[seqfan] a(n) is a divider of a[n+a(n)+1]
Eric Angelini
Eric.Angelini at kntv.be
Wed May 2 19:19:26 CEST 2012
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.
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