7/28/2019 Read Out Loud Sequence

http://slidepdf.com/reader/full/read-out-loud-sequence 1/2

A Survey of the Properties of the

“Read-Out-Loud” Sequence

Matthew Weaver

February 18, 2013

1 Introduction

We define the “Read-Out-Loud” Sequence as follows:

For any given seed n ∈ N, each new generation is created by taking theprevious generation and, starting at the left, counting how many common digitscoincide. Once a total is reached, the count followed by the digit is added tothe right of the new generation string. This process is repeated until the entireprevious generation string is read and transcribed into the new generation.

Example of seed = 1:

1, 11, 21, 1211, 111221, 312211, 12112221, 1112213211, . . .

This sequence is called the Read-Out-Loud Sequence because a more intuitiveway of understanding it is that each generation is the previous generation liter-ally read out-loud and then transcribed to paper:

1 is “one 1.” 11 is “two 1’s.” 21 is “one 2 one 1”. . .

2 Properties

Theorem 2.1. Only the digits  1, 2, and  3 can be generated in the series.

Proof.

Lemma 2.2. Digits ≥ 4 cannot be generated in the sequence.

Proof. Assume a 4 exists somewhere within a string of the sequence.

. . . 4 . . . (1)

This yields two possibilities: Either the four was present in the previous string(i.e. “one 4”), or there were four of a number in a row (i.e. “four 1’s”). Only

1

7/28/2019 Read Out Loud Sequence

http://slidepdf.com/reader/full/read-out-loud-sequence 2/2

the latter is of interest, as a 4 cannot be present in the string without beinggenerated at some point unless the original seed contains a 4, which is irrelevant

to this proof. Now, going backwards in the sequence from the 4, we get thefollowing:

. . .xxxx. . .→ . . . 4x . . . (2)

Going back another generation, we get,

. . . x1 . . . xxx1 . . . xx → . . .xxxx. . .→ . . . 4x . . . (3)

However, if this were the case,

. . . x1 . . . xxx1 . . . xx → . . . (2x)x . . . (4)

and . . . (2x)x . . . does not become . . . 4x . . . in the following generation. There-fore, 4 can never be generated in this sequence.

Lemma 2.3. The digits  1, 2, and  3 can be generated in the sequence.

Proof. This is easiest to prove by example. Start with the seed 1. the Sequencesgoes...

1, 11, 21, 1211, 111221, 312211 . . . (5)

Notice how new ones, twos, and a three were introduced into the sequence.Therefore, it is possible to generate 1, 2, and 3.

Combining Lemma 2.2 and Lemma 2.3, we can conclude that exclusively thedigits 1, 2, and 3 can be generated in the sequence.

Theorem 2.4. There will always be an even number of digits in every element 

of the series.Proof. Assume there are an odd number of digits in the element, n:

x1x2 . . . xn−1xn (6)

For this number to be generated, there were x1 many x2’s, x3 many x4’s , . . . ,and xn−2 many xn−1s. However, there is no way for the last digit, xn, to begenerated. Therefore there must always be an even number of digits, as futureelements are defined by pairs of digits.

Corollary 2.5. The sum of the odd digits (i.e. the first, third, fifth, etc. . . digits beginning on the left) must always be even.

Proof.

1

1I would like to continue analyzing this series, but unfortunately, I have to study for my

math midterm. So for now, this project is To Be Continued. . .

2