The whole point of this proof is that it can only be proven by trying. Like Goldbach's conjecture, i.e, all even numbers can be written as the sum of two even numbers (8=3+5, 16=3+13, 30=13+17, etc). No matter what you try, it must be directly shown.

Cosmic honeymoon creating an offspring which is this Universe. LOL! Sorry Cosmology joke.

But if this is evidence of a multiverse than the first law of thermodynamics is not applicable to a Universe. — TheQuestion

This process can be repeated over and over. Every time a low entropy can be created from nothing (the two empty parallel branes) from two entropically maximized universe on both branes. Which is similar to two universes forming on two sides of a wormhole, accelerating away from each other, after which two new universes are created (like two branes periodically meeting, residing, approaching, meeting, etc.).

All universes must obey the laws of thermodynamics, i.e, where time moves forward. Be it the many worlds in the MWI or whatever. There could be local reversals of time, like gas collecting in one corner, but these are due to conscious intervention. The initial state can't be such that this occurs later. Only in the very universe there could be local patches were time might seem to run backwards, but these are always embedded in the larger whole of forward time and can't lead to such time reversals in later times. Which is not to say that order can't occur.

This discussion is often heard in the context of Boltzmann brains. This clearly violates the second law, and in an infinite universe, occupied by homogeneous matter, this should eventually happen. A whole universe could evolve like this. It is said that these fluctuations are temporary, like the local patches of time reversal in the early universe. The point is though that these kinds of local ordering can only appear in the very early universe, to be overtaken by the forward march of time. If they appear later this will need initial fine-tuning, while this fine-tuning is not required for ordered life to appear. All initial random configurations will inevitably lead to life, and any deviation from randomness will be flattened out fast.

What do you think of the fact that 3n+13n+1 is a straight line and 2m2m is an exponential function? As far as I know, a straight line and an exponential graph intersects at a maximum of only two points. Is this relevanat? :chin: — TheMadFool

You can compare only the functions 3x+1 and 2exp(x). For comparing 3n+1 and 2exp(m) there are only the solutions I mentioned:


For example 1, 5, 21, 85, 341, 1365, etc. Look at the differences: 2exp2, 2exp4, 2exp6, 2exp8, 2exp10, etc. A very small subset of the integers (but infinite!).

In general: x(k)=sum(0 to k)2exp(2k). This gives rise to the row 1, 21, 85, etc.

Once in a while (for larger numbers), we also hit a number that if we divide by 2 we can keep doing that until we reach — Benkei

This is not how it works. All even numbers I mentioned in the previous comment, will orbit back to 9, 12, or 15. But they can't be part of a series except as a starting number.

There are no orbits that include these, except as starting numbers. The chances you hit a power of 2 grow smaller for increasing n. For 21 you already hit it the first time. There are very few numbers. So every backward part of orbits that hits 21 hits 64. You can backwards hit 5, giving 16. But only very few hit one as part of a forward path. So going backwards you have to hit a power of 2, while in going forwards this holds for limited numbers for starter values. Usually 5 gets hit. Which goes from 16 to 8, 4, 2, 1. When numbers grow, 21, 85, 341, 1365, etc. can be hit backwards, giving rise to a direct line to 1 (so they can't be produced forwards).

In fact, the whole problem is reduced to show the conjecture for only the numbers 9x2exp(n), 12x2exp(n), and 15x2exp(n). Still an infinity.

Correction. The last chain of numbers must be:

15, 30, 60, 120, 240, 480, 960, 1920, 3840, 7680.....

This only gives an unlimited but partial set of valid values of x. The only values for which 3x+1=2exp(n) are x(k)=sum(0 to k)2exp(2k).

For example 1, 5, 21, 85, 341, 1365, etc. Look at the differences: 2exp2, 2exp4, 2exp6, 2exp8, 2exp10, etc. A very small subset of the integers (but infinite!).

If you could show that for every x the serie following ends up at a value smaller than x, the conjecture would be proven. All integers up to 100 converge to 1.so if the 101 orbit ends on a number between 1 and 100 it will do. 101 goes to 304, which goes to 152, which goes to 76, so yes, it delivers. 102 goes 307, which goes 922, which goes 411, 1234, 617, 1852, 926, 463, 1390, 695, 2086, 1043, 3130, 1565, 4696, 2348, 1174, 587, 1762, 881, 2644, 1322, 661, 1984, 992, 496, 248, 124, 62, 31, 94, 47, 142, 71, 214, 107, 322, 161, 484, 242, 121, 364, 182, 91, 274, 137, 412, 206, 103, 310, 155, 466, 233, 700, 350, 175, 526, 263, 790, 345, 1036, 518, 259, 778, 389, 1268, 634, 317, 952, 476, 238, 119, 358, 179, 538, 269, 808, 404, 202, 101, heeee, we had that already, so it delivers!

But now the crucial point:

9, 18, 36, 72, 144, 288, 596, 1192, 2384, 4768,.....
and
12, 24, 48, 96, 192, 384, 768, 1536, 3072, 6144...
and
15, 30, 60, 120, 320, 640, 1280, 2560, 5120...,

will never be part of a series running backwards nor forwards, except as starting point (you can check this by looking at the numbers in the row above, and I used this in fact to correct a number I calculated wrongly). So you always have to check these numbers separately. Which looks like a considerable reduction, but still are infinite numbers....