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.

A353656 Number of terms in the Lucas-Fibonacci representation of n.

Original entry on oeis.org

1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 3, 2, 3, 4, 2, 3, 4, 1, 2, 2, 2, 3, 2, 3, 4, 2, 3, 4, 3, 3, 2, 3, 4, 3, 3, 1, 2, 2, 2, 3, 2, 3, 4, 2, 3, 4, 3, 3, 2, 3, 4, 3, 3, 4, 4, 3, 2, 3, 4, 3, 3, 4, 4, 3, 1, 2, 2, 2, 3, 2, 3, 4, 2, 3, 4
Offset: 1

Views

Author

Clark Kimberling, May 04 2022

Keywords

Comments

The Lucas-Fibonacci representation of n, denoted by LF(n), is defined for n>=1 as the sum t(1) + t(2) + ... + t(k), where t(1) is the greatest Lucas number (A000032(n), with n >= 1) that is <= n, and t(2) is the greatest Fibonacci number (A000045(n), with n >= 2) that is <= n - t(1), and so on; that is, the greedy algorithm is applied to find successive greatest Lucas and Fibonacci numbers, in alternating order, with sum n. (See Example.)

Examples

			   n     LF(n)
   1  =  1
   2  =  1 + 1
   3  =  3
   4  =  4
   5  =  4 + 1
   6  =  4 + 2
  17  =  11 + 5 + 1
  66  =  47 + 13 + 4 + 2
		

Crossrefs

Programs

  • Mathematica
    z = 120; fib = Map[Fibonacci, Range[2, 51]];
    luc = Map[LucasL, Range[1, 50]];
    t = Map[(n = #; fl = {}; f = 0; l = 0;
         While[IntegerQ[l], n = n - f - l;
          f = fib[[NestWhile[# + 1 &, 1, fib[[#]] <= n &] - 1]];
          l = luc[[NestWhile[# + 1 &, 1, luc[[#]] <= n - f &] - 1]];
          AppendTo[fl, {f, l}]];
         {Total[#], #} &[Select[Flatten[fl], IntegerQ]]) &, Range[z]];
    u = Take[Map[Last, t], z];
    u1 = Map[Length, u]  (* A353655 *)
    t = Map[(n = #; lf = {}; f = 0; l = 0;
         While[IntegerQ[f], n = n - l - f;
          l = luc[[NestWhile[# + 1 &, 1, luc[[#]] <= n &] - 1]];
          f = fib[[NestWhile[# + 1 &, 1, fib[[#]] <= n - l &] - 1]];
          AppendTo[lf, {l, f}]];
         {Total[#], #} &[Select[Flatten[lf], IntegerQ]]) &, Range[z]];
    v = Take[Map[Last, t], z];
    v1 = Map[Length, v]   (* A353656 *)
    u1 - v1  (* A353657 *)
    (* Peter J. C. Moses, May 04 2022 *)

A353657 a(n) = A353655(n)- A353656(n).

Original entry on oeis.org

0, -1, 0, 1, -1, 0, 2, -1, 0, 1, 1, 0, -1, 0, 0, 0, -1, 2, 1, 0, -1, -1, 1, -1, -2, 1, 0, -2, 2, 1, 1, 0, 0, -1, -1, -1, 0, -1, -1, 0, -1, 1, 0, -1, -1, 0, 2, 1, 2, 1, 1, 0, 0, -1, -1, -1, -1, -1, -1, 1, 0, -2, 0, 0, -1, -2, 0, 1, 0, 0, 0, 1, -2, -1, 0, 2, 2
Offset: 1

Views

Author

Clark Kimberling, May 04 2022

Keywords

Comments

Conjectures: a(n) = 0 for infinitely many n, and (a(n)) is unbounded below and above.

Examples

			a(7) because A353655(u) = 3 and A353656(7) = 1, since the Fibonacci-Lucas representation of 7 is FL(7) = 5 + 1 + 1, and the Lucas-Fibonacci representation of 7 is LF(7) = 7.
		

