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Positions of 0 in A286922; complement of A286924.
3

%I #11 Nov 19 2024 00:48:11

%S 3,7,12,16,21,25,29,34,38,43,47,51,56,60,65,69,74,78,82,87,91,96,100,

%T 104,109,113,118,122,127,131,135,140,144,149,153,157,162,166,171,175,

%U 179,184,188,193,197,202,206,210,215,219,224,228,232,237,241,246

%N Positions of 0 in A286922; complement of A286924.

%C a(n) - a(n-1) is in {4,5} for n>=2, and a(n)/n -> 3 + sqrt(2).

%H Clark Kimberling, <a href="/A286923/b286923.txt">Table of n, a(n) for n = 1..10000</a>

%F From _Ridouane Oudra_, Nov 02 2024: (Start)

%F a(n) = floor((3+sqrt(2))*n - 1).

%F a(n) = floor(sqrt(2)*n) + 3*n - 1.

%F a(n) = A001951(n) + A008585(n) - 1

%F a(n) = A187338(n) - 1. (End)

%e As a word, A286922 = 1101110111101110111101110..., in which 0 is in positions 3,7,12,16....

%p seq(floor((3+sqrt(2))*n)-1, n=1..80); # _Ridouane Oudra_, Nov 02 2024

%t s = Nest[Flatten[# /. {0 -> 1, 1 -> {1, 1, 0}}] &, {1}, 7] (* A080764 *)

%t w = StringJoin[Map[ToString, s]]

%t w1 = StringReplace[w, {"0" -> "01"}]

%t st = ToCharacterCode[w1] - 48 ; (* A286922 *)

%t Flatten[Position[st, 0]]; (* A286923 *)

%t Flatten[Position[st, 1]]; (* A286924 *)

%Y Cf. A080764, A286922, A286924, A001951, A187338, A008585.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, May 17 2017