A205378 Least k such that n divides s(k)-s(j) for some j<k, where s(j)=(2j-1)^2.

2, 2, 3, 2, 4, 3, 5, 2, 5, 4, 7, 3, 8, 5, 6, 3, 10, 5, 11, 4, 7, 7, 13, 3, 8, 8, 8, 5, 16, 6, 17, 5, 9, 10, 9, 5, 20, 11, 10, 4, 22, 7, 23, 7, 10, 13, 25, 4, 11, 8, 12, 8, 28, 8, 11, 5, 13, 16, 31, 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 - 1)^2; z1 = 600; z2 = 60;
Table[s[n], {n, 1, 30}]     (* A016754, odd squares *)
u[m_] :=  u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]     (* A205376 *)
%/8                         (* A049777 *)
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}]     (* A205377 *)
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}]     (* A205378 *)
Table[j[n], {n, 1, z2}]     (* A205379 *)
Table[s[k[n]], {n, 1, z2}]  (* A205380 *)
Table[s[j[n]], {n, 1, z2}]  (* A205381 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]      (* A205382 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}]  (* A205383 *)

Crossrefs

Cf. A016754, A204892, A205383.

Sequence in context: A361511 A345147 A214126 * A323889 A286378 A331745

Adjacent sequences: A205375 A205376 A205377 * A205379 A205380 A205381

Keyword

nonn

Author

Clark Kimberling, Jan 26 2012

Status

approved