%I
%S 0,2,5,6,3,5,4,14,14,8,6,6,7,4,14,30,9,14,10,13,5,6,12,14,13,8,41,12,
%T 15,14,16,62,6,9,18,14,19,10,7,14,21,5,22,6,14,12,24,46,25,13,14,10,
%U 27,41,8,26,14,16,30,14,31,16,25,126,8,6,34,10,14,18,36
%N a(n) is the least k >= 0 such that n divides C(k) (where C(k) are the Catalan numbers A000108).
%C The sequence is well defined:
%C  if k has t+1 ones in binary representation, 2^t divides C(k),
%C  for any odd prime number p: if k has e digits (p+1)/2 in base p, p^e divides C(k),
%C  for any n with prime factorisation 2^t * Prod_{i=1..o} p_i^e_i (where p_i are distinct odd prime numbers),
%C  by the Chinese remainder theorem, there is a number N
%C ending with t+1 ones in base 2
%C and ending with e_i digits (p_i+1)/2 in base p_i for i = 1..o,
%C  C(N) is a multiple of n
%C  and a(n) <= N.
%C As a consequence, A309200 is a permutation of the positive integers (since for any n > 0, we have infinitely many multiples of n among the Catalan number, and then the argument used to prove that A111273 is a permutation completes the proof).
%H Rémy Sigrist, <a href="/A309364/b309364.txt">Table of n, a(n) for n = 1..10000</a>
%F a(p) = (p+1)/2 for any prime number p > 3.
%F a(C(k)) = k for k <> 1.
%o (PARI) a(n) = for (k=0, oo, my (c=binomial(2*k, k)/(k+1)); if (c%n==0, return (k)))
%Y Cf. A000108, A309200.
%K nonn
%O 1,2
%A _Rémy Sigrist_, Jul 25 2019
