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Positions of 1 in A285076; complement of A285077.
3

%I #11 Sep 24 2017 02:07:28

%S 1,4,6,9,11,14,16,18,21,23,26,28,30,33,35,38,40,43,45,47,50,52,55,57,

%T 59,62,64,67,69,72,74,76,79,81,84,86,88,91,93,96,98,100,103,105,108,

%U 110,113,115,117,120,122,125,127,129,132,134,137,139,142,144,146

%N Positions of 1 in A285076; complement of A285077.

%C Conjecture: -1 + sqrt(2) < n*r - a(n) < sqrt(2) for n>=1, where r = 1 + sqrt(2).

%C Appears to differ from A285075 only at a(1). - _R. J. Mathar_, Apr 24 2017

%C Both these conjectures can be proved (for n>=2), see the Comments of A285073 and A285075 - _Michel Dekking_, Sep 24 2017

%H Clark Kimberling, <a href="/A285078/b285078.txt">Table of n, a(n) for n = 1..10000</a>

%e As a word, A285076 = 100101001010..., in which 1 is in positions 1,4,6,9,11,...

%t s = Nest[Flatten[# /. {0 -> {1, 0}, 1 -> {0, 1, 0}}] &, {0}, 13]; (* A285076 *)

%t Flatten[Position[s, 0]]; (* A285077 *)

%t Flatten[Position[s, 1]]; (* A285078 *)

%Y Cf. A285074, A285076, A285078.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Apr 19 2017