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%I #31 May 24 2019 11:23:22
%S 2,5,9,12,16,20,23,27,30,34,38,41,45,49,52,56,59,63,67,70,74,77,81,85,
%T 88,92,96,99,103,106,110,114,117,121,125,128,132,135,139,143,146,150,
%U 153,157,161,164,168,172,175,179,182,186,190,193,197,200,204,208
%N Positions of 1 in A284749.
%C This sequence and A214971 and A284625 partition the positive integers into sequences with slopes t = (5+sqrt(5))/2, u = (5+sqrt(5))/2, v = sqrt(5), where 1/t + 1/u + 1/v = 1. The positions of 0 in A284749 are given by A214971, and of 2, by A284625.
%H Clark Kimberling, <a href="/A284624/b284624.txt">Table of n, a(n) for n = 1..10000</a>
%H J.-P. Allouche, F. M. Dekking, <a href="https://arxiv.org/abs/1809.03424">Generalized Beatty sequences and complementary triples</a>, arXiv:1809.03424v3 [math.NT], 2018-2019.
%F Conjectured: a(n) = 1 + A214971(n).
%F Proof of this conjecture: it is easy to see that the infinite Fibonacci word A003849 is a concatenation of the words 01 and 001. So if we replace all 001 by 2, only the 01 remain, i.e., every 0 is directly followed by 1. - _Michel Dekking_, Aug 19 2018
%F a(n) = floor((n-1)*phi) + 2*(n-1) + 2 (Theorem 31 in Allouche and Dekking). - _Michel Dekking_, Oct 08 2018
%e As a word, A284749 = 012012201201220122..., in which 1 is in positions 2,5,9,12,...
%t s = Nest[Flatten[# /. {0 -> {0, 1}, 1 -> {0}}] &, {0}, 13] (* A003849 *)
%t w = StringJoin[Map[ToString, s]]
%t w1 = StringReplace[w, {"001" -> "2"}]
%t st = ToCharacterCode[w1] - 48 (* A284749 *)
%t Flatten[Position[st, 0]] (* A214971 conjectured *)
%t Flatten[Position[st, 1]] (* A284624 *)
%t Flatten[Position[st, 2]] (* A284625 *)
%Y Cf. A214971, A284625, A284749.
%K nonn,easy
%O 1,1
%A _Clark Kimberling_, May 02 2017