OFFSET
1,2
COMMENTS
Let s(n)=F(n+1), where F=A000045 (Fibonacci numbers), so that s=(1,2,3,5,8,13,21,...). If c is a positive integer, there are infinitely many pairs (k,j) such that c divides s(k)-s(j). The set of differences s(k)-s(j) is ordered as a sequence at A204922. Guide to related sequences:
c....k..........j..........s(k)-s(j)....[s(k)-s(j)]/c
EXAMPLE
The first six terms match these differences:
s(3)-s(1) = 3-1 = 2 = 2*1
s(4)-s(1) = 5-1 = 4 = 2*2
s(4)-s(3) = 5-3 = 2 = 2*1
s(5)-s(2) = 8-2 = 6 = 2*3
s(6)-s(1) = 13-1 = 12 = 2*6
s(6)-s(3) = 13-3 = 10 = 2*5
MATHEMATICA
s[n_] := s[n] = Fibonacci[n + 1]; z1 = 400; z2 = 60;
f[n_] := f[n] = Floor[(-1 + Sqrt[8 n - 7])/2];
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] = Delete[w[n], Position[w[n], 0]]
c = 2; t = d[c] (* A205556 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 t[[n]] - 1])/2]
j[n_] := j[n] = t[[n]] - f[t][[n]] (f[t[[n]]] + 1)/2
Table[k[n], {n, 1, z2}] (* A205837 *)
Table[j[n], {n, 1, z2}] (* A205838 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A205839 *)
Table[(s[k[n]] - s[j[n]])/c, {n, 1, z2}] (* A205840 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Clark Kimberling, Feb 01 2012
STATUS
approved