A356447 Integers k such that (k+1)*(2*k-1) does not divide the central binomial coefficient B(k) = binomial(2*k,k) = A000984(k).

2, 5, 8, 11, 14, 26, 29, 32, 35, 38, 41, 80, 83, 86, 89, 92, 95, 107, 110, 113, 116, 119, 122, 242, 245, 248, 251, 254, 257, 269, 272, 275, 278, 281, 284, 323, 326, 329, 332, 335, 338, 350, 353, 356, 359, 362, 365, 728, 731, 734, 737, 740, 743, 755, 758, 761

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

Table of n, a(n) for n=1..56.

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

Cf. A000108, A000984, A073076, A096304,

Sequence in context: A275604 A173698 A162938 * A353985 A118518 A329604

Adjacent sequences: A356444 A356445 A356446 * A356448 A356449 A356450

Keyword

nonn,easy

Author

Valerio De Angelis, Aug 07 2022

Status

approved