|
Search: id:A071096
|
|
|
| A071096 |
|
Number of ways to tile hexagon of edges n, n+1, n+2, n, n+1, n+2 with diamonds of side 1. |
|
+0
1
|
|
| 1, 10, 490, 116424, 133613766, 739309710568, 19702998159210080, 2527580342020127455360, 1560172391098377453031770400, 4632518859090968506120863642225000, 66153724447703043353053979949899667187500, 4542776083800437392420665771479758969781250000000, 1499928882906010042230116408158354282455601808812500000000
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
REFERENCES
|
J. Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see page 261).
|
|
LINKS
|
J. Propp, Updated article
J. Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics
|
|
FORMULA
|
Product_{i=0..a-1} Product_{j=0..b-1} Product_{k=0..c-1} (i+j+k+2)/(i+j+k+1) with a=n, b=n+1, c=n+2.
a(n)=(-1)^floor((n+1)/2)*det(M(n+1)) where M(n) is the n X n matrix m(i, j)=C(2n, i+j), with i and j ranging from 1 to n. - Benoit Cloitre (benoit7848c(AT)orange.fr), Jan 31 2003
a(n) = (1/2)*Product[Product[Product[(i+j+k-1)/(i+j+k-2),{i,1,n+1}],{j,1,n+1}],{k,1,n+1}]. a(n) = A008793[n+1]/2. - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 10 2006
|
|
MATHEMATICA
|
Table[Product[Product[Product[(i+j+k-1)/(i+j+k-2), {i, 1, n+1}], {j, 1, n+1}], {k, 1, n+1}], {n, 0, 5}]/2 - Alexander Adamchuk (alex(AT)kolmogorov.com), Jul 10 2006
|
|
PROGRAM
|
(PARI) a(n)=abs(matdet(matrix(n, n, i, j, binomial(2*n, i+j))))
|
|
CROSSREFS
|
Cf. A008793.
Sequence in context: A001327 A159533 A035320 this_sequence A039835 A127947 A095232
Adjacent sequences: A071093 A071094 A071095 this_sequence A071097 A071098 A071099
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com), May 28 2002
|
|
|
Search completed in 0.002 seconds
|
