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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).

%I A316587 #34 Nov 05 2020 06:47:15
%S A316587 1,3,10,27,69,161,361,767,1578,3134,6064,11432,21105,38175,67863,
%T A316587 118658,204455,347439,583063,966952,1586231,2575474,4141832,6600731,
%U A316587 10430455,16349788,25434178,39280676,60250276,91810915,139034070,209294256,313269591,466343647
%N 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).
%C A316587 Let S be a fixed matching of size n in a complete graph G with >= 4n vertices. Given T,T' (also matchings of size n), define the equivalence relation where T ~ T' if and only if there exists an automorphism of G that maps edges in T to edges in T' while mapping edges in S to edges in S. Then the number of equivalence classes is a(n).
%C A316587 a(n) is the number of partitions of 2n with 4 kinds of parts (types 1,2,3,4) where (i) all parts of types 1,2 are odd and all parts of types 3,4 are even; and (ii) the number of type 1 and type 2 parts are equal.
%H A316587 Yu Hin Au, Nathan Lindzey, and Levent Tunçel, <a href="https://arxiv.org/abs/2008.08628">Matchings, hypergraphs, association schemes, and semidefinite optimization</a>, arXiv:2008.08628 [math.CO], 2020.
%e A316587 To see a(2)=10, let S = {{1,2},{3,4}}. Then a representative from each of the 10 equivalence classes are
%e A316587   1. {{1,2}, {3,4}}
%e A316587   2. {{1,3}, {2,4}}
%e A316587   3. {{1,5}, {3,4}}
%e A316587   4. {{1,3}, {4,5}}
%e A316587   5. {{1,2}, {5,6}}
%e A316587   6. {{1,3}, {5,6}}
%e A316587   7. {{1,5}, {2,6}}
%e A316587   8. {{1,5}, {3,6}}
%e A316587   9. {{1,5}, {6,7}}
%e A316587   10. {{5,6}, {7,8}}
%Y A316587 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.
%K A316587 nonn
%O A316587 0,2
%A A316587 _Yu Hin Au_, Aug 31 2018