%I #48 Jan 01 2021 12:15:25
%S 1,3,7,5,11,8,17,10,22,13,27,15,31,18,37,20,41,23,47,25,51,28,57,30,
%T 62,33,67,35,71,38,77,40,82,43,87,45,91,48,97,50,102,53,107,55,111,58,
%U 117,60,121,63,127,65,131,68,137,70,142,73,147,75,151,78,157
%N Lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.
%C The graph appears to consist of two lines whose slopes are approximately equal to 1.25 and 2.5.
%C Conjecture from _N. J. A. Sloane_, Nov 04 2019: (Start)
%C a(2t) = floor((5t+1)/2) for t >= 1 (essentially A047218),
%C a(4t+1) = 10t+1(+1 if binary expansion of t ends in odd number of 0's) for t >= 0 (essentially A297469),
%C a(4t+3) = 10t+7 for t >= 0.
%C These formulas explain all the known terms.
%C One could also say that a(4t+1) = 10t+1+A328979(t+1) for t >= 0.
%C There is a similar conjecture for A328196.
%C 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.
%C 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)
%C Comment from _N. J. A. Sloane_, Nov 06 2019: (Start)
%C 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.
%C 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.
%C 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.
%C But this is not yet a proof that S IS the lexicographically earliest such sequence. (End)
%C 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
%H Peter Kagey, <a href="/A328190/b328190.txt">Table of n, a(n) for n = 1..10000</a>
%e a(1) = 1.
%e a(2) != 1 because a(1) = 1,
%e a(2) != 2 because then a(2) - a(1) = a(1), so
%e a(2) = 3.
%e The first eight terms of this sequence and first seven terms of its first differences are
%e [1, 3, 7, 5, 11, 8, 17, 10] and
%e [2, 4, -2, 6, -3, 9, -7] respectively, and these sequences have no common terms.
%Y Cf. A005228, A047218, A080426, A297469, A327460, A328196 (first differences), A328979.
%Y See A328984 and A328985 for simpler sequences which almost have the properties of A329190 and A328196. - _N. J. A. Sloane_, Nov 07 2019
%K nonn
%O 1,2
%A _Peter Kagey_, Oct 06 2019