cp's OEIS Frontend

A105774 A "fractal" transform of the Fibonacci numbers: a(1)=1; then if F(n) < k <= F(n+1), a(k) = F(n+1) - a(k - F(n)) where F(n) = A000045(n).

%I A105774 #35 Jun 23 2024 10:32:59
%S A105774 1,1,2,4,4,7,7,6,12,12,11,9,9,20,20,19,17,17,14,14,15,33,33,32,30,30,
%T A105774 27,27,28,22,22,23,25,25,54,54,53,51,51,48,48,49,43,43,44,46,46,35,35,
%U A105774 36,38,38,41,41,40,88,88,87,85,85,82,82,83,77,77,78,80,80,69,69,70,72,72
%N A105774 A "fractal" transform of the Fibonacci numbers: a(1)=1; then if F(n) < k <= F(n+1), a(k) = F(n+1) - a(k - F(n)) where F(n) = A000045(n).
%C A105774 Let tau = (1+sqrt(5))/2; then the missing numbers 3,5,8,10,13,16,18,21,... are given by round(tau^2*k) for k > 0 (A004937).
%C A105774 Indices n such that a(n) = a(n+1) are given by floor(tau^2*k) - 1 for k > 0 (A003622).
%C A105774 Numbers n such that a(n) differs from a(n+1) are given by floor(tau*k+1/tau) for k > 0 (A022342).
%C A105774 Indices n giving isolated terms (a(n) differs from a(n-1) and a(n+1)) are given by floor(tau*floor(tau^2*k)) for k > 0 (A003623).
%C A105774 Remove 0's from the first differences of sorted values; then you get a version of the infinite Fibonacci word (A001468). I.e., sorted values are 1,1,2,4,4,6,7,7,9,9,11,12,12,..., first differences are 0,1,2,0,2,1,0,2,0,2,1,0,2,0,1,...; removing 0's gives 1,2,2,1,2,2,1,2,1,2,2,1,2,1,2,2,1,2,... #{ k : a(k)=k}=infty.
%H A105774 Antti Karttunen, <a href="/A105774/b105774.txt">Table of n, a(n) for n = 1..10946</a>
%H A105774 Benoit Cloitre and Jeffrey Shallit, <a href="https://arxiv.org/abs/2312.11706">Some Fibonacci-Related Sequences</a>, arXiv:2312.11706 [math.CO], 2023.
%H A105774 Jeffrey Shallit, <a href="https://flat.fc.up.pt/2024-05-08-JeffreyShallit-using_automata_to_prove_theorems_about_sequences/">Using automata to prove theorems about sequences</a>, One FLAT World Seminar, 2024.
%F A105774 a(A000045(n)) = A006498(n-1) for n >= 1. - Typo corrected by _Antti Karttunen_, Mar 17 2017
%F A105774 limsup a(n)/n = tau and liminf a(n)/n = (tau+2)/5 where tau = (1+sqrt(5))/2. - Corrected by _Jeffrey Shallit_, Dec 17 2023
%F A105774 a(n) mod 2 = A085002(n) - _Benoit Cloitre_, May 10 2005
%F A105774 a(1) = 1; for n > 1, a(n) = A000045(2+A072649(n-1)) - a(n-A000045(1 + A072649(n-1))). - _Antti Karttunen_, Mar 17 2017
%e A105774 For 1 = F(2) < k <= F(3) = 2 the rule gives a(2) = 2 - a(1) = 1 ... if 5 = F(5) < k <= F(6) = 8 the rule forces a(6) = 8 - a(6-5) = 8 - a(1) = 7; a(7) = 8 - a(2) = 7; a(8) = 8 - a(3) = 6.
%o A105774 (Scheme, with memoization-macro definec)
%o A105774 (definec (A105774 n) (if (= 1 n) n (- (A000045 (+ 2 (A072649 (- n 1)))) (A105774 (- n (A000045 (+ 1 (A072649 (- n 1)))))))))
%o A105774 ;; _Antti Karttunen_, Mar 17 2017
%Y A105774 Cf. A000045, A001468, A003622, A003623, A004937, A006498, A022342, A072649, A085002, A105669, A105670, A105672, A093347, A093348, A283766.
%K A105774 nonn
%O A105774 1,3
%A A105774 _Benoit Cloitre_, May 04 2005