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%I #14 Jun 09 2017 03:36:45
%S 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,
%T 83,79,76,74,70,67,63,61,58,56,143,138,134,130,126,123,121,117,113,
%U 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
%N The infinite trunk of Zeckendorf (Fibonacci) beanstalk, with reversed subsections.
%C 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).
%H Antti Karttunen, <a href="/A261076/b261076.txt">Table of n, a(n) for n = 0..11817; rows 0 .. 23 of the irregular table.</a>
%H Indranil Ghosh, <a href="/A261076/a261076.txt">Python program to generate the sequence</a>
%F 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)).
%F As a composition:
%F a(n) = A219648(A261102(n)).
%e As an irregular table, the sequence looks like:
%e 0;
%e 1;
%e 2;
%e 4;
%e 7, 5;
%e 12, 9;
%e 20, 17, 14;
%e 33, 29, 27, 24, 22;
%e 54, 50, 47, 45, 42, 40, 37, 35;
%e ...
%e After zero, each row n is A261091(n) elements long.
%o (Scheme, with memoization-macro definec)
%o (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)))))
%Y Cf. A000045, A000071, A007895, A072649, A219641, A219648, A261091, A261102.
%Y Cf. A218616 (analogous sequence for base-2).
%K nonn,tabf
%O 0,3
%A _Antti Karttunen_, Aug 09 2015
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