A316587 a(n) = [x^(2n)y^n] Product_{i>=1} 1/((1-x^(2i-1)y^i)(1-x^(2i-1)y^(i-1))(1-x^(2i)y^i)^2).
1, 3, 10, 27, 69, 161, 361, 767, 1578, 3134, 6064, 11432, 21105, 38175, 67863, 118658, 204455, 347439, 583063, 966952, 1586231, 2575474, 4141832, 6600731, 10430455, 16349788, 25434178, 39280676, 60250276, 91810915, 139034070, 209294256, 313269591, 466343647
Offset: 0
Keywords
Examples
To see a(2)=10, let S = {{1,2},{3,4}}. Then a representative from each of the 10 equivalence classes are 1. {{1,2}, {3,4}} 2. {{1,3}, {2,4}} 3. {{1,5}, {3,4}} 4. {{1,3}, {4,5}} 5. {{1,2}, {5,6}} 6. {{1,3}, {5,6}} 7. {{1,5}, {2,6}} 8. {{1,5}, {3,6}} 9. {{1,5}, {6,7}} 10. {{5,6}, {7,8}}
Links
- Yu Hin Au, Nathan Lindzey, and Levent Tunçel, Matchings, hypergraphs, association schemes, and semidefinite optimization, arXiv:2008.08628 [math.CO], 2020.
Crossrefs
If the equivalence relation is defined as T~T' if and only if there exists an automorphism of G mapping union of S,T to union of S,T' (i.e., the map does not necessarily fix edges in S), then we obtain A305168.
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