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A302919 The number of ways of placing 2n-1 white balls and 2n-1 black balls into unlabeled bins such that each bin has both an odd number of white balls and black balls.

%I A302919 #22 Aug 12 2022 09:18:30
%S A302919 1,2,4,12,32,85,217,539,1316,3146,7374,16969,38387,85452,187456,
%T A302919 405659,866759,1830086,3821072,7894447,16148593,32723147,65719405,
%U A302919 130871128,258513076,506724988,985968770,1904992841,3655873294,6970687150,13208622956,24879427889,46593011280,86773920240,160742462714,296227087942,543183754454,991213989213
%N A302919 The number of ways of placing 2n-1 white balls and 2n-1 black balls into unlabeled bins such that each bin has both an odd number of white balls and black balls.
%H A302919 Marko Riedel, <a href="https://math.stackexchange.com/a/2737549/121988">How many ways to get an odd number of each color in each bin?</a>, Mathematics Stack Exchange.
%e A302919 For n = 3 the a(3) = 4 ways to place five white and five black balls are (wwwwwbbbbb), (wwwbbb)(wb)(wb), (wwwb)(wbbb)(wb), and (wb)(wb)(wb)(wb)(wb).
%t A302919 nmax = 15; p = 1; Do[Do[p = Expand[p*(1 - x^(2*i - 1)*y^(2*j - 1))]; p = Select[p, (Exponent[#, x] <= 2*nmax - 1) && (Exponent[#, y] <= 2*nmax - 1) &], {i, 1, 2*nmax - 1}], {j, 1, nmax}]; p = Expand[Normal[Series[1/p, {x, 0, 2*nmax - 1}, {y, 0, 2*nmax - 1}]]]; p = Select[p, Exponent[#, x] == Exponent[#, y] &]; Flatten[{1, Table[Coefficient[p, x^(2*n - 1)*y^(2*n - 1)], {n, 1, nmax}]}] (* _Vaclav Kotesovec_, Apr 16 2018 *)
%Y A302919 Cf. A090806, A108469.
%K A302919 nonn
%O A302919 1,2
%A A302919 _Peter Kagey_, Apr 15 2018
%E A302919 More terms from _Vaclav Kotesovec_, Apr 16 2018