A204924 Least k such that n divides s(k)-s(j) for some j in [1,k), where s(k)=A000045(k+1) (Fibonacci numbers).
2, 3, 4, 4, 5, 5, 5, 6, 7, 6, 6, 6, 7, 9, 12, 7, 9, 7, 7, 7, 8, 10, 12, 12, 9, 8, 9, 10, 8, 17, 8, 8, 8, 9, 21, 12, 14, 10, 18, 17, 11, 9, 10, 10, 12, 12, 9, 12, 13, 9, 17, 9, 9, 9, 10, 17, 12, 12, 25, 22
Offset: 1
Keywords
Examples
1 divides s(2)-s(1), so a(1)=2 2 divides s(3)-s(1), so a(2)=3 3 divides s(4)-s(2), so a(3)=4 9 divides s(7)-s(3), so a(9)=7
Links
- G. C. Greubel, Table of n, a(n) for n = 1..1000
Programs
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Mathematica
s[n_] := s[n] = Fibonacci[n + 1]; z1 = 300; z2 = 60; Table[s[n], {n, 1, 30}] u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]] Table[u[m], {m, 1, z1}] (* A204922 *) 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}] (* A204923 *) 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}] (* A204924 *) Table[j[n], {n, 1, z2}] (* A204925 *) Table[s[k[n]], {n, 1, z2}] (* A204926 *) Table[s[j[n]], {n, 1, z2}] (* A204927 *) Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A204928 *) Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204929 *)
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