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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).

%I A356447 #64 Jul 30 2025 09:45:42
%S A356447 2,5,8,11,14,26,29,32,35,38,41,80,83,86,89,92,95,107,110,113,116,119,
%T A356447 122,242,245,248,251,254,257,269,272,275,278,281,284,323,326,329,332,
%U A356447 335,338,350,353,356,359,362,365,728,731,734,737,740,743,755,758,761
%N 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).
%C A356447 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).
%C A356447 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.
%C A356447 Other equivalent conditions are:
%C A356447 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.
%C A356447 2) the base 3 expansion of k+1 is either u0 or u20, where u has no 2's.
%H A356447 Mathematics Stack Exchange, <a href="https://math.stackexchange.com/questions/4494256/factors-of-central-binomial-coefficient/4501152#4501152">Factors of central binomial coefficient</a>
%F A356447 a(n) = 3*A096304(n) - 1.
%e A356447 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.
%e A356447 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).
%e A356447 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).
%t A356447 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 *)
%o A356447 (PARI) isok(k) = Mod(binomial(2*k,k), (k+1)*(2*k-1)) != 0; \\ _Michel Marcus_, Aug 10 2022
%o A356447 (Python)
%o A356447 def A356447(n):
%o A356447     a, b = divmod(n,3)
%o A356447     return (int(bin(a)[2:],3)*3+b)*3-1 # _Chai Wah Wu_, Jul 29 2025
%Y A356447 Cf. A000108, A000984, A073076, A096304,
%K A356447 nonn,easy
%O A356447 1,1
%A A356447 _Valerio De Angelis_, Aug 07 2022