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WOWZY! COOL MATH!
CURIOUS MATHEMATICS FOR FUN AND JOY
JANUARY 2014
PROMOTIONAL CORNER: Have
you an event, a workshop, a website,
some materials you would like to share
with the world? Let me know! If the work
is about deep and joyous and real
mathematical doing I would be delighted
to mention it here. ***
Every Monday evening at 4pm
pacific/7pm eastern Illustrative
Mathematics hosts discussion on the
previous Tuesday’s posted task. Join a conversation with other teachers,
mathematicians, mathematics education
experts; voice queries and ideas; learn
nuances of the Common Core. Go to
www.illustrativemath.org and click on
NEW! PROFESSIONAL
DEVELOPMENT and then CONTINUE
THE CONVERSATION for details.
PUZZLER: It is possible to sprinkle
plus and minus sign throughout a list of
the first seven counting numbers to get a
sum of zero:
1 2 3 4 5 6 7 0+ + − + − − + = .
The same distribution of plus and minus
signs among the first seven square
numbers also gives zero. Whoa! 2 2 2 2 2 2 21 2 3 4 5 6 7 0+ + − + − − + = .
Is there a number N and some
distribution of plus and minus signs that,
when applied to the list of the first N
numbers AND when applied to the list
of the first N square numbers AND
when applied to the list of the first N
cube numbers gives a sum of zero each
time?
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
THE PROUHET-THUE-MORSE
SEQUENCE
In 1851 mathematician Eugène Prouhat
discovered a sequence with surprising
mathematical properties. This sequence
was rediscovered, and described
explicitly, in 1906 by Axel Thue, and
then made famous to the world by
Marston Morse in 1921 with its
applications to advanced calculus and
geometry. I cannot resist introducing you
to this famous sequence for the start of
the New Year.
Here’s the sequence:
Start by writing each numbers in base
two (use a 1 2← machine from
gdaymath.com/courses/exploding-dots/!)
and look at whether the number of 1s
in each expression is even or odd:
The sequence words of “even” and
“odd” one obtains is the Prouhet-Thue-
Morse sequence. That’s it!
If we write 0 for “even” and 1 for
“odd,” then the sequence appears:
011010011001011010010110011010011001011001...
If we let Nt denote the N th entry in this
sequence ( 0N ≥ ), then 0Nt = if there
are an even number of 1s in the binary
representation of N , it is 1 otherwise.
THE OPENING PUZZLER
Write the first eight counting numbers
(starting with zero) and match them with
the first eight terms of the Prouhet-Thue-
Morse sequence.
Assign + to each term that sits with a 1 and − to each term that sits with a 0 . Notice:
1 2 3 4 5 6 7 0+ + − + − − + =
and
2 2 2 2 2 2 21 2 3 4 5 6 7 0+ + − + − − + = .
Whoa!
Do the same for the first sixteen terms:
This time notice that the same
phenomenon works up to the cubes:
2 2 2 2 2 2 2 2 2
2 2 2 2 2 2
3 3 3 3 3 3 3 3 3
3 3 3 3 3 3
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 0
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 0
1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 0
+ − + − − + + −
− + − + + − =
+ − + − − + + −
− + − + + − =
+ − + − − + + −
− + − + + − = The first 32 counting numbers with
signs given by the Prouhet-Thue-Morse
sequence give sums that work up to
fourth-power sums, the first 64 up to
fifth-power sums, and so on.
In math speak:
( )2 1
0
1 0
N
nt k
n
n−
=
− =∑
for 0,1,2,..., 1k N= − . Mega-Whoa!
The goal of this essay is to prove this
remarkable property of the sequence.
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
ALTERNATIVE CONSTRUCTIONS
Before we start, here are some
alternative ways to construct the
Prouhet-Thue-Morse sequence, each
curious in its own right.
Alternative Construction 1:
Look at the first four entries of the
sequence followed by the next four:
Or the first two entries followed by the
next two:
Or the first eight followed by the next
eight:
It seems that the second block of entries
is the “dual” of the first: it is identical to
the first half, except it has 1s for 0 s, and
0 s for 1s. If this is true, then we can
construct the Prouhet-Thue-Morse
sequence as follows:
Stage 1: Write 0 .
Any other Stage: Repeat what
you currently have, but swap the
1s for 0 s, and 0 s for 1s.
This gives:
Reason Why this is Happening:
If we think of the first four numbers,
0,1,2,3, as each having a base two
representations two digits long,
00, 01, 10, 11 , then placing a 1 in front
of each of these,100, 101, 110, 111, gives
the base two representation of the next
four numbers, 4,5,6,7 . The count of 1s
in each representation has increased by
one.
In general, if k is a number between 0
and 2 1N − , inclusive, then 2N k+ is a
number between 2N and 12 1N + −
inclusive. Each representation in the
second block has one more 1 than its
matching representation in the first
block. The evenness/oddness of the
count of 1s are thus opposite for these
two numbers. So after the first 2N
entries in the Prouhet-Thue-Morse
numbers, the next 2N entries are indeed
opposite in value, in turn, of those first
entries.
