A simple fractal (and lynchean) sequence

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 12, 13, 2, 14, 15, 5, 16, 17, 18, 19, 20, 6, 21, 22, 23, 24, 25, 26, 3, 27, 28, 29, 7, …

 

We want to build a simple fractal sequence S starting with a(1) = 1.

We also want every natural number to appear at least once in S.

Here is our lynchean technique :

Start with a(1) = 1.

Now erase a(n) terms immediately after a(n), if this a(n) term has not been erased before.

This operation must leave the sequence unchanged. 

Let's see how the above S was computed:

S = 1, a, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

As a(1) = 1, we must erase 1 term immediately to the right of a(1) —> in yellow here:

S = 1, a, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

As this term is erased, the next term will survive (in blue here):

S = 1, a, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

The first term will never disappear (because it has no predecessor) —> also in blue, then:

S = 1, a, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

Now, what is the value of a?

It could be anything, because this a will disappear, 2023 for example!

But, no... as we want S to be the lexicographically earliest fractal sequence with the above lynchean property, we cannot have a = 2023. Indeed, though a is erased now, it will pop up somewhere else. Better try first a = 1:

S = 1, 1, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

If a = 1 , then b = 1 too (because the blue sequence must be exactly the same as S); we have:

S = 1, 1, 1, c, d, e, f, g, h, i, j, k, l, m, n, ...

.... we see that the choice of a = 1 will lead to an S made by only 1s... which is, strictly speaking, not a fractal sequence.

Let's try a = 2:

S = 1, 2, b, c, d, e, f, g, h, i, j, k, l, m, n, ...

Again, as the blue sequence must be the same as the original S one, we have to assign the value 2 to b:

S = 1, 2, 2, c, d, e, f, g, h, i, j, k, l, m, n, ...

(We see here that the value 2023 would have built a later sequence than S.) Now, what about the values of c, d, e...?

The integers c and d will be erased [because of a(3) = 2 and David Lynch], thus we must have e = 2 (in order to have a blue sequence that looks the same, so far, as S) :

S = 1, 2, 2, c, d, 2, f, g, h, i, j, k, l, m, n, ...

Remember our will to introduce every natural number as quickly as possible in S; this leads to c = 3 and d = 4 (in yellow, as they are erased);

S = 1, 2, 2, 3, 4, 2, f, g, h, i, j, k, l, m, n, ...

Yellow for f and g too [because a(6) = 2 erases both], and blue for h (as h will survive):

S = 1, 2, 2, 3, 4, 2, f, g, h, i, j, k, l, m, n, ...

As h is the 4th term of the blue sequence, we must do h = 3 (because the 4th term of S is equal to 3):

S = 1, 2, 2, 3, 4, 2, f, g, 3, i, j, k, l, m, n, ...

And we assign the values 5 and 6 to f and g, for the same reason as before (we want to introduce quickly, when there are no other constraints, the smallest integers not yet present in S):

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, i, j, k, l, m, n, ...

The blue 3 erases the next 3 terms – leading to a blue l:

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, i, j, k, l, m, n, ...

As l is the 5th blue term, we must assign to l the value of the 5th term of S:

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, i, j, k, 4, m, n, ...

And the remaining yellow letter-terms will be given the values 7, 8 and 9:

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, m, n, ... Etc.

____________________

May 21 update:

This is now https://oeis.org/A363100 in the OEIS: thank you Allan W.!

(The above image is explained here)

S = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 12, 13, 2, 14, 15, 5, 16, 17, 18, 19, 20, 6, 21, 22, 23, 24, 25, 26, 3, 27, 28, 29, 7, 30, 31, 32, 33, 34, 35, 36, 8, 37, 38, 39, 40, 41, 42, 43, 44, 9, 45, 46, 47, 48, 49, 50, 51, 52, 53, 4, 54, 55, 56, 57, 10, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 11, 68, ...

The above blue sequence is the same as the sequence S.

But what about the variant V, where digits are erased – instead of terms?  Here is an attempt:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11,...

The sequence V diverges here from S, as 10 and 11 use 4 digits we must erase. The next blue integer will be 2, as before:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, ...

And the next 2 digits that will be erased will belong to 12, the smallest available integer not yet present in V:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, ...

Now comes the blue integer 5, as before:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, 5, ... 

And here we have a problem : how should we extend V ? What integers must we use now – whose digits will be erased? A solution would be 13 and 100, as they use a total of 5 digits:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, 5, 13, 100, ... 

And we proceed with the next blue integer, the 6:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, 5, 13, 100, 6, ... 

We have to erase the next 6 digits – no problemo, we extend V with 14, 15 and 16:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, 5, 13, 100, 6, 14, 15, 16, ... 

Now comes the blue integer 3, which is odd:

V = 1, 2, 2, 3, 4, 2, 5, 6, 3, 7, 8, 9, 4, 10, 11, 2, 12, 5, 13, 100, 6, 14, 15, 16, 3, ... 

(to be continued -- dinner time here in Brussels !-)

2*, 3, 4, 2*, 5, 6, 3*, 7, 8, 9, 4*, 10, 11, 2*, 12, 5*, 13, 14, 1/6*, 15, 17, 18, 3*, 19, 2/7*, 20, 21, 22, 2/8*, 23, 24, 25, 26, 9*, 29, 30, 31, 32, 3/4*, 33, 35, 10*, 36, 37, 38, 39, 40, 41, 11*, 42, 43, 44, 45, 46, 5/2*, 47, 12*, 48, 49, 50, 51, 53, 54, 5*, 55, 56, 1/13*, 57, 58, 59, 60, 61, 62, 1/14* 63, 64, 65, 66, 67, 68, 69, 6*, 70, 71, 72, 15*, 73, 74, 75, 76, 77, 78, 79, 1/17*, 80, 81, 82, 83, 84, 85, 86, 87, 1/18*, 88, 89, 90, 91, 92, 93, 94, 95, 96, 3*, 98, 9/7* 8* 9*