A Pattern I Randomly Found

I don't quite remember how I found this, but when you put in:

f(n)=10n+818f(n) = \frac{10^{n}+8}{18}

You get the pattern:

1=11 = 1

2=62 = 6

3=563 = 56

4=5564 = 556

5=55565 = 5556

etc etc

Having changed it a bit, I changed it to:

f(n)=10n+1+8181f(n) = \frac{10^{n+1}+8}{18}-1

And you get the pattern:

1=51 = 5

2=552 = 55

3=5553 = 555

4=55554 = 5555

5=555555 = 55555

etc etc

I just wondered if this pattern has any significance...

#NumberTheory

Note by Luke Corey
2 years, 7 months ago

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Comments

If you modify the second f(n)f(n), you end up with

f(n)=10n+11018=5(10n1)9. f(n) = \frac{10^{n+1}-10}{18} = \frac{5(10^n-1)}{9}. .

As

10n1=9i=0n110i, 10^n - 1 = 9 \sum_{i=0}^{n-1} 10^i,

you have that

f(n)=5×i=0n110i. f(n) = 5 \times \sum_{i=0}^{n-1} 10^i.

In decimal form, this represents a string of nn 55's, which explains the phenomenon you see.

To explain the first f(n)f(n), rewrite it as

f(n)=10n+11018+1 f(n) = \frac{10^{n+1}-10}{18} + 1

and apply the same principle as above. Long story short: what you have found is not a random pattern but rather formulas that turn out to have interesting decimal representations.

A Former Brilliant Member - 2 years, 7 months ago
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