A204979 Least k such that n divides 2^(k-1)-2^(j-1) for some j satisfying 1<=j<k.

2, 3, 3, 4, 5, 4, 4, 5, 7, 6, 11, 5, 13, 5, 5, 6, 9, 8, 19, 7, 7, 12, 12, 6, 21, 14, 19, 6, 29, 6, 6, 7, 11, 10, 13, 9, 37, 20, 13, 8, 21, 8, 15, 13, 13, 13, 24, 7, 22, 22

Offset

1,1

Comments

See A204892 for a discussion and guide to related sequences.

Links

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

Example

1 divides 2^2-2^1, so a(1)=2
2 divides 2^3-2^2, so a(2)=3
3 divides 2^3-2^1, so a(3)=3
4 divides 2^4-2^3, so a(4)=4

Mathematica

s[n_] := s[n] = 2^(n - 1); z1 = 800; z2 = 50;
Table[s[n], {n, 1, 30}]       (* A000079 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]       (* A130328 *)
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}]       (* A204939 *)
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}]       (* A204979 *)
Table[j[n], {n, 1, z2}]       (* A001511 ? *)
Table[s[k[n]], {n, 1, z2}]    (* A204981 *)
Table[s[j[n]], {n, 1, z2}]    (* A006519 ? *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204983 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204984 *)

Crossrefs

Cf. A000079, A204892.

Sequence in context: A244929 A302920 A280386 * A243351 A071585 A328801

Adjacent sequences: A204976 A204977 A204978 * A204980 A204981 A204982

Keyword

nonn

Author

Clark Kimberling, Jan 21 2012

Status

approved