[seqfan] a(n) is a divider of a[n+a(n)+1]

[Eric Angelini]

Hello Seqfans,
we could describe A000027 (the Naturals) like this:

a(n) is a divider of a[n+a(n)], and a(n) is "the smallest
available integer not yet in the sequence and not leading 
to a contradiction".

Keeping the same prescription, but with a(n) being a divider
of [n+a(n)+1] would lead to S:

S = 1,2,3,4,6,5,9,7,8,10,11,30,12,13,14,21,18,16,15,17,20,19,22,...

Explanation:

We start S with 1:
S = 1, . . . . . 

We jump over one "term" and put a Post-It underneath the 
place where we land. This Post-It says: "Remember, the
term here must have a(1) as divider":

S = 1, . , . , . , . , .
PostIt     1     

The leftmost free place must be filled with "the smallest
available integer not yet present in S and not leading 
to a contradiction". This is "2" (and its according Post-It):

S = 1, 2 , . , . , . , .
PostIt     1       2

The leftmost free place will obviously be filled by "3"
(which has "1" as a divider, of course):

S = 1, 2 , 3 , . , . , . , . , . ,
PostIt     1       2       3

The leftmost etc. , will host "4":

S = 1, 2 , 3 , 4 , . , . , . , . , . ,
PostIt     1       2       3       4

The next term of S must be the smallest one not yet present 
and having "2" as divider; this is "6", of course:

S = 1, 2 , 3 , 4 , 6 , . , . , . , . , . , . , . , . ,
PostIt     1       2       3       4           6

The smallest available integer in order to extend S is "5";
and this "5" adds a constraint on the "6" Post-It:

S = 1, 2 , 3 , 4 , 6 , 5 , . , . , . , . , . , . , . 
PostIt     1       2       3       4           6
                                               5

Yes, the (12th) term of S will at least have 6 and 5 as
dividers. We go on like this -- "9" being the "smallest
available integer etc. having "3" as divider":

S = 1, 2 , 3 , 4 , 6 , 5 , 9 , . , . , . , . , . , . , . , . , . , 
PostIt     1       2       3       4           6               9
                                               5

The next terms of S will be:

S = 1, 2 , 3 , 4 , 6 , 5 , 9 , 7 , 8 , 10 , 11 , 30 , . , . , . , . , 
PostIt     1       2       3       4              6               9
                                                  5

etc.
I guess S will then be the lexicographically first 
permutation of the Naturals.

----------

More such sequences can be computed if we assign the 
values [2,3,4,5,6,7,...] to k in the hereunder rule:

"a(n) is a divider of a[n+a(n)+k], and a(n) is "the smallest
available integer not yet in the sequence and not leading 
to a contradiction".

Best,
É.