A205000 Least k such that n divides s(k)-s(j) for some j satisfying 1<=j<k, where s(j)=3^j-2^j.

2, 2, 3, 2, 4, 3, 3, 4, 3, 4, 6, 4, 7, 3, 4, 4, 6, 3, 6, 4, 5, 6, 4, 5, 6, 7, 9, 7, 13, 4, 9, 4, 6, 6, 5, 8, 9, 6, 8, 6, 7, 5, 12, 6, 8, 4, 8, 5, 7, 6, 7, 8, 16, 9, 6, 7, 9, 13, 13, 4

Offset

1,1

Comments

See A204892 for a discussion and guide to related sequences.

Links

Table of n, a(n) for n=1..60.

Mathematica

s[n_] := s[n] = 3^n - 2^n; z1 = 500; z2 = 60;
Table[s[n], {n, 1, 30}]      (* A001047 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]      (* A205105 *)
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}]  (* A205106 *)
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}]                  (* A205000 *)
Table[j[n], {n, 1, z2}]                  (* A205107 *)
Table[s[k[n]], {n, 1, z2}]               (* A205108 *)
Table[s[j[n]], {n, 1, z2}]               (* A205109 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A205110 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205111 *)

Crossrefs

Cf. A001047, A204892, A205111.

Sequence in context: A023522 A366311 A355271 * A355270 A066241 A334857

Adjacent sequences: A204997 A204998 A204999 * A205001 A205002 A205003

Keyword

nonn

Author

Clark Kimberling, Jan 22 2012

Status

approved