Kimberlike sequence (for k=10)

[hv at crypt.org]

"Eric Angelini" <Eric.Angelini at kntv.be> wrote:
: 
:I wrote:
:
:> Looks like, yes, it is always possible to fill the "holes"...
:
:... mmmmh, not so sure now; Gilles Sadowski computed 5000 terms
:for k=10 and integers 2, 4, 6, 8, 9, ... are still missing!
:... this pb is like packing combs of k=2, 3, 4... teeth in the most
:economical way... Wish I was bald...
:Best,
:É.
:(see bottom of page for k=10):
:http://www.cetteadressecomportecinquantesignes.com/Kimberlike02.htm

I have every confidence that every n will appear in the sequence for any k,
but no idea how you might prove it.

I knocked up a program to find where 2 would get inserted for various k:

 k  n   used   shown  passed     seq
 2  2      1       1       0       1
 3  2     11      24       1      23
 4  2     18      48       5      60
 5  2      5      12       3      36
 6  2    346    1384      14    1785
 7  2   5725   26336      70   34759
 8  2   9833   50715     180   69998
 9  2  38238  217258     734  311739
10  2  33637  211951    1238  313806

For k=3, these mean that when the first 2 appeared there were 24 numbers
to its left in the sequence ('shown'), and 23 numbers (or holes) already
known to its right ('seq'); 11 different numbers appeared before the
first 2 ('used'), and 1 number had been 'passed' (ie out of 1..13, one
number has not yet been used).

For k = 10, n in (1 .. 100), here are the 'used' values:

(n=1..10) 0 33637 1 79609 5 48647 2 37597 133669 87742
(n=11..20) 9 49117 27191 58164 3 83523 91922 4 13 120918
(n=21..30) 67986 98032 6 82594 38993 43316 7 8 104427 42940
(n=31..40) 10 102548 80105 15 104550 52375 58202 96393 22 42760
(n=41..50) 48638 11 66769 19 73314 28 80498 18 26 14
(n=51..60) 36 12 92532 83185 16 26380 81768 38333 20 103525
(n=61..70) 37957 30080 64703 92274 38747 23 98316 75075 31 32
(n=71..80) 34 38622 17 67139 42371 96870 48 24 25 91412
(n=81..90) 91652 73714 46186 81232 27 30 21 58 33 46
(n=91..100) 65989 98462 65864 63 66631 29 64 39 82235 83836

It took about 17 minutes to find them all; last to appear was 9:
 k   n    used   shown  passed      seq
10   9  133669  840571    9207  1257313
and peak values of 'passed' and 'seq' (within the 100 results captured)
appear at the previous number:
10  20  120918  764934   21958  1268672

Hope this helps,

Hugo