Challenge: Explain why the first 4 , the
first16 , the first 64 , and, in general, the
first 4N entries of the sequence are
palindromes. Also, explain why the first
4N entries have the form: B B B B
where B is a sequence of 0 s and 1s,
and B is its dual.
Challenge: Fold a strip of paper, blue
(B) on one side and red (R) on the other,
into quarters as shown.
The colours that appear on the top
surface of each layer are: BRRB.
Now fold this folded strip into quarters
again, following the same pattern.
The sequence of colors is
BRRBRBBRRBBRBRRB. If we repeat
this folding over and over again, we get
the Prouhet-Thue-Morse sequence. Can
you explain why?
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
Alternative Construction 2:
Look at alternate terms in the Prouhet-
Thue-Morse sequence:
The entries in even positions (starting at
zero) are: 0110100110010110.... , the
original sequence!
The entries in the odd positions (starting
at one) are: 1001011001101001... , the
dual of the original sequence!
It seems:
The Prouhet-Thue-Morse sequence is
itself and its dual interwoven. The
entry in an even position 2N is the
same as the entry in positionN , and
the entry in odd positon 2 1N + is the
dual of this entry.
That is:
2N Nt t= and 2 1N Nt dual of t+ = .
Why this is so: Given the base two
representation of the number N , we
obtain base two representation of 2N
simply by placing a 0 at its end. The
number of 1s does not change, so:
2N Nt t= .
We obtain the base two representation of
2 1N + by doubling and adding a one,
which is the same as placing a1 at the
end of the representation. This changes
the evenness/oddness of the count of 1s
and so:
2 1 1N N Nt dual of t t+ = = − .
That explains it.
So if we would like to compute, say, the
301st entry of the sequence, which is
300t , then just notice:
( )300 150 75 37 18
9 4 2 1
1 1 1
1 1 1 0.
t t t t t
t t t t
= = = − = − −
= = − = − = − =
Alternative Construction 3:
Start with 0 and then perform the
following iterative process:
For any given sequence of 0 s and 1s,
replace each 0 with the pair 01 , and
each 1 with the pair 10 .
This gives:
0 01 0110 0110 10 01
0110 10 0110 01 0110
→ → →
→
→L
Notice: If you start with 1 instead we
seem to obtain the dual of the Prouhet-
Thue-Morse sequence:
1 10 1001 1001 0110
10 01 0110 0110 10 01
→ → →
→
→L
Reason Why This Works:
This replacement operation doubles the
number of entries in any portion of the
sequence. So as we perform this
operation, an entry with N terms to its
left (in position N ) is bumbed to
position 2 .N Since 0 is replaced with
01 and 1 with 10 , if this entry is a 0 is
remains a zero and remains a 1 if it was
a 1. We get:
2N Nt t= .
The entry just to the right of this, after
the replacements are done, is the
opposite:
2 1 2N N Nt dual of t dual of t+ = = .
This replacement operation thus
generates a sequence with exactly the
same recursion relations as approach 2,
and so must produce the same Prouhet-
Thue-Morse sequence.
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
Alternative Construction 4:
Expand the infinite product:
( ) ( )( ) ( )( )2 2 4 81 1 1 1 1x x x x x− − − − − L
You get: 2 3 4 5 6 7 81 x x x x x x x x− − + − + + − − +L
The sequence of plus/minus signs is the
Prouhet-Thue-Morse sequence!
The Reason Why:
We have:
1 1x x− = −
( ) ( ) ( )2 2
2 3
1 1 1 1
1
x x x x x
x x x
− − = − − −
= − − +
( ) ( )( )
( )
2 4
2 3 4 2 3
2 3 4 5 6 7
1 1 1
1 1
1
x x x
x x x x x x x
x x x x x x x
− − −
= − − + − − − +
= − − + − + + −
and
( ) ( )( )( )2 4 81 1 1 1x x x x− − − −
equals the previous answer together with 8x− times the previous answer. This will
give us the terms 8 15,...,x x but with dual
distribution of signs. We see alternative
construction 1 in play!
PROVING THE PUZZLER
(WARNING: HARD!)
Consider the first 2N counting numbers:
0,1, 2,3,..., 2 1N − . Let A be subset of
these which have a matching Prouhet-
Thue-Morse term that is 0 , and B the
subset with matching terms of 1:
{ {0,1,.., 2 1}: 0}
{ {0,1,.., 2 1}: 1}
N
a
N
b
A a t
B b t
= ∈ − =
= ∈ − =
Since + signs are assigned to elements of one set and − to elements of the other, we want to show:
k k
a A b B
a b∈ ∈
=∑ ∑ for 0,1,2,..., 1k N= − .
For example, for the first eight terms
( 3N = ) we had:
Thus {0,3,5,6}A = and {1,2,4,7}B = .
