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A226083 Smallest element of the set of largest prime powers p^k dividing C(2*n,n), where p is any prime factor of C(2*n,n).

%I A226083 #23 Jun 29 2014 19:43:24
%S A226083 2,2,4,2,4,3,3,2,4,4,3,4,7,8,5,2,4,3,3,4,3,3,13,4,8,8,16,5,3,7,7,2,3,
%T A226083 3,7,4,7,3,11,4,5,5,7,7,5,5,5,4,8,8,11,8,5,3,3,8,3,3,5,7,7,7,3,2,4,3,
%U A226083 3,4,7,8,11,4,8,8,5,5,5,7,7,4,5,5,3,7,5,5,3,3,9,11,7,3,7,7,13,4,8,8,3,3
%N A226083 Smallest element of the set of largest prime powers p^k dividing C(2*n,n), where p is any prime factor of C(2*n,n).
%H A226083 Alois P. Heinz, <a href="/A226083/b226083.txt">Table of n, a(n) for n = 1..10000</a>
%F A226083 a(n) = min_{p prime, p|C(2n,n)} max_{k, p^k|C(2n,n)} p^k.
%F A226083 a(n) = min_{k=0..A067434(n)-1} A226078(n,k).
%e A226083 a(89) = 9: C(2*89,89) = 2^4 * 3^2 * 5^3 * 7^2 * 11^1 * ... * 173^1, the smallest prime power is 3^2 = 9.  3^2 is the largest prime power for prime 3 dividing C(2*89,89).
%e A226083 a(9993) = 59: 59^1 is the largest power of 59 dividing C(2*9993,9993), it is smaller than the largest powers of all other prime factors.
%p A226083 a:= proc(n) local h, i, m, p;
%p A226083       p:=1; m:=infinity;
%p A226083       while p < m do p:= nextprime(p); i:= 0;
%p A226083          h:= 2*n; while h>0 do h:=iquo(h, p); i:=i+h od;
%p A226083          h:= n;   while h>0 do h:=iquo(h, p); i:=i-2*h od;
%p A226083          if i>0 then m:= min(m, p^i) fi
%p A226083       od; m
%p A226083     end:
%p A226083 seq(a(n), n=1..100);
%Y A226083 Cf. A000040, A000961, A000984, A007318, A226047 (row maxima of A226078).
%K A226083 nonn,look
%O A226083 1,1
%A A226083 _Alois P. Heinz_, May 25 2013