cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

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

Views

Author

Clark Kimberling, Feb 01 2012

Keywords

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

Examples

			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
		

Crossrefs

Programs

  • 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 *)

A205839 s(k)-s(j), where the pairs (k,j) are given by A205837 and A205838.

Original entry on oeis.org

2, 4, 2, 6, 12, 10, 8, 20, 18, 16, 8, 32, 26, 54, 52, 50, 42, 34, 88, 86, 84, 76, 68, 34, 142, 136, 110, 232, 230, 228, 220, 212, 178, 144, 376, 374, 372, 364, 356, 322, 288, 144, 608, 602, 576, 466, 986, 984, 982, 974, 966, 932, 898, 754, 610, 1596
Offset: 1

Views

Author

Clark Kimberling, Feb 01 2012

Keywords

Comments

For a guide to related sequences, see A205840.

Examples

			(See the example at A205587.)
		

Crossrefs

Programs

  • Mathematica
    (See the program at A205587.)

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

Views

Author

Clark Kimberling, Feb 01 2012

Keywords

Comments

For a guide to related sequences, see A205840.

Examples

			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
		

Crossrefs

Programs

  • 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 *)
Showing 1-3 of 3 results.