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%I #8 Feb 03 2016 11:48:51
%S 2,3,12,10,3160
%N Least k such that prime(n) is the smallest odd prime factor of C(2k,k).
%C If n>0, then a(n+2) >= A129489(n) = least k>1 such that binomial(2k,k) is not divisible by any of the first n odd primes.
%C It is not known whether any more terms exist. See A129489 for bounds, comments and references.
%e C(2,1) = 2, C(4,2) = 6 = 2 * 3, C(6,3) = 20 = 2^2 * 5, and 3 = prime(2), 5 = prime(3), so a(2) = 2 and a(3) = 3.
%o (PARI) valp(n, p)=my(s); while(n\=p, s+=n); s
%o a(n)=my(q=prime(n),k=1); while(k++, forprime(p=3,q-1, if(valp(2*k, p)>2*valp(k, p), next(2))); if(valp(2*k, q)>2*valp(k, q), return(k))) \\ _Charles R Greathouse IV_, Feb 03 2016
%Y Cf. A000984, A129488, A129489, A030979.
%K bref,hard,more,nonn
%O 2,1
%A _Jonathan Sondow_, Jan 18 2016
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