%I #20 Mar 11 2018 10:09:06
%S 1,1,5,1,3,5,9,1,5,3,11,5,11,9,17,1,9,7,5,11,3,13,11,9,5,11,13,9,11,
%T 17,33,1,17,11,9,7,19,5,13,11,29,3,19,13,27,11,19,17,9,19,5,11,19,13,
%U 29,9,19,11,13,17,25,33,65,1,33,23,17,13,11,29,9,23,7
%N a(n) is the least positive k such that the binary representation of n appears as a substring in the binary representation of 1/k (ignoring the radix point and adding trailing zeros if necessary in case of a terminating expansion).
%C In other words, a(n) is the least k > 0 such that floor((2^i) / k) mod A062383(n) = n for some integer i >= 0.
%C This sequence is similar to A035335 for the base 2.
%C All terms are odd.
%C All terms appears infinitely many times (as a(n) equals at least a(2*n) or a(2*n + 1)).
%C See also A300475 for a similar sequence.
%H Rémy Sigrist, <a href="/A300428/b300428.txt">Table of n, a(n) for n = 1..8191</a>
%H Rémy Sigrist, <a href="/A300428/a300428.gp.txt">PARI program for A300428</a>
%F a(2^k) = 1 for any k >= 0.
%F a(2^k - 1) = 2^k + 1 for any k > 1.
%F a(A000975(k)) = 3 for any k > 2.
%F a(A033138(k)) = 7 for any k > 4.
%F a(n) <= A300475(n) for any n > 0.
%e The first terms, alongside the binary representation of 1/a(n) with the earliest occurrence of the binary representation of n in parentheses, are:
%e n a(n) bin(1/a(n))
%e -- ---- -----------
%e 1 1 (1).000...
%e 2 1 (1.0)000...
%e 3 5 0.00(11)001...
%e 4 1 (1.00)000...
%e 5 3 0.0(101)010...
%e 6 5 0.00(110)011...
%e 7 9 0.000(111)000...
%e 8 1 (1.000)000...
%e 9 5 0.001(1001)100...
%e 10 3 0.0(1010)101...
%e 11 11 0.000(1011)101...
%e 12 5 0.00(1100)110...
%e 13 11 0.000101(1101)000...
%e 14 9 0.000(1110)001...
%e 15 17 0.0000(1111)000...
%e 16 1 (1.0000)000...
%e 17 9 0.00011(10001)110...
%e 18 7 0.00(10010)010...
%e 19 5 0.001(10011)001...
%e 20 11 0.0001011(10100)010...
%o (PARI) See Links section.
%Y Cf. A000975, A033138, A035335, A062383, A300475.
%K nonn,base
%O 1,3
%A _Rémy Sigrist_, Mar 05 2018