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Previous Puzzles of the Month + Solutions

February-March 2008, Puzzle nr 116
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Puzzle # 116 Difficulty level: bulbbulbbulb, basic math knowledge.

The Parrot Sequence
Gaston owns a clever parrot called Paco which can describe out loud numbers read from a piece of paper displayed in front of him. One day Gaston tried this funny experiment: he wrote the digit 1 on a sheet and showed it to Paco, who answered “one 1”.
  Gaston wrote down the parrot’s answer “one 1” or ‘11’ on a new sheet and showed it to Paco, who readily answered “two 1’s”. Again, Gaston wrote down his reply “two 1’s” or ‘21’ on another sheet and showed it to Paco, who replied “one 2, one 1” or ‘1211’... Gaston continued this experiment until he obtained a long sequence of numbers 1, 11, 21, 1211, 111221, 312211, etc... in which each next term is obtained by describing the previous term.
  Can you say how many sheets of paper Gaston will need until:
    a) the sub-string of digits 333 appears, or
    b) the digit 4 occurs in the sequence?
    c) Can you also confirm that the last digit of any term of the parrot sequence is always 1?

Key words: Look-and-say sequence, likeness sequence, audioactive decay, Gleichniszahlen-Reihe.

Source of the puzzle:
©G. Sarcone, Focus BrainTrainer, No. 10, page 81
puzzle of the month


(Pn) is the "Parrot Sequence", and P1, P2, P3, ..., Pn are its terms.

To find:
a) If '...333...' is a positive integer containing a 333 string, then could '...333...' ∈ (Pn) ?

b) If '...4...' is a positive integer containing the digit 4, then could '...4...' ∈ (Pn) ?

c) If P1 = 1, then will all the terms of (Pn) end with the digit 1?

a) According to the sequence generation rules, the digits in the "Parrot Sequence" are paired: the ‘left’ digit is the ‘occurrence’ digit, and the right one, represents the ‘described’ digit (see diagram below).
parrot sequence

We can then pair up the digits of the string 333 as follows:
1) ...,X3,33,... or
2) ...,33,3X,... (X being any integer)
and can easily deduce that the sub-string 333 to appear must ALREADY be present in the PREVIOUS term of the sequence, as shown below:

Pn+1 < Pn
(X occurrences of 3’s, three 3’s)
< ...[3 or 33 or 333],333...
(three 3’s, three X’s)
< ...333,XXX...

(the symbol ‘<’ means here ‘comes from’)
And this leads to contradictions! (impossible self-referential loop) Thus the sub-string 333 will never occur...

b) The occurrence of 4 is possible only when the previous term contains a repetition of the same digit 4 times. A ‘3333’ sub-string cannot occur for the same reason a ‘333’ sub-string cannot. Therefore only ‘1111’ or ‘2222’ strings may generate the sub-string ‘41’ or ‘42’ respectively, and there are 2 possible ways to pair their digits:
1) ...,11,11,... or ...,22,22,...
2) ...,X1,11,1Y,... or ...,X2,22,2Y,... (X and Y being any integer)

But the possibility 1) is not conceivable due to the sequence generation rules: a previous ‘11’ string would always be described as “two 1’s” (‘21’), never as “one 1, one 1” (‘11,11’). The same thing happens with the string ‘2222’: to get the sub-string ‘22,22’, the parrot should previously say “two 2’s, two 2’s”, but this is both a self-reference and a contradiction, because according to the sequence generation rules Paco will spell “four 2’s” (‘42’) instead...

Concerning the possibility 2), the ‘X1,11,1Y’ string would surely generate a ‘41’ sub-string, if only a ‘X1,11,1Y’ string could exist! In fact, ‘X1111Y’ occurs only when a previous string is spelled erroneously. Below are all the possible ways to spell ERRONEOUSLY some sub-strings to get finally a ‘41’ sub-string:

Pn+2 < Pn+1 < Pn
124112 < 21,11,12 (instead of ‘31,12’) < ...11,1,2...
124113 < 21,11,13 (instead of ‘31,13’) < ...11,1,3...
134112 < 31,11,12 (instead of ‘41,12’) < ...111,1,2...
134113 < 31,11,13 (instead of ‘41,13’) < ...111,1,3...

(the symbol ‘<’ means here ‘comes from’)
We encounter the same problem as above with the ‘X2,22,2Y’ string. Thus to get a 4, or even a 1111, or a 2222 sub-string within any term of the sequence, it should ALREADY appear in the PREVIOUS term of the sequence! And this leads to a retro-causation paradox.

c) If the sequence starts with 1 (seed number), then the last digit of the terms will always be 1. In fact, the last digit of the terms will always be the digit we started with, because being the first term of the sequence it will always be the ‘described’ digit of the last pair of digits, as shown below:
(N = any digit)


cup winnerThe 5 Winners of the Puzzle of the Month are:
Chris Peterson, USA flag usa
Fabio Cirigliano, Italy italian flag
Alessio Medici, Italy italian flag
Alessandro Tornago, Brazil brazil
Amaresh G. S., India indian flag



© 2004 G. Sarcone,
You can re-use content from Archimedes’ Lab on the ONLY condition that you provide credit to the authors (© G. Sarcone and/or M.-J. Waeber) and a link back to our site. You CANNOT reproduce the content of this page for commercial purposes.

You're encouraged to expand and/or improve this article. Send your comments, feedback or suggestions to Gianni A. Sarcone. Thanks!

More Math Facts behind the puzzle

The ‘Parrot Sequence’, also know as the ‘look-and-say sequence’, was first described in the early 1980’s by the German mathematician Mario Hilgemeier (“Die Gleichniszahlen-Reihe”, in ‘Bild der Wissenschaft’ #12, 194-196, Dec. 1986).

The mathematician J. H. Conway showed that the ratio of lengths of consecutive terms of the sequence approaches a constant:

arrow ∞
  P(n+1)  = λ = 0.303577269...

λ is in fact the unique positive real root of a 71-degree polynomial.

Some curiosities:
- If the ‘Parrot Sequence’ started with 22 (seed number), then all the terms of the sequence would be 22.
- The sum of the digits in each term of the ‘Parrot Sequence’ gives the (n+1)th Fibonacci number!

To end, here is an interesting variant of the ‘Parrot Sequence’:

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