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Previous
Puzzles of the Month + Solutions
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| February-March
2008, Puzzle nr 116 |
 Back to Puzzle-of-the-Month
page | Home  |
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!
|
© 2004 G.
Sarcone, www.archimedes-lab.org
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:
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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