Cool Math Essay_January 2014_THUE SEQUENCE - Thinking

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WOWZY! COOL MATH!

CURIOUS MATHEMATICS FOR FUN AND JOY

JANUARY 2014

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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. ***

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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

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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.



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?




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

= = = − = − −

= = − = − = − =


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.




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 − .)



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.



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

[email protected]

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Cool Math Essay_January 2014_THUE SEQUENCE - Thinking