# Symmetry and floor function

We all know symmetrical sequences. could be fascinating at times. we will be reading 'bout symmetrical formulas.

## symmetries

a really common symmetry is in the question below

we can clearly see that each number is going from an ascending order to a nu mber and then descends until 1. \(1234321\) starts in natural order{1,2,3] and reaches 4. than it goes in the reverse order. we can tell the answer is \(123454321\). but we dont have a formula to prove it, or do we!?????## find the next next number in the following sequence

\[1,121,12321,1234321\]

## the formula and the proof

from the question above, we get that
\[0^2=0, 1^2=1, 11^2= 121, 111^2= 12321, 1111^2 =1234321\]
and so on.... notice that

⌊\(\frac{10}{9}\)⌋=11

⌊\(\frac{100}{9}\)⌋=111

⌊\(\frac{1000}{9}\)⌋=1111.

we get that ⌊\(\frac{10^N}{9}\)⌋=n-digited-111.....

so since n-digited \(1111....^2\) makes the Nth term in the sequence,we conclude the formula:

Nth-term=\(⌊\frac{10^{N}}{9}⌋^{2} \)

## the tricky one

the previous one didn't really need much of a formula to determine the next term. but this will

well, the answer guessed by you is probably 12345678. but it is \(12345679\) it has not much discussion, but it cant be called a symmetry, since numbers are only ascending.. well try it and the formula is## find the next next number in the following sequence

\[1,12,123,1234,12345,123456,1234567 \]

Nth-term=\(⌊\frac{10^{N}}{81}⌋ \)so the 9th term is\(\boxed{12345679}\)

## similar symmetries

well i have shown you the basics , now try makin' a formula for the following

## find the formula for the Nth term

a)9,1089,110889,11108889.... b)36,4356,443556,44435556...

**Cite as:**Symmetry and floor function.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/symmetry-and-floor-function/