A205386 Least k such that n divides s(k)-s(j) for some j<k, where s(j)=(1/2)C(2j,j).

2, 2, 3, 4, 4, 6, 3, 4, 3, 7, 7, 6, 5, 6, 7, 4, 4, 9, 7, 8, 6, 7, 10, 6, 4, 9, 6, 6, 5, 7, 11, 4, 7, 4, 7, 9, 9, 7, 8, 8, 5, 6, 15, 9, 9, 10, 13, 6, 7, 8

Offset

1,1

Comments

See A204892 for a discussion and guide to related sequences.

Links

Table of n, a(n) for n=1..50.

Mathematica

s[n_] := s[n] = (1/2) Binomial[2 n, n]; z1 = 700; z2 = 50;
Table[s[n], {n, 1, 30}]    (* A001700 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]     (* A205384 *)
v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]]
Table[d[n], {n, 1, z2}]     (* A205385 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
Table[k[n], {n, 1, z2}]        (* A205386 *)
Table[j[n], {n, 1, z2}]        (* A205387 *)
Table[s[k[n]], {n, 1, z2}]     (* A205388 *)
Table[s[j[n]], {n, 1, z2}]     (* A205389 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]       (* A205390 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}]   (* A205391 *)

Crossrefs

Cf. A001700, A204892, A205391.

Sequence in context: A321424 A289677 A113967 * A206922 A276775 A271169

Adjacent sequences: A205383 A205384 A205385 * A205387 A205388 A205389

Keyword

nonn

Author

Clark Kimberling, Jan 27 2012

Status

approved