%I #5 Mar 30 2012 18:58:10

%S 2,3,4,3,3,4,4,5,4,5,8,7,6,4,8,5,5,4,4,8,7,8,5,7,5,6,7,7,6,8,12,5,8,5,

%T 8,10,10,8,9,9,6,7,16,8,10,11,8,7,8,5,7,10,8,7,10,7,8,6,20,8

%N Least k such that n divides s(k)-s(j) for some j satisfying 1<=j<k, where s(j)=C(2j-2,j-1).

%C See A204892 for a discussion and guide to related sequences.

%t s[n_] := s[n] = Binomial[2 (n - 1), n - 1];

%t z1 = 700; z2 = 60;

%t Table[s[n], {n, 1, 30}] (* A000984 *)

%t u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]

%t Table[u[m], {m, 1, z1}] (* A205008 *)

%t v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]

%t w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]

%t d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]]

%t Table[d[n], {n, 1, z2}] (* A205009 *)

%t k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]

%t m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]

%t j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2

%t Table[k[n], {n, 1, z2}] (* A205010 *)

%t Table[j[n], {n, 1, z2}] (* A205011 *)

%t Table[s[k[n]], {n, 1, z2}] (* A205012 *)

%t Table[s[j[n]], {n, 1, z2}] (* A205013 *)

%t Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A205014 *)

%t Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205015 *)

%Y Cf. A000984, A204892, A205008.

%K nonn

%O 1,1

%A _Clark Kimberling_, Jan 22 2012