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A261076
The infinite trunk of Zeckendorf (Fibonacci) beanstalk, with reversed subsections.
4
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
0,3
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).
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)).
EXAMPLE
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.
PROG
(Scheme, with memoization-macro definec)
(definec (A261076 n) (cond ((<= n 2) n) ((A219641 (A261076 (- n 1))) => (lambda (maybe_next) (if (= 1 (A007895 (+ 1 maybe_next))) (+ -1 (A000045 (+ 3 (A072649 (+ 1 maybe_next))))) maybe_next)))))
CROSSREFS
Cf. A218616 (analogous sequence for base-2).
Sequence in context: A244591 A377351 A299324 * A302991 A015791 A335818
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Aug 09 2015
STATUS
approved