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Turtle soup
THE MOCK TURTLE'S SONG

Beautiful Soup, so rich and green,
Waiting in a hot tureen!
Who for such dainties would not stoop!
Soup of the evening, beautiful Soup!
Soup of the evening, beautiful Soup!
Beau-ootiful Soo-oop!
Beau-ootiful Soo-oop!
Soo-oop of the e-e-evening,
Beautiful, beautiful Soup!
Beautiful Soup! Who cares for fish,
Game, or any other dish!
Who would not give all else for two
pennyworth only of beautiful Soup!
Pennyworth only of beautiful Soup!
Beau-ootiful Soo-oop!
Beau-ootiful Soo-oop!
Soo-oop of the e-e-evening,
Beautiful, beauti-FUL SOUP!
- Lewis Carol, 'Alice's Adventures in Wonderland'
(see the video!)

# Previous Puzzles of the Month + Solutions

February-March 2008, Puzzle nr 116
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Puzzle # 116 Difficulty level: , 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. - Gastone possiede un bel pappagallo addomesticato di nome Paco, il quale può descrivere ad alta voce dei numeri letti da un foglio di carta.   Un giorno, Gastone fece questo divertente sperimento: segnò la cifra 1 su di un foglietto e lo mostrò a Paco, il quale rispose perentoriamente “un 1!”. Gastone riportò la risposta del pappagallo “un 1”, ossia ‘11’, su di un nuovo foglio di carta e lo fece vedere a Paco, che prontamente gracchiò “due 1!”. Di nuovo, Gastone annotò la risposta “due 1”, ossia ‘21’, su di un altro foglietto, che mostrò a Paco, ottenendo come risposta “un 2, un 1”, ossia ‘1211’… Gastone procedette con questo sperimento fino ad ottenere una lunga sequenza di numeri 1, 11, 21, 1211, 111221, 312211, ecc… nella quale ogni termine della lista è la descrizione ‘a voce’ del termine precedente.   Però, quanti fogli di carta pensi che Gastone debba scarabocchiare prima che appaia: a) la stringa di 3 cifre 333, oppure b) la cifra 4, nella sequenza?   E poi, c) sapresti dire se la cifra 1 termina sempre tutti i numeri della sequenza? Parole chiavi: Costante di Conway, audioactive decay, Gleichniszahlen-Reihe. Suggerisci un'altra soluzione Chiudi - Gaston possède un perroquet savant nommé Paco qui peut décrire à haute voix des chiffres lus sur une feuille de papier disposée face à lui. Un jour, Gaston tenta cette amusante expérience : il écrivit le chiffre 1 sur une feuille de papier et la montra à Paco, qui répondit « un 1 ». Gaston écrivit la réponse du perroquet « un 1 » soit '11' sur une nouvelle feuille de papier et la montra à Paco, qui répondit promptement « deux 1 ». A nouveau, Gaston écrivit sa réponse « deux 1 » ou '21' sur une autre feuille de papier et la montra à Paco, qui répondit : « un deux, un 1 » soit '1211'… Gaston poursuivit cette expérience jusqu'à obtenir une longue suite de nombres 1, 11, 21, 1211, 111221, 312211, etc… dans laquelle chaque terme suivant est obtenu par la description 'vocale' du terme précédent.   Pouvez-vous dire combien de feuilles de papier Gaston aura besoin de griffonner avant de voir apparaître :     a) un série de trois 3, soit 333 ? ou     b) le chiffre 4 dans la séquence ?     c) Pouvez-vous confirmer également que le dernier chiffre de chaque terme de la 'suite du perroquet' est toujours 1 ? Mots clés: Commentaire numérique, suite de Conway, audioactive decay, Gleichniszahlen-Reihe. Propose une autre solution Fermer Source of the puzzle: ©G. Sarcone, Focus BrainTrainer, No. 10, page 81

 (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).

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 ...,X3,33,... (X occurrences of 3’s, three 3’s) < ...[3 or 33 or 333],333... ...,33,3X,... (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
1N
11,1N
21,1N
12,21,1N
11,22,21,1N
etc...
(N = any digit)

The 5 Winners of the Puzzle of the Month are:
Chris Peterson, USA
Fabio Cirigliano, Italy
Alessio Medici, Italy
Alessandro Tornago, Brazil
Amaresh G. S., India

Congratulations!

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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:

 lim n  ∞ P(n+1) = λ = 0.303577269... Pn

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’:
OE
ONE O ONE E
ONE O ONE N ONE E ONE O ONE O ONE N ONE E ONE E
etc...

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