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Numbers k such that 3k does not divide (6k-4)!/((3k-2)!*(3k-1)!).
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%I #27 Sep 11 2022 09:31:59

%S 1,2,3,4,5,9,10,11,12,13,14,27,28,29,30,31,32,36,37,38,39,40,41,81,82,

%T 83,84,85,86,90,91,92,93,94,95,108,109,110,111,112,113,117,118,119,

%U 120,121,122,243,244,245,246,247,248,252,253,254,255,256,257,270,271

%N Numbers k such that 3k does not divide (6k-4)!/((3k-2)!*(3k-1)!).

%C Equivalently, members of A019469 divisible by 3, divided by 3.

%C _Ralf Stephan_'s formula is that terms k written in ternary have an arbitrary least significant digit and above that only 0's and 1's (per A340051). - _Kevin Ryde_, May 22 2021

%C {3a(n)-1:n>=1} is the set of positive integers k such that the k-th central binomial coefficient is not divisible by (k+1)*(2k-1). Such integers k are characterized by the following property: k is congruent to 2 (mod 3), and at least one of k-1, k+1 has no 2's in its base-3 expansion. - _Valerio De Angelis_, Aug 08 2022

%H Harvey P. Dale, <a href="/A096304/b096304.txt">Table of n, a(n) for n = 1..600</a>

%H Kevin Ryde, <a href="/A096304/a096304.txt">Proof of Ralf Stephan's formula</a>

%H Math Stackexchange, <a href="https://math.stackexchange.com/questions/4494256/factors-of-central-binomial-coefficient/4501152#4501152">Factors of central binomial coefficient</a>

%H <a href="/index/Ar#3-automatic">Index entries for 3-automatic sequences</a>.

%F a(n) = 9 * A005836(floor(n/6)) + (n mod 6) (conjectured) (confirmed, see links).

%F G.f.: x*(1+2*x)/(1-x^3) + 3/(1-x) * Sum_{i>=0} 3^i * x^(3*2^i) / (1 + x^(3*2^i)). - _Kevin Ryde_, May 22 2021

%t Select[Range[300],Mod[(6#-4)!/((3#-2)!(3#-1)!),3#]!=0&] (* _Harvey P. Dale_, Jun 11 2019 *)

%o (PARI) for(n=1,300,if(((6*n-4)!/(3*n-2)!/(3*n-1)!)%(3*n),print1(n",")))

%o (PARI) a(n) = my(r);[n,r]=divrem(n,3); fromdigits(concat(binary(n),r), 3); \\ _Kevin Ryde_, May 22 2021

%Y Cf. A340051 (ternary digits), A005836, A019469, A187358.

%K nonn,easy

%O 1,2

%A _Ralf Stephan_, Aug 03 2004