A206828 Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k

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, 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, 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

Offset: 2

Author

Clark Kimberling, Feb 15 2012

Keywords

• nonn

Comments

For a guide to related sequences, see A206588.

Examples

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

Crossrefs

Cf. A206588.

Programs

Mathematica

s[k_] := Binomial[2 k, k];
f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
a[n_] := Count[Flatten[t[n]], 1]
Table[a[n], {n, 2, 120}]  (* A206828 *)