A205845 [s(k)-s(j)]/3, where the pairs (k,j) are given by A205842 and A205843, and s(k) denotes the (k+1)-st Fibonacci number.

1, 2, 1, 4, 6, 11, 7, 18, 14, 7, 29, 28, 27, 47, 41, 77, 76, 75, 48, 125, 124, 123, 96, 48, 203, 199, 192, 185, 328, 322, 281, 532, 528, 521, 514, 329, 861, 857, 850, 843, 658, 329, 1393, 1392, 1391, 1364, 1316, 1268, 2254, 2248, 2207, 1926, 3648

Offset

1,2

Comments

For a guide to related sequences, see A205840.

The first six terms match these differences:

s(4)-s(2) = 5-2 = 3 = 3*1

s(5)-s(2) = 8-2 = 6 = 3*2

s(5)-s(4) = 8-5 = 3 = 3*1

s(6)-s(1) = 13-1 = 12 = 3*4

s(7)-s(3) = 21-3 = 18 = 3*6

s(8)-s(1) = 34-1 = 33 + 3*11

(See the program at A205842.)

Links

Table of n, a(n) for n=1..53.

Example

The first six terms match these differences:
s(4)-s(2) = 5-2 = 3 = 3*1
s(5)-s(2) = 8-2 = 6 = 3*2
s(5)-s(4) = 8-5 = 3 = 3*1
s(6)-s(1) = 13-1 = 12 = 3*4
s(7)-s(3) = 21-3 = 18 = 3*6
s(8)-s(1) = 34-1 = 33 + 3*11

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 = 3; t = d[c]       (* A205841 *)
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}]      (* A205842 *)
Table[j[n], {n, 1, z2}]      (* A205843 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}] (* A205844 *)
Table[(s[k[n]] - s[j[n]])/c, {n, 1, z2}] (* A205845 *)

Crossrefs

Cf. A204890, A205842, A205845.

Sequence in context: A366043 A054408 A285637 * A034424 A095012 A192781

Adjacent sequences: A205842 A205843 A205844 * A205846 A205847 A205848

Keyword

nonn

Author

Clark Kimberling, Feb 01 2012

Status

approved