%I #17 Feb 29 2024 08:52:53
%S 1,0,0,1,0,0,0,1,2,1,1,1,0,0,0,6,3,1,2,2,1,3,2,3,6,2,3,5,5,2,3,6,4,6,
%T 5,10,12,6,4,10,8,3,8,5,8,5,3,5,11,4,4,10,7,8,9,19,21,12,8,7,8,4,3,17,
%U 13,19,23,10,11,13,16,18,14,9,11,19,13,9,16,16,25,21,17,15,18,13,16,17,19
%N a(n) = number of 0<=k<=n such that n+k divides binomial(n,k).
%H Robert Israel, <a href="/A051556/b051556.txt">Table of n, a(n) for n = 1..10000</a>
%e For n=9, 0<=k<=n, n+k divides C(n,k) only when k=3 and k=5, so a(9)=2.
%p f:= proc(n)
%p nops(select(t -> binomial(n,t) mod (n+t) = 0, [$0..n]))
%p end proc:
%p map(f, [$1..100]); # _Robert Israel_, Feb 27 2024
%t A051556[n_] := Count[Divisible[Binomial[n, Range[0, n]], Range[n, 2*n]], True];
%t Array[A051556, 100] (* _Paolo Xausa_, Feb 29 2024 *)
%o (PARI) a(n) = sum(k=0, n, (binomial(n,k) % (n+k)) == 0); \\ _Michel Marcus_, May 18 2014
%Y Cf. A051574 (similar definition).
%K nonn
%O 1,9
%A _Leroy Quet_