OFFSET
1,1
COMMENTS
It is well known that B(k) divided by (k+1) is an integer (the Catalan numbers A000108). It is also easy to see that (2k-1) divides B(k). So we ask when the product (k+1)*(2k-1) divides B(k). The terms of this sequence are the positive integers k such that (k+1)*(2k-1) does not divide B(k).
A necessary and sufficient condition for an integer k to be a term of this sequence is: k is congruent to 2 (mod 3), and at least one of (k+1) or (k-1) has no 2's in its base-3 expansion. In particular, this sequence has density 0. This is proved in the Stack Exchange post cited below.
Other equivalent conditions are:
1) k is congruent to 2 (mod 3), and its base-3 expansion either has no 2's, or is of form u12, or u02^i for some i>=1, where u has no 2's and 2^i means a string of i consecutive 2's.
2) the base 3 expansion of k+1 is either u0 or u20, where u has no 2's.
LINKS
Mathematics Stack Exchange, Factors of central binomial coefficient
FORMULA
a(n) = 3*A096304(n) - 1.
EXAMPLE
k = 95 is a term, since it is k == 2 (mod 3) and k-1 = 94 has base-3 expansion 10111 which has no digit 2's. It can be checked that B(k) = binomial(190,95) is not divisible by (k+1)*(2*k-1) = 18144.
As another example, a(18)=107, also congruent to 2 (mod 3), and 107+1=108 has base-3 expansion 11000. It can be checked that binomial(2*107,107) is not divisible by (107+1)*(2*107-1).
125 is not a term of the sequence, because even though it's congruent to 2 (mod 3), the base-3 expansions of 125+1=126 and 125-1=124 are 11200 and 11121. It can be checked that binomial(2*125,125) is divisible by (125+1)*(2*125-1).
MATHEMATICA
kmax=762; a={}; For[k=1, k<=kmax, k++, If[Not[Divisible[Binomial[2k, k], (k+1)(2k-1)]], AppendTo[a, k]]]; a (* Stefano Spezia, Aug 12 2022 *)
PROG
(PARI) isok(k) = Mod(binomial(2*k, k), (k+1)*(2*k-1)) != 0; \\ Michel Marcus, Aug 10 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Valerio De Angelis, Aug 07 2022
STATUS
approved