Crossrefs

Programs

  • Mathematica
    z = 120; fib = Map[Fibonacci, Range[2, 51]];
    luc = Map[LucasL, Range[1, 50]];
    t = Map[(n = #; fl = {}; f = 0; l = 0;
         While[IntegerQ[l], n = n - f - l;
          f = fib[[NestWhile[# + 1 &, 1, fib[[#]] <= n &] - 1]];
          l = luc[[NestWhile[# + 1 &, 1, luc[[#]] <= n - f &] - 1]];
          AppendTo[fl, {f, l}]];
         {Total[#], #} &[Select[Flatten[fl], IntegerQ]]) &, Range[z]];
    u = Take[Map[Last, t], z];
    u1 = Map[Length, u]  (* A353655 *)
    t = Map[(n = #; lf = {}; f = 0; l = 0;
         While[IntegerQ[f], n = n - l - f;
          l = luc[[NestWhile[# + 1 &, 1, luc[[#]] <= n &] - 1]];
          f = fib[[NestWhile[# + 1 &, 1, fib[[#]] <= n - l &] - 1]];
          AppendTo[lf, {l, f}]];
         {Total[#], #} &[Select[Flatten[lf], IntegerQ]]) &, Range[z]];
    v = Take[Map[Last, t], z];
    v1 = Map[Length, v]   (* A353656 *)
    u1 - v1  (* A353657 *)
    (* Peter J. C. Moses *)

A353659 Rectangular array read by downwards antidiagonals: row k lists the numbers whose Lucas-Fibonacci representation has k terms.

Original entry on oeis.org

1, 3, 2, 4, 5, 15, 7, 6, 17, 25, 11, 8, 22, 28, 172, 18, 9, 24, 36, 174, 279, 29, 10, 27, 39, 193, 282, 1913, 47, 12, 33, 44, 195, 313, 1915, 3096, 76, 13, 35, 54, 248, 316, 1934, 3099, 21221, 123, 14, 38, 57, 250, 402, 1936, 3130, 21223, 34337
Offset: 1

Views

Author

Clark Kimberling, May 04 2022

Keywords

Comments

The Lucas-Fibonacci representation of n, denoted by LF(n), is defined for n>=1 as the sum t(1) + t(2) + ... + t(k), where t(1) is the greatest Lucas number (A000032(n), with n >= 1) that is <= n, and t(2) is the greatest Fibonacci number (A000045(n), with n >= 2) that is <= n - t(1), and so on; that is, the greedy algorithm is applied to find successive greatest Lucas and Fibonacci numbers, in alternating order, with sum n. Every positive integer occurs exactly once in this array.

Examples

			Northwest corner:
     1     3     4     7    11    18    29    47    76   123
     2     5     6     8     9    10    12    13    14    16
    15    17    22    24    27    33    35    38    40    41
    25    28    36    39    44    54    57    62    65    66
   172   174   193   195   248   250   269   271   276   278
   279   282   313   316   402   405   436   439   447   450
  1913  1915  1934  1936  2146  2148  2167  2169  2756  2758
  3096  3099  3130  3133  3473  3476  3507  3510  4460  4463
		

Crossrefs

Programs

  • Mathematica
    fib = Map[Fibonacci, Range[2, 51]];
    luc = Map[LucasL, Range[1, 50]];
    t = Map[(n = #; lf = {}; f = 0; l = 0;
         While[IntegerQ[f], n = n - l - f;
          l = luc[[NestWhile[# + 1 &, 1, luc[[#]] <= n &] - 1]];
          f = fib[[NestWhile[# + 1 &, 1, fib[[#]] <= n - l &] - 1]];
          AppendTo[lf, {l, f}]];
         {Total[#], #} &[Select[Flatten[lf], IntegerQ]]) &, Range[50000]];
    Length[t];
    u = Table[Length[t[[n]][[2]]], {n, 1, Length[t]}];
    Take[u, 150]
    TableForm[Table[Flatten[Position[u, k]], {k, 1, 11}]];
    w[k_, n_] := Flatten[Position[u, k]][[n]]
    Table[w[n - k + 1, k], {n, 11}, {k, n, 1, -1}] // Flatten
    (* Peter J. C. Moses, May 04 2022 *)
Showing 1-3 of 3 results.