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A309668
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a(n) is the least positive number of the form floor(2^k/n) for some k >= 0 not yet in the sequence.
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2
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1, 2, 5, 4, 3, 10, 9, 8, 7, 6, 11, 21, 19, 18, 17, 16, 15, 14, 13, 12, 24, 23, 22, 42, 20, 39, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 52, 25, 49, 48, 47, 46, 45, 44, 43, 85, 41, 40, 80, 78, 38, 75, 74, 73, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61
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OFFSET
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1,2
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COMMENTS
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The sequence is well defined as for any n > 0, there are infinitely many positive numbers of the form floor(2^k/n) with k >= 0.
The sequence is a permutation of the natural numbers, with inverse A309734:
- for any m > 0, floor(2^k/A300475(m)) = m for some k,
- also, for any u > 0, floor(2^(k-u)/(A300475(m)*2^u)) = m,
- so the set S_m = { v such that floor(2^k/v) = m for some k >= 0 } is infinite
- and eventually a(n) = m for some n in S_m, QED.
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LINKS
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EXAMPLE
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The first terms, alongside the binary representations of a(n) and of 1/n (with that of a(n) in parentheses), are:
-- ---- --------- ---------------------
1 1 1 (1).00000000000000...
2 2 10 0.(10)000000000000...
3 5 101 0.0(101)0101010101...
4 4 100 0.0(100)0000000000...
5 3 11 0.00(11)0011001100...
6 10 1010 0.00(1010)10101010...
7 9 1001 0.00(1001)00100100...
8 8 1000 0.00(1000)00000000...
9 7 111 0.000(111)00011100...
10 6 110 0.000(110)01100110...
11 11 1011 0.000(1011)1010001...
12 21 10101 0.000(10101)010101...
13 19 10011 0.000(10011)101100...
14 18 10010 0.000(10010)010010...
15 17 10001 0.000(10001)000100...
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PROG
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(PARI) s=1; for (n=1, 67, q=1/n; while (bittest(s, f=floor(q)), q*=2); print1 (f ", "); s+=2^f)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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