Squares of a number yields a palindrome?

Question

I was doing my statistics homework when I observed an interesting pattern:

$ 11^2 = 121 $

$ 111^2 = 12321$

$ 1111^2 = 1234321 $

$ 11111^2 = 123454321 $

$ 111111^2 = 1.234565432 \times 10^{10} $

What's really interesting about this phenomena is that the number in the middle is exactly the # of digits of 1 that you are squaring. I thought perhaps the reason for the palindrome pattern was that I was computing squares of primes, but I believe only 11 is prime.

What's even more interesting is that:

$ 11^0 = 1 $

$ 11^1 = 11 $

$ 11^2 = 121 $

$ 11^3 = 1331 $

$ 11^4 = 14641 $

This is exactly Pascal's triangle, however, this pattern breaks down for $ 11^5 = 161051 $. What I do not understand is why powers of 11 yield Pascal's triangle. I thought Pascal's triangle was related to binomial distributions/expansions, but here we have the implication that this is related to powers of a prime number. Can anyone explain why these patterns happen, and what relation (if any), exists between the two patterns? Or are these just happy coincidences?

And the more I play around with these relationships the more patterns I am finding:

$111^2 = 12321$, $111^3 = 1367631$, however this pattern for $111^n$ breaks down for $n = 4$. So it seems when we add another digit, from $11$ to $111$ the palindrome pattern breaks down earlier (for $11^n$ it breaks down at $n = 5$). Similarly, for $1111^n$ the pattern breaks down at $n = 3$, with $1111^2 = 1234321$ being the only "palindrome number" $1111^n$ will yield. However, I will raise the "conjecture" that no matter how many digits of 1 you have (take $11111111111$, for example), you will always yield a palindrome number for $n = 2$. I hope this leads to some very interesting mathematical discussions, and I hope you enjoyed reading. Also, please excuse the lack of rigor in my statements, I'm just a mere undergrad :)

Answer 1

You can write $11^n = (10^0 + 10^1)^n$ and then you will get that the digit in the $k$th place is the number of ways that you can choose $n-k$ exponents to be $0$ and $k$ exponents to be $1$. This gives you palendromes (at least for $n$ small enough) and identities similar to Pascal's triangle. However, as soon as $n$ is large enough that the "digit" you compute in the $k$th place is greater than $9$, then the palendrome phenomenon will break down. Palendrome phenomena for $111^n$ and $1111^n$ can be analyzed similarly.

Answer 2

"What I do not understand is why powers of 11 yield Pascal's triangle."

The binomial theorem states the following: $$(x+y)^n = \sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$ Using this, we can set $x=10,\ y=1$: $$11^n = (10 + 1)^n = \sum_{k=0}^n \binom{n}{k} 10^k\cdot 1^{n-k} = \sum_{k=0}^n \binom{n}{k} 10^k$$

So as long as you have an $n$ such that $\binom{n}{k} < 10\ \forall k$, you will find that the palindrome pattern works for powers of $11$.