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 :)