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Positions of 1 in A189820; complement of A003278.
3

%I #16 Dec 05 2024 15:33:37

%S 3,6,7,8,9,12,15,16,17,18,19,20,21,22,23,24,25,26,27,30,33,34,35,36,

%T 39,42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,

%U 64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,84

%N Positions of 1 in A189820; complement of A003278.

%H G. C. Greubel, <a href="/A189822/b189822.txt">Table of n, a(n) for n = 1..10000</a>

%t a[1] = 0; h = 50;

%t Table[a[3 k - 2] = a[k], {k, 1, h}];

%t Table[a[3 k - 1] = a[k], {k, 1, h}];

%t Table[a[3 k] = 1, {k, 1, h}];

%t Flatten[Position[%%, 1]]

%o (Python)

%o from gmpy2 import digits

%o def A189822(n):

%o def f(x):

%o l = (s:=digits(x-1,3)).find('2')

%o if l >= 0: s = s[:l]+'1'*(len(s)-l)

%o return n+1+int(s,2)

%o m, k = n, f(n)

%o while m != k: m, k = k, f(k)

%o return m # _Chai Wah Wu_, Dec 05 2024

%Y Cf. A081610, A189820, A003278.

%K nonn,easy,changed

%O 1,1

%A _Clark Kimberling_, Apr 28 2011