Although, some of them, it can sure feel like you’re making progress. Like, I’ve found (as I’m sure have others before me) that if there are any counterexamples to the Collatz Conjecture, they must be of the form either 12n+3 or 12n+7. And I continually feel like I’m on the verge of finding something much bigger than that, with that specific problem.
Am I actually? Probably not. I’m probably not even on the verge of something that others have already found: Realistically, the 12+3 and 12n+7 thing just might be the most exciting thing I’m ever going to find concerning Collatz.
I agree completely with your post. Incidentally, the explanation for the fact that the second digit of the powers of 2 repeats every 20 steps isn’t obvious. What is obvious is that the last two repeat every 40 steps and so the first is not astonishing. Since phi(100) = phi(25) x phi(4) = 20 x 2 = 40, it follows from Fermat’s little theorem that 2^40 = 1 (mod 100) from which the repitition every 40 is obvious. That there is such a repetition no matter what number you are raising to powers or what base you are using is also obvious.
As to the OP, best confine it to the trash heap where it belongs. There is much new mathematics being discovered every day, but not that sort of drivel.
What I wrote about is partial nonsense. Clearly 2^40 = 1 (mod 100) is false. But 2^40 = 1 (mod 25) is correct and thus repeat every 40 times (actually, every 20 since phi(25) = 20) and mod 4 all higher powers of 2 are 0. So the last two digits repeat every 20 times. Sorry about my brain cramp.
Another way to see this is to consider that there are 100 possible values for (2^n) mod 100 (n>1). But all of them have to be even (so 50 of the possibilities are eliminated) - and none of them can end with 0 - so 10 more are eliminated - leaving just 40 left. So 40 is the maximum length before you repeat a value.
This is just equal to the order of the multiplicative group of integers modulo 25, so 220 ≡ 1 (mod 25), 320 ≡ 1 (mod 25), 420 ≡ 1 (mod 25), 6 20 ≡ 1 (mod 25), and so on. [So 220 ≡ 420 ≡ 620 ≡ 76 (mod 100), 320 ≡ 1 (mod 100), etc.]
I am sure one can discover many things through playing around with a calculator (i.e., numerical + symbolic computer algebra system), but this search should be directed somehow, the result should be non-trivial and actually interesting (maybe has useful applications or generalizes some existing results), you should prove it is actually true, and of course check that it has not already been published.
The so-called Euler totient (what is a totient anyway) function phi(n), is defined for positive integers n as the number of numbers less than or equal to n whose only common divisor with n is 1. The reason for the “or equal” part is to force phi(1) = 1. There are three properties that matter. First that if m and n have no divisor > 1, then phi(mn) = phi(m)phi(n). Second that if p is prime, then phi(p^k) = (p-1)p^(k-1) and third, Fermat’s little theorem (so named to distinguish it from the famous last), if a and n have no common divisor > 1, then a^phi(n) = 1 (mod n). In the case at hand, phi(25) = 20, so 2^20 = 1 (mod 25), so that 2^(20+k) = 2^k (mod 25) for all k, while 2^(20+k) = 2^k = 0 (mod 4) for all k except k=1.
The upshot of all this–should anyone still be interested, is that the last two digits of the powers 2^k start repeating with a period of 20 starting from 2^2. But 2^21, 2^41, etc., all end in 52. For similar reasons, the last three digits repeat of 2^k with a period of 100 = phi(125) starting with 2^3.