A329190 -id:A329190 - cp's OEIS frontend

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

Peter Kagey, Oct 06 2019

nonn

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

			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.

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

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

A328196 First differences of A328190.

Original entry on oeis.org

2, 4, -2, 6, -3, 9, -7, 12, -9, 14, -12, 16, -13, 19, -17, 21, -18, 24, -22, 26, -23, 29, -27, 32, -29, 34, -32, 36, -33, 39, -37, 42, -39, 44, -42, 46, -43, 49, -47, 52, -49, 54, -52, 56, -53, 59, -57, 61, -58, 64, -62, 66, -63, 69, -67, 72, -69, 74, -72, 76

Offset: 1

Peter Kagey, Oct 07 2019

sign

Conjecture from N. J. A. Sloane, Nov 05 2019: (Start)
a(4t) = 5t+1(+1 if binary expansion of t ends in odd number of 0's) for t >= 1,
a(4t+1) = -(5t-2(+1 if binary expansion of t ends in odd number of 0's)) for t >= 1,
a(4t+2) = 5t+4 for t >= 0,
a(4t+3) = -(5t+2) for t >= 0.
These formulas explain all the known terms.
a(2t) is closely related to A298468. The expressions for a(4t) and a(4t+1) can also be written in terms of A328979.
The conjecture would establish that the terms lie on two straight lines, of slopes +-5/4.
There is a similar conjecture for A328190. (End)

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

Cf. A298468, A328190, A328979.
The negative terms are (conjecturally) listed in A329982 (see also A328983).
See A328984 and A328985 for simpler sequences which almost have the properties of A329190 and A328196. - N. J. A. Sloane, Nov 07 2019

A354282 Weird numbers k such that k+1 is the sum of a subset of the aliquot divisors of k.

Original entry on oeis.org

70, 836, 4030, 5830, 7192, 7912, 10792, 17272, 45356, 83312, 91388, 113072, 222952, 243892, 254012, 388076, 410476, 786208, 1713592, 4145216, 4199030, 4632896, 6911512, 7257530, 7354304, 7607530, 9928792, 10402490, 10580624, 11339816, 11547352, 12052390, 13086016

Offset: 1

Amiram Eldar, May 22 2022

nonn

First differs from A258250 and A329190 at n=13.

There are 17270452 weird numbers below 10^10 and only 94 are in this sequence.

			70 is a term since it is a weird number, its aliquot divisors are {1, 2, 5, 7, 10, 14, 35} and 71 = 5 + 7 + 10 + 14 + 35.

Amiram Eldar, Table of n, a(n) for n = 1..94

Subsequence of A006037.
A354283 is a subsequence.
Cf. A005835, A258250, A329190, A354281.

q[n_] := Module[{d = Most @ Divisors[n], x, s, c}, If[Plus @@ d <= n, False, s = Series[Product[1 + x^d[[i]], {i, Length[d]}], {x, 0, n + 1}]; c = SeriesCoefficient[s, #] & /@ (n + {0, 1}); c[[1]] == 0 && c[[2]] > 0]]; Select[Range[10000], q]

is(n, d=divisors(n)[^-1], s=vecsum(d))={s>n && !is_A005835(n, d, s) && is_A005835(n+1, d, s)}; \\ using is_A005835() by M. F. Hasler at A005835

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A328190 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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A328196 First differences of A328190.

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A354282 Weird numbers k such that k+1 is the sum of a subset of the aliquot divisors of k.

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