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Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k<n.
4

%I #5 Mar 30 2012 18:58:12

%S 1,1,2,1,2,1,2,2,1,1,1,1,2,2,3,2,2,1,8,2,1,1,5,2,1,2,11,1,0,1,4,1,3,4,

%T 1,2,1,1,8,1,15,1,2,12,1,1,5,2,3,0,1,3,3,3,1,0,1,1,2,1,1,0,5,2,23,1,4,

%U 0,4,1,7,3,1,12,2,24,2,1,8,3,3,1,6,0,3,1,37,1,3,26,1,1,1,0,4

%N Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k<n.

%C For a guide to related sequences, see A206588.

%e 2 divides exactly two of the numbers 20-1, 20-2, 20-6, so that a(3)-2.

%t s[k_] := Binomial[2 k, k];

%t f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];

%t t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]

%t a[n_] := Count[Flatten[t[n]], 1]

%t Table[a[n], {n, 2, 120}] (* A206828 *)

%Y Cf. A206588.

%K nonn

%O 2,3

%A _Clark Kimberling_, Feb 15 2012