The sums we consider are:
1 2 3 4 5 6 7 0+ + − + − − + =
and
2 2 2 2 2 2 21 2 3 4 5 6 7 0+ + − + − − + =
These are equivalent to:
2 2 5 2 2 2 2 2
0 3 5 6 1 2 4 7
0 3 5 6 1 2 4 7
+ + + = + + +
+ + + = + + + .
Actually notice that (0) (3) (5) (6) (1) (2) (4) (7)p p p p p p p p+ + + = + + +
holds for any polynomial of degree zero,
one, or two. (Can you see why?)
Back to the general case:
Let’s prove that ( ) ( )a A b B
p a p b∈ ∈
=∑ ∑ for
any polynomial of degree up to and
including 1N − . (Then, in particular, the
result will be true for the polynomials
( ) kp x x= , 0,1,.., 1k N= − .)
We have seen that the result is true for
3N = (above), and it is true for 1N =
and 2.N = We seem set for an argument
by induction.
Suppose the result is true for all values
up to N and let’s consider the ( 1)N + th
case.
Let A and B be the sets defined
previously for the numbers 10,1, 2,..., 2 1N+ − and let p be a
polynomial of degree at mostN .
When expanding ( ) ( )2Np x p x+ − , the
highest powers of x cancel and so
( ) ( )2Np x p x+ − is a polynomial of
lower degree. (It is of degree at most
1N − .)
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
By the induction hypothesis:
( )( )
( )( )0 2
0 2
( 2 )
( 2 )
N
N
N
a A
a
N
b B
b
p a p a
p b p b
∈
≤ <
∈
≤ <
+ −
= + −
∑
∑
Some algebra:
( ) ( )
( ) ( )
0 2 0 2
0 2 0 2
2
2
N N
N N
N
a A b B
a b
N
b B a A
b a
p a p b
p b p a
∈ ∈
≤ < ≤ <
∈ ∈
≤ < ≤ <
+ +
= + +
∑ ∑
∑ ∑
Now recall from alternative construction
1 that if a is between 0 and 2N , then
2Na + is between 2N and 12N + . So each
side of the above equation is a
summation over all the values from 0
up to 12N + .
Also, matching terms for a and 2Na +
in the Prouhet-Thue-Morse sequence
have opposite values. So if a belongs to
the set A , then 2Na + belongs to the set
B , and vice versa. So the sum above
really reads:
( ) ( )
( ) ( )
1
1
2 2 0 2
2 2 0 2
N N N
N N N
b B b B
b b
a A a A
a a
p b p b
p a p a
+
+
∈ ∈
≤ < ≤ <
∈ ∈
≤ < ≤ <
+
= +
∑ ∑
∑ ∑
That is:
( ) ( )1 10 2 0 2N N
b B a A
b a
p b p a
+ +∈ ∈
≤ < ≤ <
=∑ ∑
which is what we wanted to establish for
the induction argument.
Phew!
Comment: The Prouhet-Thue-Morse
sequence does not give all sets of
numbers with magical sums-of-powers
properties. Care to find other sets of
curious consecutive counting numbers?
RESEARCH CORNER:
Connections to Paper Folding
The appearance of the Prouhet-Thue-
Morse sequence through paper-folding
reminds of the curious mathematics
found via crease-marks through simpler
folding. (This mathematics is described
in detail in chapter 8 of Mathematics
Galore! published by the MAA in 2012.)
Briefly:
Take a strip of paper and imagine its left
end is taped to the ground. Pick up the
right end, fold the strip in half, and
unfold it. This leaves a valley crease in
its center. Label it 1.
Suppose we perform two folds, lifting
the right end up over to the left end.
When we unfold the paper, we find three
creases: two valley creases and one
mountain crease. Label the mountain
crease as 0 .
For three folds we get:
The sequence for four fold turns out to
be 110110011100100.
© James Tanton 2014
www.jamestanton.com and www.gdaymath.com
Look at the sequences we have for one,
two, three, and four folds:
1
1 1 0
110 1 100
1101100 1 1100100
Each begins with the entire previous
sequence, and the latter portion of the
sequence is the “reverse dual” of the
first. These are all centered about a
central column of 1s.
It is easy to continue this pattern to
generate the paper-folding crease
sequence.
This got me thinking on yet another way
to construct the Prouhet-Thue-Morse
sequence.
Alternative Construction 5: Write the first one, three, seven, fifteen,
entries of the Prouhet-Thue-Morse
sequence in the same layout as above:
0
0 1 1
011 0 001
0110001 1 0110110
Research Question 1: It seems we have
a central column that alternates 0 and 1
this time. Also, the second block of
entries, it seems, is just the dual (not
reverse dual) of the first block, and the
first block is a repeat of the previous
line.
Do these patterns persist? Is this indeed
a valid alternative method for generating
the Prouhet-Thou-Morse sequence?
Research Question 2: What is the
sequence of crease patterns that arise in
the Prouhet-Thue-Morse folding?
What mathematical properties does this
crease sequence possess? If given just
this crease sequence, is it clear how to
construct the Prouhet-Thue-Morse
sequence from it?
© 2014 James Tanton