A328190 - OEIS

%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