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%I #7 Oct 20 2019 17:46:37
%S 0,0,1,3,6,13,25,45,91,175,322,645,1245,2325,4651,9031,17061,34123,
%T 66547,126763,253527,496063,950818,1901637,3730293,7184421,14368843,
%U 28243063,54604081,109208163,215008363,416990563,833981127
%N Number of nonempty subsets of {1,2,...,n} in which exactly 1/3 of the elements are <= n/3.
%C Comment from _Jinyuan Wang_, Oct 19 2019 (Start)
%C This is also the number of nonempty subsets of {1,2,...,n} in which exactly 1/3 of the elements are <= (n+1)/3.
%C Proof: Let b(n) = number of nonempty subsets of {1,2,...,n} in which exactly 1/3 of the elements are <= (n+1)/3.
%C We only need to prove b(3k-1) = a(3k-1).
%C Now a(3k-1) = Sum_{m=1..k-1} binomial(k-1, m)*binomial(2k, 2m).
%C b(3k-1) = Sum_{m=1..k-1} binomial(k, m)*binomial(2k-1, 2m).
%C Because binomial(k-1, m)*binomial(2k, 2m) = binomial(k, m)*binomial(2k-1, 2m), we have b(3k-1) = a(3k-1). (End)
%K nonn
%O 1,4
%A _Clark Kimberling_
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