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A369800
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Lexicographically earliest sequence of distinct positive integers such that the parity of the successive bits of the concatenated binary expansions of the terms is equal to the parity of the terms.
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0
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1, 3, 5, 7, 2, 9, 11, 13, 15, 17, 4, 19, 6, 8, 21, 23, 10, 25, 27, 29, 31, 12, 33, 35, 37, 39, 41, 43, 14, 16, 18, 45, 47, 20, 22, 49, 24, 26, 51, 53, 55, 57, 28, 59, 30, 32, 34, 61, 36, 63, 38, 65, 67, 40, 69, 71, 73, 75, 42, 77, 44, 79, 81, 46, 48, 83, 85, 87, 50, 89, 91, 93, 95
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OFFSET
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1,2
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COMMENTS
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The concatenated binary expansions of the terms act as instructions to take the next unused odd or even number.
This is a permutation of the positive integers.
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LINKS
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EXAMPLE
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terms 1 3 5 7 2 9
binary {1}, {1, 1}, {1, 0, 1}, {1, 1, 1}, {1, 0}, {1, 0, 0, 1}
terms 1, 3, 5, 7, 2, 9, 11,13,15, 17, 4, 19, 6, 8,21
n = 1 2 3 4 5 6 7 8 9 10 11 ...
a(11) = 4 because the 11th bit in the binary expansions is 0 (from term a(5)=2 as it happens) which means a(11) must be even and 4 is the least even positive integer not yet in the sequence.
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MATHEMATICA
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a[1]=1; a[n_]:=a[n]=(k=1; While[MemberQ[Array[a, n-1], k]||Boole@OddQ@k!= Flatten[IntegerDigits[Join[Array[a, n-1], {k}], 2]][[n]], k++]; k); Array[a, 73]
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PROG
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(Python)
from itertools import count, islice
def agen(): # generator of terms
an, parity, nextk = 1, [None, 1, 1, 1], [2, 5]
yield from [1, 3]
for n in count(3):
an = nextk[parity[n]]
yield an
nextk[parity[n]] = nextk[parity[n]] + 2
parity.extend(map(int, bin(an)[2:]))
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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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