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

DAN_CYN_J dan_cyn_j at comcast.net
Wed Jul 3 14:57:33 CEST 2013



  

Thanks Lars, you explained it better then I ever could. 

  

Please enter this in  OEIS because I would probably screw it up 

and no credits necessary. 

  

Cheers,   

  

Dan 

  

  



----- Original Message -----
From: "Lars Blomberg" <lars.blomberg at visit.se> 
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu> 
Sent: Tuesday, July 2, 2013 3:27:13 PM 
Subject: [seqfan] Re: Sequence like A026272 but uses 3 integers instead of 2. 

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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