Odd/even
Langford/Skolem-like sequence
Hello
SeqFan,
I need a little help here, please.
I would like to build
another Langford/Skolem-like sequence which would go like this :
- every integer exactly
twice in sequence ;
- between to integers T (say two 5’s) exactly
T even and T odd integers (exactly 5 even and 5 odd terms between two
5’s) ;
Calculated by hand (with a so-so
algorithm), here is what I get
for the first terms :
1,2,3,1,6,4,2,5,7,3,8,9,12,10,4,11,13,6,5,14,15,16,17,7,18,19,20,8,23,21,9,22,26,24,10,25,27,12,11,28,29,30,31,13,32,33,36,34,14,35,37,15,40,38,16,39,41,17,42,43,44,18,.,.,19,.,.,20,
etc.
My so-so
algorithm goes like that :
(a)
Start with « 1 » :
1
(b)
Put the second « 1 », but leave the right number of « holes » between this « 1 » and
the first one (between two
T’s there are 2T
holes -- two times the
value of T) :
1 . . 1
(c) Fill the first « hole »
with the smallest available integer T not leading to a contradiction so
far ; when this is done, duplicate this integer T accordingly to the right (there
are 4 integers between two 2’s, 6 integers between two 3’s, etc. Half of those integers
must be odd and half must be even) ;
1 2 .
1 . . 2
(d)
back to instruction (c)
Remark :
- if
the last two consecutive holes are filled with integers
that « neglect »
a smaller one, permute those
two integers (« consecutive holes » are holes not separated by an integer).
The sequence builds on like this, step
by step :
1 2 .
1 . . 2
1 2 3 1 .
. 2 . . 3
1 2
3 1 4
. 2 . . 3 . . . 4
Now, can we put 5 in the underlined hole, like this :
1 2
3 1 4 5 2 . . 3 . . . 4
No, between the two 2’s there would be
three odd and one even terms -- which
is not allowed.
We must put 6 instead
(smallest available integer not leading to a
contradiction) :
1 2
3 1 4 6 2 . . 3 . . . 4
Is it correct now ? No, not yet -- according to the Remark, we must permute 4
and 6 because we have
« neglected » 5, which
is smaller than 6 (last two consecutive holes rule) ; we must not forget to correct the
position of the second 4 and put the second 6 :
1 2
3 1 6 4 2 . . 3 . . . . 4 . . 6
Ok, there are :
-
one even and one odd integer between the two 1’s ;
- two even and two
odd integers between the two 2’s.
The next step is
easy, 5 is the smallest unused integer so far :
1 2
3 1 6 4 2 5
. 3 . . . . 4 . . 6 5
Next step requires an odd integer : 7 fits. No
permutation 5<-->7 is made because
no integer has been neglected ;
we have now three odd and three
even integers between the two 3’s :
1 2
3 1 6 4 2 5 7 3 . .
. . 4 . . 6 5 . . . . 7
Next two steps use 8 and 9 :
1 2
3 1 6 4 2 5 7 3 8 9 .
. 4 . . 6 5 . . . . 7 . . . 8 . . 9
We must now put two even
integers in the last two consecutive holes : 10 and
12. We shall permute them because 11 has been neglected :
1 2
3 1 6 4 2 5 7 3 8 9 12
10 4 . . 6 5 . . . . 7 . . . 8 . . 9 .
. . 10
. . 12
Next two steps use 11 and 13, without
permutation and correctly fill
the gaps between the two
6’s and 5’s :
1 2
3 1 6 4 2 5 7 3 8 9 12 10 4 11
13 6 5 . . . . 7 .
. . 8 . . 9 . . . 10 . . 12 11 . . . . 13
Etc.
A
note and two questions :
Note : the sequence should
be started with the pair 0,0. We would have thus :
0,0,1,2,3,1,6,4,2,5,7,3,8,9,12,10,4,11,13,6,5,14,15,16,17,7,18,19,20,8,23,21,9,22,26,24,10,25,27,12,11,28,29,30,31,13,32,33,36,34,14,35,37,15,40,38,16,39,41,17,42,43,44,18,.,.,19,.,.,20,
etc.
Two questions :
1)
The so-so algorithm seems to work well so
far, but is it
correct ?
2) I’ve tried to build
another Langford/Skolem sequence like this
-- using primes and non-primes instead
of odd and even terms. Without success...
Best,
É.