Friedman Numbers and 'mystery'

Friedman numbers are a set of those numbers which can be expressed as a function involving the digits of that number and basic operators such as + , - , * , / , ^ , ().

For example,

25 = 5^2

125 = 5^(1+2)

Now, the 'mystery' which I want a help in resolving is that as we move forward in the number line or in other words, if the numbers tend to get bigger and bigger, then the frequency of finding such numbers also increase. In mathematical language,

\({lim}_{n->\infty} \frac{f(n)}{n}\) = 1

where f(n) is a function of n generating Friedman numbers.

It would be great if anyone knowing a proof of this limit shares it here. I am really very eager to know it. Thanks in anticipation

Note by Kartik Sharma
3 years, 11 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)


Sort by:

Top Newest

This is not a complete proof, but we can at least get an understanding of this "mystery" if we look at very large numbers. Typically, a very large number will have a roughly equal number of each of the digits \(1,2,3,4,5,6,7,8,9,0\), so, let's say we have a number that has \(10n\) digits, where \(n\) is a suitably large integer. Let's propose some kind of a mathematical expression using basic operations that uses \(10n\) digits. How many ways can the digits of that large number can be plugged into this mathematical expression? It's approximately

\( \dfrac { \left( 10n \right) ! }{ { \left( 10! \right) }^{ 10 } } \)

different ways. Now, but this number is approximately \(10\cdot { 10 }^{ \displaystyle n }\)

What's the ratio of the two, that is, this number divided by the number of ways its digits can be permutated in that mathematical expression? It's

\(10\cdot { 10 }^{\displaystyle n }\dfrac { { \left( 10! \right) }^{ 10 } }{ \left( 10n \right) ! } \)

As \(n\) increases, it first starts out as a very large number, but it then very quickly drops down towards zero. That is, there can be far more permutations possible than the number itself! Thus, the odds of one of those permutations yielding exactly that number rises exponentially for large numbers, to where it becomes nearly a certainty.

Michael Mendrin - 3 years, 11 months ago

Log in to reply

@Calvin Lin @Michael Mendrin

Kartik Sharma - 3 years, 11 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...