A309364 a(n) is the least k >= 0 such that n divides C(k) (where C(k) are the Catalan numbers A000108).

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

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 factorization 2^t * Product_{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)))
(Python)
from itertools import count
def A309364(n):
    if n == 1: return 0
    c = 1
    for k in count(1):
        if not c%n: return k
        c = c*((k<<1)+1<<1)//(k+2) # Chai Wah Wu, May 04 2023

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