

A309364


a(n) is the least k >= 0 such that n divides C(k) (where C(k) are the Catalan numbers A000108).


2



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, 15, 14, 16, 62, 6, 9, 18, 14, 19, 10, 7, 14, 21, 5, 22, 6, 14, 12, 24, 46, 25, 13, 14, 10, 27, 41, 8, 26, 14, 16, 30, 14, 31, 16, 25, 126, 8, 6, 34, 10, 14, 18, 36
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

The sequence is well defined:
 if k has t+1 ones in binary representation, 2^t divides C(k),
 for any odd prime number p: if k has e digits (p+1)/2 in base p, p^e divides C(k),
 for any n with prime factorisation 2^t * Prod_{i=1..o} p_i^e_i (where p_i are distinct odd prime numbers),
 by the Chinese remainder theorem, there is a number N
ending with t+1 ones in base 2
and ending with e_i digits (p_i+1)/2 in base p_i for i = 1..o,
 C(N) is a multiple of n
 and a(n) <= N.
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).


LINKS

Rémy Sigrist, Table of n, a(n) for n = 1..10000


FORMULA

a(p) = (p+1)/2 for any prime number p > 3.
a(C(k)) = k for k <> 1.


PROG

(PARI) a(n) = for (k=0, oo, my (c=binomial(2*k, k)/(k+1)); if (c%n==0, return (k)))


CROSSREFS

Cf. A000108, A309200.
Sequence in context: A340859 A336817 A340858 * A062825 A154925 A154962
Adjacent sequences: A309361 A309362 A309363 * A309365 A309366 A309367


KEYWORD

nonn


AUTHOR

Rémy Sigrist, Jul 25 2019


STATUS

approved



