A205130 Least k such that n divides s(k)-s(j) for some j satisfying 1<=j<k, where s(j)=2*j^2-j, the j-th hexagonal number.

2, 3, 3, 5, 2, 5, 3, 9, 3, 5, 4, 6, 4, 3, 5, 17, 5, 7, 6, 6, 6, 4, 7, 11, 7, 11, 4, 11, 8, 5, 9, 33, 9, 14, 8, 9, 10, 6, 5, 13, 11, 12, 12, 5, 7, 7, 13, 18, 9, 14, 6, 12, 14, 8, 11, 13, 8, 23, 16, 6

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] = 2 n^2 - n; z1 = 500; z2 = 60;
Table[s[n], {n, 1, 30}] (* A000384, hexagonal numbers *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}] (* A205128 *)
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}]        (* A205129 *)
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}]        (* A205130 *)
Table[j[n], {n, 1, z2}]        (* A205131 *)
Table[s[k[n]], {n, 1, z2}]     (* A205132 *)
Table[s[j[n]], {n, 1, z2}]     (* A205133 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A205134 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A205135 *)

Crossrefs

Cf. A000384, A204892, A205135.

Sequence in context: A341701 A049272 A181483 * A324799 A069461 A329071

Adjacent sequences: A205127 A205128 A205129 * A205131 A205132 A205133

Keyword

nonn

Author

Clark Kimberling, Jan 25 2012

Status

approved