[seqfan] Re: Sequence like A026272 but uses 3 integers instead of 2.

Tue Jul 2 21:27:13 CEST 2013

Hello,

An implementation according to the rules:
1. All the natural numbers must occur exactly 3 times.
2. The number of terms between the 3 occurrences of the number k must be 
(k,k), (k,k+1) or (k+1,k+1)
3. When no previously used number is available for a new term, it is set to 
the smallest not yet used number

The result agrees with Dan's up to term a[116]=28,
but a[117] must be 37 (not 53) to match a[78]=37.

1,2,1,3,1,2,4,3,5,2,6,3,4,7,5,8,9,6,4,10,5,7,11,12,6,8,9,13,14,7,10,15,16,17,11,8,9,12,18,19,20,10,13,14,21,22,11,15,23,16,24,12,17,25,26,27,28,
13,18,14,19,20,29,15,30,31,16,21,22,32,33,17,23,34,35,24,36,37,18,25,38,19,26,20,27,28,39,40,41,42,21,22,29,43,44,30,23,31,45,46,24,47,32,48,
33,25,49,50,34,51,26,35,52,27,36,28,37,53,54,38,55,56,29,57,58,59,30,39,40,31,41,60,42,61,62,32,63,43,33,44,64,65,66,34,45,67,46,68,35,47,69,
70,36,48,71,37,49,72,38,50,73,51,74,75,76,52,77,78,39,40,79,53,41,54,80,42,55,81,56,82,83,43,57,58,44,59,84,85,86,...

The program has built a sequence which includes all the triplets up to 
23730,
followed by 7383 double entries and 14044 single entries.

It surprises me that no irresolvable conflict has yet occurred, that is,
where two numbers k1 and k2 both have their previous entries with a gap k1+1 
and k2+1, respectively.
Perhaps someone with a little mathematical skill can prove it.

Regards,
Lars B

-----Ursprungligt meddelande----- 
From: DAN_CYN_J
Sent: Monday, July 01, 2013 9:22 PM
To: Sequence Fanatics Discussion
Subject: [seqfan] Sequence like A026272 but uses 3 integers instead of 2.

Hi again seq. fans,

Can this sequence continue --->oo ?

A complex sequence where all integers are represented just 3 times,
The problem arises when two integers need to share the same index.
In that case the smaller integer wins out and the gap that represents
that larger integer is increased by (1) for that entry and the
same gap for the next entry.

In the sequence below (1) can be repeated (3) times with a (1) gap because 
it has
precedent over (2).

(2) gap therefore has to be increased by (1) for that
entry and the same gap for the next entry. if another conflict arises
then the one with no change in its gap adds one more to its gap size.
In other words no gap can exceed it integer equivalent >1.

(14) is and example below where it starts with a gap of (14) but has to 
change
the gap to (15) because (18) already has a (19) gap assigned to it.

A similar sequence that produces each integer (twice) instead of 3 times
is in OEIS as A026272

Is an algorithm with the above rules to create this sequence possible?

I believe there is but I lack the sophistication to create one.

I do believe this to be correct for the first 3/4 of the terms!
(remember the latter terms always need adjusting as each integer set is 
completed)

(n) = completed integer represented 3 times.

1,2,1,3,(1),2,4,3,5,(2),6,(3),4,7,5,8,9,6,(4),10,(5),7,11,12,(6),8,9,13,14,(7),10,15,16,17,11,(8),(9),12,18,19,20,(10),13,14,21,22,(11),15,23,16,24,(12),17,25,26,27,28,(13),18,(14),19,20,29,(15),30,31,(16),21,22,32,33,(17),23,34,35,24,36,37,(18),25,38,(19),26,(20),27,28,39,40,41,42,(21),(22),29,43.44,30,(23),31,45,46,(24),47,32,48,33,(25),49,50,34,51,(26),35,52,(27),36,(28),53,54,55,56,57,58,(29),59,60,61,(30),62,63,(31),64,65,66,67,68,(32),69,70,(33),71,72,73,74,(34),75,76,77,78,(35),79,80,81,(36),... 
 --->00??

This sequence has to be continually adjusted for at least the last 1/4 of 
the terms.
So the latter part of this sequence contain the wrong first integer added
which will ultimately be less for that particular index.

(*) where (1) is added to the gap of certain integers and continues with the 
same gap for
the last integer.

1=1 gap
2=3 gap *
3=3 gap
4=5 gap *
5=5 gap
6=6 gap
7=7 gap
8=9 gap *
9=9 gap
10=10 gap
11=11 gap
12=13 gap *
13=14 gap *
14=14 gap and between the 2nd (14) and 3rd (14) 14= a 15 gap *
15=15 gap
16=16 gap
17=18 gap *
18=19 gap *
19=20 gap *
20=20 gap  and between the 2nd (20) and 3rd (20) 20= a 21 gap *
21=22 gap *
22=22 gap
23=23 gap
24=24 gap
25=25 gap
26=27 gap *
27=28 gap *
28=28 gap and between the 2nd (28) and 3rd (28) 28= a 29 gap *
29=29 gap
30=30 gap
31=31 gap
32=32 gap
33=33 gap
34=34 gap
35=36 gap *
36=37 gap *
...
I hope there are no errors up to 3/4 the way threw this sequence because

that would screw up the whole sequence.

This was all done by brute force so any errors are possible!

Cheers,

Dan

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