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Positions of zeros in A065344, i.e., binomial(2n,n) mod ((n+1)*(n+2)) = 0.
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%I #20 Dec 12 2024 14:53:20

%S 3,5,8,9,11,12,15,16,17,18,19,20,21,22,23,24,26,27,29,32,33,35,36,38,

%T 39,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,63,

%U 64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,80,81,83,84,86,87,89,90

%N Positions of zeros in A065344, i.e., binomial(2n,n) mod ((n+1)*(n+2)) = 0.

%H Harry J. Smith, <a href="/A065347/b065347.txt">Table of n, a(n) for n = 1..1000</a>

%H C. Pomerance, <a href="https://math.dartmouth.edu/~carlp/catalan5.pdf">Divisors of the middle binomial coefficient</a>, Amer. Math. Monthly, 112 (2015), 636-644.

%t ris = {}; Do[If[Mod[Binomial[2 n, n], (n + 1) (n + 2)] == 0, AppendTo[ris, n]], {n, 100}]; ris (* _Bruno Berselli_, Jan 06 2014 *)

%o (PARI) isok(k) = { binomial(2*k, k) % ((k + 1)*(k + 2)) == 0 } \\ _Harry J. Smith_, Oct 17 2009

%Y Cf. A000108, A065344, A065345, A065346, A065348, A065349.

%K nonn

%O 1,1

%A _Labos Elemer_, Oct 30 2001