A205138 Least k such that n divides s(k)-s(j) for some j satisfying 1<=j<k, where s(j)=j(3j-1)/2, the j-th pentagonal number.

2, 2, 4, 2, 4, 5, 3, 6, 10, 4, 3, 7, 5, 8, 5, 6, 4, 10, 7, 8, 4, 8, 5, 7, 6, 13, 28, 9, 6, 5, 11, 22, 9, 5, 7, 10, 13, 18, 6, 8, 8, 11, 15, 15, 13, 6, 9, 7, 13, 6, 13, 15, 10, 29, 10, 9, 8, 7, 11, 15

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] = n (3 n - 1)/2; z1 = 500; z2 = 60;
Table[s[n], {n, 1, 30}]   (* A000326, pentagonal *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]    (* A205136 *)
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}]    (* A205137 *)
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}]        (* A205138 *)
Table[j[n], {n, 1, z2}]        (* A205139 *)
Table[s[k[n]], {n, 1, z2}]     (* A205140 *)
Table[s[j[n]], {n, 1, z2}]     (* A205141 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]      (* A205142 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}]  (* A205143 *)

Crossrefs

Cf. A000326, A204892, A205143.

Sequence in context: A323381 A060609 A330882 * A233763 A109526 A309894

Adjacent sequences: A205135 A205136 A205137 * A205139 A205140 A205141

Keyword

nonn

Author

Clark Kimberling, Jan 25 2012

Status

approved