3 ms·

Right. The theory that the parent came up with does seem like a good idea. However, you can't really expect to have been the first one to come up with this for

Right. The theory that the parent came up with does seem like a good idea. However, you can't really expect to have been the first one to come up with this for a problem with as many iq-eyeball-hours as the collatz. Indeed, some googling suggests that this approach of matching powers of 2s and powers of 3s is one existing path for collatz adventurers: https://terrytao.wordpress.com/2011/08/25/the-collatz-conjecture-littlewood-offord-theory-and-powers-of-2-and-3/ https://terrytao.wordpress.com/2011/08/25/the-collatz-conjec... Although I don't understand it, a skim of that Terry Tao post suggests that if the statement about powers of 2 and 3 getting getting farther apart is actually true, it would be a big step towards proving the (weak) collatz.

LegionMammal978 4y ago
Unfortunately, the true property of getting farther apart would have to be more sophisticated than the version the parent used. You can easily find powers of 2 and 3 arbitrarily close on the logarithmic scale by generating the convergents of log 2/log 3.
- dankearney 4y ago
  Got an example? From my exploration, the distance between a power of 3 and the nearest power of 3 tends to grow.
  - LegionMammal978 4y ago
    Did you not see the n = 1000000001, e = 10439860591, o = 6586818670 example in my earlier post? (I edited it in a while after initially posting it.)
    - eru 4y ago
      Slight tangent: I wonder if there's a limit to how close they can get in an absolute sense (instead of in a logarithmic sense)?
      - LegionMammal978 4y ago
        Yes; see, e.g., the answer at https://mathoverflow.net/a/116852 https://mathoverflow.net/a/116852, adjacent to the one cited by the parent. This gives a hard lower bound of |2^x - 3^y| ≥ max(2^x, 3^y)/((e max(x, y))^821013301). (But it's not sufficient for the parent's argument, since (3 + 1/n₁)^o = 3^o·(1 + 1/(3n₁))^o grows larger than 3^o by an exponential factor, so we can always increase (e, o) until the interval is wide enough.)
dankearney 4y ago
The powers of 2 and 3 getting further away bit is not correct as you mention but it isn't actually all that important. My Github page is a WIP so bear with me.. If a cycle does exist for some large-ish cycle length, let’s say 100 odd numbers, then it means that (3+1/x1)(3+1/x2)…(3+1/x100) exactly equals some power of 2. Since (3+1/xn) rapidly approaches 3.0, for any large x1, the product of (3+1/x1)(3+1/x2) etc is going to be pretty close to a power of 3, but it needs to be much larger than that to equal a power of 2, since powers of 2 and 3 are always somewhat far apart. That means that x1, x2 etc must be relatively small so that the product of the (3+1/x) can actually be substantially larger than a power of 3. That puts a hard floor on the possible cycle values, and that floor actually trends smaller as the cycle gets larger since the distance between a power of 3 and the nearest power of 2 also trends larger.
- LegionMammal978 4y ago
  > since powers of 2 and 3 are always somewhat far apart. This is the part where your argument fails. On a logarithmic scale, powers of 2 and powers of 3 can become arbitrarily close together. To see this, notice that for any ε > 0, we can create a rational approximation x/y such that 0 < ln 2/ln 3 - x/y ≤ ε. Multiplying by y ln 3, we have 0 < y ln 2 - x ln 3 ≤ εy ln 3, or equivalently, x ln 3 < y ln 2 ≤ (x + εy) ln 3. Finally, we take the exponential to get 3^x < 2^y ≤ 3^(x + εy) = 3^(x(1 + εy/x)). As a consequence of Dirichlet's approximation theorem, εy/x can become arbitrarily small, so our upper bound 3^(x(1 + εy/x)) can become arbitrarily close to 3^x on a logarithmic scale.

LegionMammal978 4y ago

Unfortunately, the true property of getting farther apart would have to be more sophisticated than the version the parent used. You can easily find powers of 2 and 3 arbitrarily close on the logarithmic scale by generating the convergents of log 2/log 3.

dankearney 4y ago
Got an example? From my exploration, the distance between a power of 3 and the nearest power of 3 tends to grow.
- LegionMammal978 4y ago
  Did you not see the n = 1000000001, e = 10439860591, o = 6586818670 example in my earlier post? (I edited it in a while after initially posting it.)
  - eru 4y ago
    Slight tangent: I wonder if there's a limit to how close they can get in an absolute sense (instead of in a logarithmic sense)?
    - LegionMammal978 4y ago
      Yes; see, e.g., the answer at https://mathoverflow.net/a/116852 https://mathoverflow.net/a/116852, adjacent to the one cited by the parent. This gives a hard lower bound of |2^x - 3^y| ≥ max(2^x, 3^y)/((e max(x, y))^821013301). (But it's not sufficient for the parent's argument, since (3 + 1/n₁)^o = 3^o·(1 + 1/(3n₁))^o grows larger than 3^o by an exponential factor, so we can always increase (e, o) until the interval is wide enough.)

dankearney 4y ago

The powers of 2 and 3 getting further away bit is not correct as you mention but it isn't actually all that important. My Github page is a WIP so bear with me.. If a cycle does exist for some large-ish cycle length, let’s say 100 odd numbers, then it means that (3+1/x1)(3+1/x2)…(3+1/x100) exactly equals some power of 2. Since (3+1/xn) rapidly approaches 3.0, for any large x1, the product of (3+1/x1)(3+1/x2) etc is going to be pretty close to a power of 3, but it needs to be much larger than that to equal a power of 2, since powers of 2 and 3 are always somewhat far apart. That means that x1, x2 etc must be relatively small so that the product of the (3+1/x) can actually be substantially larger than a power of 3. That puts a hard floor on the possible cycle values, and that floor actually trends smaller as the cycle gets larger since the distance between a power of 3 and the nearest power of 2 also trends larger.

LegionMammal978 4y ago
> since powers of 2 and 3 are always somewhat far apart. This is the part where your argument fails. On a logarithmic scale, powers of 2 and powers of 3 can become arbitrarily close together. To see this, notice that for any ε > 0, we can create a rational approximation x/y such that 0 < ln 2/ln 3 - x/y ≤ ε. Multiplying by y ln 3, we have 0 < y ln 2 - x ln 3 ≤ εy ln 3, or equivalently, x ln 3 < y ln 2 ≤ (x + εy) ln 3. Finally, we take the exponential to get 3^x < 2^y ≤ 3^(x + εy) = 3^(x(1 + εy/x)). As a consequence of Dirichlet's approximation theorem, εy/x can become arbitrarily small, so our upper bound 3^(x(1 + εy/x)) can become arbitrarily close to 3^x on a logarithmic scale.