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%I #11 Sep 21 2018 22:37:00
%S 2,2,2,3,2,2,3,2,2,2,3,2,4,2,3,2,2,3,4,2,2,3,2,2,4,3,2,5,2,2,3,2,2,4,
%T 2,3,2,2,2,3,5,2,4,2,2,3,2,2,2,2,3,2,4,2,2,5,3,2,2,2,6,2,3,2,4,2,2,2,
%U 3,2,2,2,5,2,3,4,2,2,2,3,2,2,2,6,2,3,4,2,2,2,5,2,3,2,2,2,2,3,2
%N Binomial(A319500(n),a(n)) = A006917(n) with 2 <= a(n) <= A319500(n)/2.
%C First differs from A022912 at n=18.
%H Robert Israel, <a href="/A318955/b318955.txt">Table of n, a(n) for n = 1..10000</a>
%e a(3) = 6 because A006987(3) = 15 = binomial(6,2).
%p N:= 10000: # for binomial(n, k) values <= N
%p S:= {}:
%p for n from 2 while n*(n-1)/2 <= N do
%p for k from 2 to n/2 do
%p v:= binomial(n, k);
%p if v > N then break fi;
%p if not member(v,S) then
%p S:= S union {v};
%p K[v]:= k;
%p fi
%p od od:
%p A006987:= sort(convert(S,list)):
%p seq(K[A006987[i]],i=1..nops(A006987));
%Y Cf. A006987, A022912, A319500.
%K nonn
%O 1,1
%A _Robert Israel_, Sep 20 2018
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