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A328190
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Lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.
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9
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1, 3, 7, 5, 11, 8, 17, 10, 22, 13, 27, 15, 31, 18, 37, 20, 41, 23, 47, 25, 51, 28, 57, 30, 62, 33, 67, 35, 71, 38, 77, 40, 82, 43, 87, 45, 91, 48, 97, 50, 102, 53, 107, 55, 111, 58, 117, 60, 121, 63, 127, 65, 131, 68, 137, 70, 142, 73, 147, 75, 151, 78, 157
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OFFSET
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1,2
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COMMENTS
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The graph appears to consist of two lines whose slopes are approximately equal to 1.25 and 2.5.
a(2t) = floor((5t+1)/2) for t >= 1 (essentially A047218),
a(4t+1) = 10t+1(+1 if binary expansion of t ends in odd number of 0's) for t >= 0 (essentially A297469),
a(4t+3) = 10t+7 for t >= 0.
These formulas explain all the known terms.
One could also say that a(4t+1) = 10t+1+A328979(t+1) for t >= 0.
There is a similar conjecture for A328196.
Call the three sets of conjectured terms S0, S1, and S3. The terms in S0 are == 0 or 3 mod 5; those are in S1 are == 1 or 2 mod 10; and those in S3 are == 7 mod 10. So the sets are disjoint, as required by the definition.
This conjecture would imply that the points a(2t) lie on a line of slope 5/4 and the points a(2t+1) on a line of slope 5/2, as conjectured by Peter Kagey. (End)
Let us DEFINE a sequence S by the conjectured formulas given here, and a sequence T by the conjectured formulas given in A328196. Then it is not difficult to prove that the first differences of S are given by T, and that the terms of S and T are disjoint.
So S is certainly a candidate for the lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.
Furthermore Peter Kagey's b-files for this sequence and A328196 show that the first 10000 terms of S are indeed the first 10000 terms of the lexicographically earliest such sequence.
But this is not yet a proof that S IS the lexicographically earliest such sequence. (End)
To construct the bisection a(2n-1), start with [4]. Apply the substitution rule 4 -> 46, 5 -> 46, 6 -> 55. Prepend [1, 6] to the resulting list, then take partial sums. - John Keith, Dec 31 2020
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LINKS
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EXAMPLE
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a(1) = 1.
a(2) != 1 because a(1) = 1,
a(2) != 2 because then a(2) - a(1) = a(1), so
a(2) = 3.
The first eight terms of this sequence and first seven terms of its first differences are
[1, 3, 7, 5, 11, 8, 17, 10] and
[2, 4, -2, 6, -3, 9, -7] respectively, and these sequences have no common terms.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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