A205122 Least k such that n divides s(k)-s(j) for some j satisfying 1<=j<k, where s(j)=j*(2^(j-1)).

2, 3, 2, 3, 4, 5, 4, 3, 6, 4, 3, 5, 7, 4, 6, 5, 5, 6, 5, 4, 10, 8, 7, 5, 9, 7, 11, 4, 9, 6, 4, 6, 8, 5, 10, 6, 7, 5, 11, 6, 13, 10, 16, 8, 6, 7, 6, 5, 9, 9, 8, 7, 10, 11, 13, 6, 14, 9, 8, 6, 16, 8, 10, 7, 17, 9, 15, 5, 12, 10

Offset

1,1

Comments

See A204892 for a discussion and guide to related sequences.

Links

Table of n, a(n) for n=1..70.

Mathematica

s[n_] := s[n] = n*2^(n - 1); z1 = 250; z2 = 70;
Table[s[n], {n, 1, 30}]   (* A001787 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]   (* A205120 *)
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}]   (* A205121 *)
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}]            (* A205122 *)
Table[j[n], {n, 1, z2}]            (* A205123 *)
Table[s[k[n]], {n, 1, z2}]         (* A205124 *)
Table[s[j[n]], {n, 1, z2}]         (* A205125 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A205126 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205127 *)

Crossrefs

Cf. A001787, A204892, A205127.

Sequence in context: A138554 A063772 A209302 * A174863 A064289 A078759

Adjacent sequences: A205119 A205120 A205121 * A205123 A205124 A205125

Keyword

nonn

Author

Clark Kimberling, Jan 25 2012

Status

approved