A328190 Lexicographically earliest infinite sequence of distinct positive integers such that the sequence and its first differences have no values in common.

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

Offset: 1

Author

Peter Kagey, Oct 06 2019

Keywords

• nonn

Comments

The graph appears to consist of two lines whose slopes are approximately equal to 1.25 and 2.5.

Conjecture from N. J. A. Sloane, Nov 04 2019: (Start)

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)

Comment from N. J. A. Sloane, Nov 06 2019: (Start)

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

Examples

			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.
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A005228, A047218, A080426, A297469, A327460, A328196 (first differences), A328979.

See A328984 and A328985 for simpler sequences which almost have the properties of A329190 and A328196. - N. J. A. Sloane, Nov 07 2019