A181631 Triangle by rows, number of leading 1's in the maximal Fibonacci representation (A104326).

1, 1, 2, 1, 2, 3, 1, 1, 2, 3, 4, 1, 1, 1, 2, 2, 3, 4, 5, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7, 8

Offset

1,3

Comments

Row sums = A001911: (1, 3, 6, 11, 19, 32, ...).

A112310 = number of 1's in the maximal Fibonacci representation, which has headings of (..., 8, 5, 3, 2, 1) filling entries from the right to left; as opposed to the minimal Fibonacci representation (A014417) which starts from the left. For example, 8 in maximal = 1011 = (5 + 2 + 1) whereas in minimal = (10000) = (8).

Rows have (1, 2, 3, 5, 8, ...) terms.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Formula

a(n) = A090996(A003754(n+1)). - Amiram Eldar, May 31 2025

Example

First few rows of the triangle:
  1;
  1, 2;
  1, 2, 3;
  1, 1, 2, 3, 4;
  1, 1, 1, 2, 2, 3, 4, 5;
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 6;
  ...
Example: a(14) = 1 since 14 in the maximal Fibonacci representation is 10111.

Mathematica

f[s_] := Module[{i = FirstPosition[s, 0]}, If[MissingQ[i], Length[s], i[[1]] - 1]]; f /@ Select[IntegerDigits[#, 2] & /@ Range[300], SequencePosition[#, {0, 0}] == {} &] (* Amiram Eldar, May 31 2025 *)

Crossrefs

Cf. A000045 (row lengths), A003754, A001911 (row sums), A014417, A090996, A104326, A112310.

Sequence in context: A322984 A277822 A327616 * A348390 A133674 A348245

Adjacent sequences: A181628 A181629 A181630 * A181632 A181633 A181634

Keyword

nonn,tabf

Author

Gary W. Adamson, Nov 02 2010

Extensions

More terms from Amiram Eldar, May 31 2025

Status

approved