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Positions of 1 in A285384; complement of A072939.
3

%I #14 Jan 29 2025 18:21:45

%S 1,2,4,5,6,8,10,12,13,14,16,17,18,20,21,22,24,26,28,29,30,32,34,36,37,

%T 38,40,42,44,45,46,48,49,50,52,53,54,56,58,60,61,62,64,65,66,68,69,70,

%U 72,74,76,77,78,80,81,82,84,85,86,88,90,92,93,94,96,98

%N Positions of 1 in A285384; complement of A072939.

%H Clark Kimberling, <a href="/A285385/b285385.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A171946(n) for n>=2.

%F a(n + 1) = a(n) + A096268(a(n) - 1) + 1 for all n > 0. (conjecture) - _Velin Yanev_, Mar 23 2019

%e As a word, A285384 = 11011101..., in which the positions of 1 are 1,2,4,5,6,8,...

%t s = Nest[Flatten[# /. {0 -> {1, 1}, 1 -> {0, 1}}] &, {0}, 16] (* A285384 *)

%t Flatten[Position[s, 0]] (* A072939 *)

%t Flatten[Position[s, 1]] (* A285385 *)

%o (Python)

%o def A285385(n):

%o def f(x):

%o c, s = n, bin(x-1)[2:]

%o l = len(s)

%o for i in range(l&1,l,2):

%o c += int(s[i])+int('0'+s[:i],2)

%o return c

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

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

%o return m # _Chai Wah Wu_, Jan 29 2025

%Y Cf. A072939, A171946, A285384.

%K nonn,easy,changed

%O 1,2

%A _Clark Kimberling_, Apr 26 2017