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Numbers k such that there is a m with 2^m/(m+1) < binomial(m,k) <= 2^m/m and k < m/2.
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%I #4 Oct 04 2013 15:36:10

%S 1,1,6,23,35,53,57,126,150,233,323,359,426,436,446,498,598,645,744,

%T 809,904,918,1063,1186,1348,1467,1756,2008,2028,2109,2150,2339,2382,

%U 2469,2491,2670,2808,3069,3442,3675,3888,3915,4302,4735,5095,5786,5916,6415,6517

%N Numbers k such that there is a m with 2^m/(m+1) < binomial(m,k) <= 2^m/m and k < m/2.

%H T. D. Noe, <a href="/A229486/b229486.txt">Table of n, a(n) for n = 1..104</a> (for m less than 10^5)

%Y Cf. A229484.

%K nonn

%O 1,3

%A _T. D. Noe_, Oct 04 2013