%I #23 Feb 01 2016 05:23:51
%S 0,1,2,1,1,2,2,2,3,1,1,2,1,1,3,3,3,3,5,5,6,3,3,3,3,2,2,1,1,5,1,1,2,4,
%T 4,2,1,1,4,1,1,5,5,5,4,4,4,4,4,3,2,2,2,5,5,2,2,2,3,3,2,2,2,3,3,3,3,5,
%U 5,5,3,3,3,3,6,6,6,7,5,5,5,1,1,5,1,1,6,6,6,6,1,1,6,1,1,7,7,7,3,3,3
%N Index of largest primorial factor of binomial(2n,n).
%C For n > 0, binomial(2n,n) is even, so a(n) >= 1.
%C Is a(n) unbounded? (The largest value for n <= 100000 is a(45416) = 43.)
%C From _Robert Israel_, Jan 28 2016: (Start)
%C a(n) = A000720(p)-1 where p is the least prime that does not divide A000984(n).
%C Equivalently, p is the least prime such that the base-p representation of n has all digits < p/2.
%C a(primorial(k)-1) >= k. In particular the sequence is unbounded. (End)
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Lucas%27_theorem">Lucas' theorem</a>
%F a(A267823(n)) >= n.
%F min{k : a(k) >= n} = A267823(n).
%e Binomial(16,8) = 12870 is divisible by primorial(3) = 2*3*5 = 30, but not by prime(4) = 7, so a(8) = 3.
%t PrimorialFactor[n_] := (k = 0; While[Mod[n, Prime[k + 1]] == 0, k++]; k);
%t Table[PrimorialFactor[Binomial[2 n, n]], {n, 0, 100}]
%o (PARI) pf(n) = {my(k = 0); while (n % prime(k+1) == 0, k++); k;}
%o a(n) = pf(binomial(2*n, n)); \\ adapted from Mathematica; _Michel Marcus_, Jan 29 2016
%Y Cf. A000720, A000984, A002110, A226078, A267823.
%K nonn
%O 0,3
%A _Jonathan Sondow_, Jan 27 2016