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%I #13 Nov 06 2022 07:48:08
%S 0,1,1,1,2,1,3,3,4,4,5,6,6,5,7,7,8,8,9,9,10,10,11,11,12,12,13,13,14,
%T 15,15,15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,
%U 26,26,27,28,28,28,29,29,30,29,31,31,32,32,33,33,34,34,35,35,36,36,37,37
%N Greatest k <= n/2 such that n divides binomial(n,k).
%H Reinhard Zumkeller, <a href="/A123621/b123621.txt">Table of n, a(n) for n = 1..1000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BinomialCoefficient.html">Binomial Coefficient</a>
%H <a href="/index/Pas#Pascal">Index entries for triangles and arrays related to Pascal's triangle</a>
%F A007318(n,a(n)) mod n = 0.
%F a(p) = floor(p/2) for primes p.
%t gk[n_]:=Module[{k=Floor[n/2]},While[!Divisible[Binomial[n,k],n],k--];k]; Array[gk,80] (* _Harvey P. Dale_, Nov 22 2013 *)
%o (PARI) a(n) = if (n>1, my(k=n\2); while (binomial(n,k) % n, k--); k, 0); \\ _Michel Marcus_, Nov 06 2022
%K nonn
%O 1,5
%A _Reinhard Zumkeller_, Oct 03 2006
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