |
|
+0
3
|
|
| 1, 1, 0, 0, 2, 2, 0, 0, 12, 12, 0, 0, 120, 120, 0, 0, 1680, 1680, 0, 0, 30240, 30240, 0, 0, 665280, 665280, 0, 0, 17297280, 17297280, 0, 0, 518918400, 518918400, 0, 0, 17643225600, 17643225600, 0, 0, 670442572800, 670442572800, 0, 0, 28158588057600, 28158588057600, 0, 0, 1295295050649600
(list; graph; listen)
|
|
|
OFFSET
|
0,5
|
|
|
COMMENT
|
Number of solutions to the rook problem on an n X n board having a certain symmetry group (see Robinson for details).
|
|
REFERENCES
|
R. W. Robinson, Counting arrangements of bishops, pp. 198-214 of Combinatorial Mathematics IV (Adelaide 1975), Lect. Notes Math., 560 (1976).
|
|
FORMULA
|
For asymptotics see the Robinson paper.
|
|
MAPLE
|
R:=proc(n) local m; if n mod 4 = 2 or n mod 4 = 3 then RETURN(0); fi; m:=floor(n/4); (2*m)!/m!; end;
For Maple program see A000903.
|
|
CROSSREFS
|
If the duplicates and zeros are omitted we get A001813.
Cf. A000898, A000899, A000900, A000901, A000902, A000903.
Sequence in context: A167291 A063865 A037224 this_sequence A087637 A052458 A004586
Adjacent sequences: A122667 A122668 A122669 this_sequence A122671 A122672 A122673
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com), Sep 23 2006
|
|
|
Search completed in 0.002 seconds
|
