Hello SeqFan and Math-Fun,

(dont know if this is of interest)

1, 2, 1, 3, 2, 4, 5, 3, 6, 7, 4, 8, 5, 9, 10, 6, 11, 7, 12, 13, 8, 14, 15, 9, 16, 10, 17, 18, 11, 19, 20, 12, 21, 13, ...

This sequence displays every positive integer exactly twice, and the gap between the two occurrences of n contains exactly n other values. The first occurrence of n precedes the first occurrence of n+1. (cont.) [This is OEIS A026272 by Clark Kimberling]

What about a similar sequence displaying every positive integer exactly three times? We must drop the constraint The first occurrence of n precedes the first occurrence of n+1 and replace it by Always fill the first hole with the smallest available integer not used so far.

Lets start with the three 1s (a dot represents a hole which will be filled in the future):

1 . 1 . 1 . . . . . . . . .

Can we fill the first hole with a 2? No -- we would bump immediately into a 1:

1 2 1 . 1 . . . . . . . . .

Can we fill the first hole with a 3 instead? Yes, there is room for the three 3s:

1 3 1 . 1 3 . . . 3 . . . .

Can we fill the first hole with a 2? (smallest available integer not used so far)? No -- we would bump into a 3 at the end:

1 3 1 2 1 3 2 . . 3 . . . .

Can we fill the first hole with a 4 instead? Yes, there is room for the three 4s:

1 3 1 4 1 3 . . 4 3 . . . 4

Can we fill the first hole with a 2? (smallest available integer not used so far)? No -- we would bump immediately into a 3:

1 3 1 4 1 3 2 . 4 3 . . . 4

Can we fill the first hole with a 5 instead? Yes, there is room for the three 5s:

1 3 1 4 1 3 5 . 4 3 . . 5 4 . . . . 5

Can we fill the first hole with a 2? (smallest available integer not used so far)? No -- we would bump into a 4 at the end:

1 3 1 4 1 3 5 2 4 3 2 . 5 4 . . . . 5

Can we fill the first hole with a 6 instead? Yes, there is room for the three 6s:

1 3 1 4 1 3 5 6 4 3 . . 5 4 6 . . . 5 . . 6

Can we fill the first hole with a 2? (smallest available integer not used so far)? No -- we would bump immediately into a 4:

1 3 1 4 1 3 5 6 4 3 2 . 5 4 6 . . . 5 . . 6

Can we fill the first hole with a 7 instead? No, we would bump immediately into a 5:

1 3 1 4 1 3 5 6 4 3 7 . 5 4 6 . . . 5 . . 6

Can we fill the first hole with a 8 instead? Yes, there is room for the three 8s:

1 3 1 4 1 3 5 6 4 3 8 . 5 4 6 . . . 5 8 . 6 . . . . . . 8

Can we fill the first hole with a 2 or a 7 or a 9? No, but with a 10 instead, yes. Etc.

After a few more steps the sequence will look like this, if I didnt mistake (on the waiting shelf is 15 -- note that 2 and 7 have found their places):

1 3 1 4 1 3 5 6 4 3 8 10 5 4 6 7 9 11 5 8 13 6 10 7 2 12 9 2 8 11 2 7 14 10 13 16 9 . 12 . . 11 . . . . . 14 13 . . 12 16 . . .

Questions:

- Will all triplets of integers appear sooner or later in the sequence?

- If we define k as the number of occurrences of an integer, the above sequence could be called the 3-Kimberlike sequence (it deals with triplets), and A026272 the 2-Kimberlike sequence (it deals with doublets); is there a k for which the according k-Kimberlike sequence would not show all k-plets of integers? If yes, what is the smallest such k? And what are the missing integers of that sequence? If, no matter the value of k, a sequence can always be build leaving no integers behind, it would be interesting to show the array produced by, say, the 20 first values of k.

Best,

Ι.