A261076 The infinite trunk of Zeckendorf (Fibonacci) beanstalk, with reversed subsections.

0, 1, 2, 4, 7, 5, 12, 9, 20, 17, 14, 33, 29, 27, 24, 22, 54, 50, 47, 45, 42, 40, 37, 35, 88, 83, 79, 76, 74, 70, 67, 63, 61, 58, 56, 143, 138, 134, 130, 126, 123, 121, 117, 113, 110, 108, 104, 101, 97, 95, 92, 90, 232, 226, 221, 217, 213, 209, 205, 201, 198, 193, 189, 185, 181, 178, 176, 172, 168, 165, 163, 159, 156, 152, 150, 147, 145

Offset: 0

Author

Antti Karttunen, Aug 09 2015

Keywords

• nonn
• tabf

Comments

This can be viewed as an irregular table: after the initial zero on row 0, start each row n with k = F(n+2)-1 and subtract repeatedly the number of "1-fibits" (number of terms in Zeckendorf expansion of k) from k to get successive terms, until the number that has already been listed (which is always F(n+1)-1) is encountered, which is not listed second time, but instead, the current row is finished and the next row starts with (F(n+3))-1, with the same process repeated. Here F(n) = the n-th Fibonacci number, A000045(n).

Examples

			As an irregular table, the sequence looks like:
  0;
  1;
  2;
  4;
  7, 5;
  12, 9;
  20, 17, 14;
  33, 29, 27, 24, 22;
  54, 50, 47, 45, 42, 40, 37, 35;
  ...
After zero, each row n is A261091(n) elements long.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..11817; rows 0 .. 23 of the irregular table.
• Indranil Ghosh, Python program to generate the sequence

Crossrefs

Cf. A000045, A000071, A007895, A072649, A219641, A219648, A261091, A261102.

Cf. A218616 (analogous sequence for base-2).

Formula

For n <= 2, a(n) = n; for n >= 3, if A219641(a(n-1)) = F(k)-1 [i.e., one less than some Fibonacci number F(k)] then a(n) = F(k+2)-1, otherwise a(n) = A219641(a(n-1)).

As a composition:

a(n) = A219648(A261102(n)).