login
Positions of 1 in A116178; complement of A189636.
5

%I #18 Jun 20 2022 13:57:54

%S 3,6,8,9,12,15,17,18,21,23,24,26,27,30,33,35,36,39,42,44,45,48,50,51,

%T 53,54,57,60,62,63,66,68,69,71,72,75,77,78,80,81,84,87,89,90,93,96,98,

%U 99,102,104,105,107,108,111,114,116,117,120,123,125,126,129,131,132,134,135,138,141,143,144,147,149,150,152,153,156,158,159,161,162,165,168,170,171,174

%N Positions of 1 in A116178; complement of A189636.

%C See A116178.

%H Michael De Vlieger, <a href="/A189637/b189637.txt">Table of n, a(n) for n = 1..10000</a>

%H F. M. Dekking, <a href="https://arxiv.org/abs/2001.08915">Permutations of N generated by left-right filling algorithms</a>, arXiv:2001.08915 [math.CO], 2020.

%F A340407(a(n)) = A254046(a(n)). Discovered with help from sequencedb.net. - _Thomas Scheuerle_, Jun 20 2022

%t t = Nest[Flatten[# /. {0->{0,0,1}, 1->{0,1,1}}] &, {0}, 5] (* A116178 *)

%t f[n_] := t[[n]]

%t Flatten[Position[t, 0]] (* A189636 *)

%t Flatten[Position[t, 1]] (* A189637 *)

%t s[n_] := Sum[f[i], {i, 1, n}]; s[0] = 0;

%t Table[s[n], {n, 1, 120}] (* A189638 *)

%Y Cf. A116178, A189636, A189638, A254046, A340407.

%K nonn

%O 1,1

%A _Clark Kimberling_, Apr 24 2011