A205556 -id:A205556 - cp's OEIS frontend

A205840 [s(k)-s(j)]/2, where the pairs (k,j) are given by A205837 and A205838.

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

Clark Kimberling, Feb 01 2012

nonn

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
...A205837....A205838....A205839......A205840
...A205842....A205843....A205844......A205845
...A205847....A205848....A205849......A205850
...A205852....A205853....A205854......A205855
...A205857....A205858....A205859......A205860
...A205862....A205863....A205864......A205865
...A205867....A205868....A205869......A205870
...A205872....A205873....A205874......A205875
..A205877....A205878....A205879......A205880

			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

Cf. A204890, A205587, A205839.

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 *)

A205837 Numbers k for which 2 divides s(k)-s(j) for some j

Original entry on oeis.org

3, 4, 4, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16

Offset: 1

Clark Kimberling, Feb 01 2012

nonn

For a guide to related sequences, see A205840.

			The first six terms match these differences:
s(3)-s(1) = 3-1 = 2
s(4)-s(1) = 5-1 = 4
s(4)-s(3) = 5-3 = 2
s(5)-s(2) = 8-2 = 6
s(6)-s(1) = 13-1 = 12
s(6)-s(3) = 13-3 = 10

Cf. A204892, A205840, A205558.

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 *)

cp's OEIS Frontend

A205840 [s(k)-s(j)]/2, where the pairs (k,j) are given by A205837 and A205838.

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