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A205840
[s(k)-s(j)]/2, where the pairs (k,j) are given by A205837 and A205838.
45
1, 2, 1, 3, 6, 5, 4, 10, 9, 8, 4, 16, 13, 27, 26, 25, 21, 17, 44, 43, 42, 38, 34, 17, 71, 68, 55, 116, 115, 114, 110, 106, 89, 72, 188, 187, 186, 182, 178, 161, 144, 72, 304, 301, 288, 233, 493, 492, 491, 487, 483, 466, 449, 377, 305, 798, 797, 796, 792, 788
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
2....A205837....A205838....A205839......A205840
3....A205842....A205843....A205844......A205845
4....A205847....A205848....A205849......A205850
5....A205852....A205853....A205854......A205855
6....A205857....A205858....A205859......A205860
7....A205862....A205863....A205864......A205865
8....A205867....A205868....A205869......A205870
9....A205872....A205873....A205874......A205875
10...A205877....A205878....A205879......A205880
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