Voozh

A206828

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

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

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OFFSET

2,3

COMMENTS

For a guide to related sequences, see A206588.

LINKS

Table of n, a(n) for n=2..96.

EXAMPLE

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

MATHEMATICA

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

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