Établissement
Colegiul Ortodox "Mitropolitul Nicolae Colan" (Cluj-Roumanie)
Année
2023-2024
Résumé
A substitution is an operation which consists of replacing each digit by a sequence of numbers, according to predefined rules.
For example, consider the substitution 1 -> 123; 2 -> 13; 3 -> 2. If we apply it to 123, we get 123 13 2, we get 123 13 2. if we apply it again, we get 123 13 2 123 2 13.
We can see that each "number" we obtain is twice as long as the previous one (why?), and that it starts with a "number" and starts with this one. If we repeat this process indefinitely, we obtain an infinite sequence:
1231321232131231321312321231321232131232…
What happens if we apply the substitution to this sequence?
The sequence obtained has the following remarkable property: it never contains two consecutive identical blocks, for example 1232 1232. We can try a proof by the absurd, but it's a bit long because you have to study several cases. I propose to do this with another substitution, which has the same property: 1 -> 12; 2 -> 34; 3 -> 14; 4 -> 32.
Construct other sequences with other substitutions. For example, can we make the following sequences?
0110100110010110100101100110100110010110…
2100100001000000100000000100000000001000…
With the substitution 1 -> 12; 2 -> 1, starting from 1, we obtain the "numbers" 12, 121, 12112, etc. We write down their lengths: 1, 2, 3, 5, 8, etc. Is there a way of calculating these lengths without using substitution (which saves a lot of time)?
For example, consider the substitution 1 -> 123; 2 -> 13; 3 -> 2. If we apply it to 123, we get 123 13 2, we get 123 13 2. if we apply it again, we get 123 13 2 123 2 13.
We can see that each "number" we obtain is twice as long as the previous one (why?), and that it starts with a "number" and starts with this one. If we repeat this process indefinitely, we obtain an infinite sequence:
1231321232131231321312321231321232131232…
What happens if we apply the substitution to this sequence?
The sequence obtained has the following remarkable property: it never contains two consecutive identical blocks, for example 1232 1232. We can try a proof by the absurd, but it's a bit long because you have to study several cases. I propose to do this with another substitution, which has the same property: 1 -> 12; 2 -> 34; 3 -> 14; 4 -> 32.
Construct other sequences with other substitutions. For example, can we make the following sequences?
0110100110010110100101100110100110010110…
2100100001000000100000000100000000001000…
With the substitution 1 -> 12; 2 -> 1, starting from 1, we obtain the "numbers" 12, 121, 12112, etc. We write down their lengths: 1, 2, 3, 5, 8, etc. Is there a way of calculating these lengths without using substitution (which saves a lot of time)?
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