IBM July 2020
[url]http://www.research.ibm.com/haifa/ponderthis/challenges/July2020.html[/url]

Does anyone of you know, how the new puzzlemaster reacts to submissions that aren‘t correct? Oded has answered every time very fast  that was a good way to help. But I guess that he wasn‘t obliged to do so.

I got an answer after few days that my answer is correct

Spinoff
A friend of mine suggested a nice "spinoff" puzzle.
If you solved July 2020 main problem by using three letters or more, and your dictionary includes letters "I", "B", and "M", then how many times does substring "IBM" appear in your solution? Let IBM(n) be the number of occurrences within nth string of your solution. Does this sequence follow some nice pattern? A trivial upper bound is IBM(n) <= LENGTH(n)/3 = FIBONACCI(n)*FIBONACCI(n+1)/3. A slightly sharper bound is given by RARE(n), the minimum among the occurrences of letters "I", "B", and "M". What about your ratio IBM(n) / LENGTH(n)? What about your ratio IBM(n) / RARE(n)? 
As LENGTH(n) grows exponentially, I think that it makes sense to compute the ratios for n up to 10, or their limits as n grows to infinity (only for the brave)
For my currently best solution, IBM / LENGTH > 0.2 
Spinoff solution
The target sequence G(n) = F(n)*F(n+1) satisfies an order3 homogeneous linear recurrence:
G(n) = 2*G(n1) + 2*G(n2)  G(n3), so we need a homogeneous linear system of order at least three to generate it, which means a dictionary of at least three letters. Three letters actually suffice; I found three such solutions. Dictionary 1: {"A": "BC", "B": "AAB", "C": "ABC"}. Dictionary 2: {"A": "AB", "B": "AAAC", "C": "AAC"}. Dictionary 3: {"A": "AB", "B": "AABC", "C": "B"}. In each case, the game starts with letter "A"; letter "C" is always the rarest one, at least for index n>2. Dieter proved a nice closedform representation for lettercounting functions. Dictionary 1: A(n) = F(n+1)*F(n2); B(n) = F(n)*F(n1); C(n) = F(n1)^2. Dictionary 2: A(n) = F(n)^2; B(n) = F(n1)^2; C(n) = F(n1)*F(n2). Dictionary 3: A(n) = F(n)*F(n1) + F(n2)^2; B(n) = F(n)*F(n1); C(n) = F(n1)*F(n2). Dictionary 1 seems the most promising one, as letter "C" occurs slightly more often (by a factor F(n1)/F(n2), which converges to the golden ratio phi=1.618...). We can change the three spellings in 2*3*6 = 36 ways. We can replace letters "A","B","C" with letters "I","B","M" in 6 ways. Among the 216 possible variations, I chose the following one: {"M": "IB", "I": "MIM", "B": "IBM"} The game starts with letter "M"; letter "B" is the rarest one for n>2, so the upper bound IBM(n) <= B(n) is sharp. This solution almost reaches such bound: IBM(n) = B(n) if n is odd, IBM(n) = B(n)1 if n is even. Proof. In each string of the sequence, we can only find the twoletter substrings "BM", "IB", "IM", and "MI". The sequence starts with a oneletter string, so the claim is trivially true for n=1. The given list contains all the twoletter substrings of the chosen spellings, so the claim is also true for n=2. Then, only substrings "BM" and "MI" can arise from a permitted concatenation of spellings: "BM": "IB[COLOR="Red"]MI[/COLOR]B", "IB": "MI[COLOR="Red"]MI[/COLOR]BM", "IM": "MI[COLOR="red"]MI[/COLOR]B", "MI": "I[COLOR="red"]BM[/COLOR]IM", so the claim is also true for every index n>2 by induction. In particular, letter "B" is always preceded by letter "I" and followed by letter "M", unless it occurs at the beginning or at the end of a string, so the lower bound IBM(n) >= B(n)2 holds. For index n>1, it's easy to check that the first few string characters follow a regular pattern with period 2: IB... > MIM... > IB... > MIM... and the same holds for the last few string characters: ...IB > ...IBM > ...IB > ...IBM so letter "B" never occurs at the begin of a string, but it occurs at the end of evenindexed strings. For the given solution, the ratio IBM(n)/G(n) converges to the value 1/phi^3 ~ 0.236 
Very impressive Oscar:smile::smile::smile:

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