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A354125
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Lexicographically earliest sequence of distinct nonnegative integers such that for any n >= 0, the binary expansions of a(n) and a(n + a(n)) have no 1's in common.
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2
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0, 1, 2, 3, 4, 5, 8, 6, 9, 7, 10, 11, 12, 16, 17, 13, 24, 18, 14, 15, 20, 19, 32, 21, 33, 22, 23, 25, 34, 35, 26, 36, 48, 27, 64, 37, 28, 29, 30, 31, 96, 38, 39, 40, 42, 41, 43, 65, 44, 72, 45, 46, 66, 47, 67, 49, 68, 70, 50, 51, 100, 52, 69, 53, 128, 54, 98
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OFFSET
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0,3
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COMMENTS
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This sequence is a permutation of the nonnegative integers with inverse A354126:
- the sequence is well defined as we can always extend it with a power of 2,
- for any n > 0, let f(n) = n + a(n),
- suppose that there are only finitely many integers not in the image of f: say s_1 < ... < s_k,
- as the present sequence diverges, for some m > s_k, a(n) > k for any n > m,
- for any i = 1..k:
- let o_i be the orbit of s_i under repeated applications of f: o_i = {s_k, f(s_k), f(f(s_k)), ...},
- let t_i be the least integer > m in the orbit of o_i,
- let u = max(t_1, ..., t_k),
- the interval I = u+1..u+k contains k terms,
- each o_i has at most one element in common with I,
- and any orbit o_i containing u has no element in common with I,
- so by the pigeonhole principle, some element of I, say w, does not belong to any of the orbits o_i,
- so w > s_k does not belong to the image of f, a contradiction,
- so there are infinitely many integers not of the form n + a(n),
- each time we encounter such an integer, we can extend the sequence with the least unused integer, and eventually every integer will appear in the sequence.
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LINKS
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EXAMPLE
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The first terms, alongside the binary expansions of a(n) and a(n + a(n)) are:
n a(n) bin(a(n)) bin(a(n+a(n)))
-- ---- --------- --------------
0 0 0 0
1 1 1 10
2 2 10 100
3 3 11 1000
4 4 100 1001
5 5 101 1010
6 8 1000 10001
7 6 110 10000
8 9 1001 10010
9 7 111 11000
10 10 1010 10100
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PROG
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(PARI) See Links section.
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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