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%I #19 Feb 28 2024 00:59:28
%S 1,1,1,1,2,1,1,1,2,2,3,1,3,1,1,1,3,1,5,3,2,1,1,1,2,2,2,4,7,1,2,2,5,6,
%T 3,1,6,3,2,1,4,1,2,1,2,2,2,1,4,5,9,5,8,4,7,3,9,6,8,2,6,3,1,4,11,5,9,4,
%U 5,1,4,1,7,4,2,4,8,3,4,1,6,11,15,3,7,5,4,5,9,1,5,3,2,2,1,1,4,2,7,7,19,8
%N a(n) = number of k, 1<=k<=n, such that (n*k) divides binomial(n,k).
%C k = 1 always works, so a(n) >= 1. a(n) = 1 for n = 1, 2, 3, 4, 6, 7, 8, 12, 14, 15, 16, 18, 22, 23, 24, 30, 36, 40, 42, 44, 48, 63, 70, 72, 80, 90, 95, 96, 120, 240. Are there any others? - _Robert Israel_, Feb 27 2024
%H Robert Israel, <a href="/A051574/b051574.txt">Table of n, a(n) for n = 1..10000</a>
%e a(11)=3 since k=1, k=3, k=6 are only solutions to 11*k divides binomial(11,k).
%p f:= proc(n) nops(select(t -> binomial(n,t) mod (n*t) = 0, [$1..n])) end proc:
%p map(f, [$1..200]); # _Robert Israel_, Feb 27 2024
%t nkb[n_]:=Total[Table[If[Divisible[Binomial[n,k],n*k],1,0],{k,n}]]; Array[ nkb,110] (* _Harvey P. Dale_, Jul 06 2020 *)
%o (PARI) a(n) = sum(k=1, n, (binomial(n,k) % (n*k)) == 0); \\ _Michel Marcus_, May 18 2014
%Y Cf. A051556 (similar definition).
%K nonn
%O 1,5
%A _Leroy Quet_, Dec 11 1999
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