Number of palindromes made of n digits

[archaic]

Hello! A while I tried to find a formula that allows finding the number of palindromes and I actually came up with something :
So let's say P is the number of palindromes made of n digits,
if n is even then $$P = 9 * 10^\frac{n-2}{2}$$
else if n is odd and =/= 1 then $$P = 9 * 10^\frac{n-1}{2}$$
This probably exists somewhere on the internet, but who knows ..
See ya!

 

[phinds]

does it work for n=1 and n=2?

 

[archaic]

does it work for n=1 and n=2?

Thanks for your answer, I didn't consider n = 1 ..
It works for n = 2 though, 9 * 10⁰ = 9 (11,22,33,44,55,66,77,88,99).

 

[mfb]

Where is the problem with n=1? All 1-digit numbers are trivial palindromes, and the formula gives 9.

There are 9 options for the firs digit (no 0) and 10 options for each following digit up to the center of the number. For even n this means a total of n/2 numbers you are free to choose, for odd n it is (n+1)/2. Taking out the first one, we have (n-2)/2 and (n-1)/2 digits with 10 choices and 1 digit with 9 choices. The rest of the number follows from these choices. And that leads to the formulas in the first post.
Nice observation!

 

[archaic]

Where is the problem with n=1? All 1-digit numbers are trivial palindromes, and the formula gives 9.

There are 9 options for the firs digit (no 0) and 10 options for each following digit up to the center of the number. For even n this means a total of n/2 numbers you are free to choose, for odd n it is (n+1)/2. Taking out the first one, we have (n-2)/2 and (n-1)/2 digits with 10 choices and 1 digit with 9 choices. The rest of the number follows from these choices. And that leads to the formulas in the first post.
Nice observation!

Thank you!
I went to check and actually it states here that 0 is considered a palindrome.

 

[mfb]

It is a palindrome, but is it a 1-digit number?
A matter of definition I guess.