I really enjoy Numberphile videos. This one has be a bit perplexed -- not by the problem itself, but rather the intoned mystery around it. When expressed with binary digits, a clear pattern emerges. https://youtu.be/5mFpVDpKX70
-
Näytä tämä ketju
-
Here's some properties explaining why you always end up at 1. A) 1 left shifted by multiples of two are always one greater than a multiple of three. 3, 15, 63, 255, etc. B) The x/2 step just eats 0s on the right. (ie, right shift the 0s away)
1 vastaus 0 uudelleentwiittausta 0 tykkäystäNäytä tämä ketju -
C) The +1 step converts groups of 1s on the right into groups of 0s on the right (which get eaten by x/2). D) x3 can also be though of as x2+x, which is a left shift of x + x. You can think of this as a way of filling "middle" zeros with 1s.
1 vastaus 0 uudelleentwiittausta 0 tykkäystäNäytä tämä ketju -
The steps can be thought of like this: 1) Sprinkle in some 1s (left shift and add self, or the x3 part) 2) Convert all contiguous 1s on the right into 0s (and the "blocking" 0 into a 1. 3) Right shift away the newly created zeroes.
1 vastaus 0 uudelleentwiittausta 0 tykkäystäNäytä tämä ketju -
I suppose though, seeing the pattern isn't enough for math (which isn't really my field anyway). There have to be proofs expressed mathematically. Still, it's curious to me that it's presented so mysteriously.
1 vastaus 0 uudelleentwiittausta 0 tykkäystäNäytä tämä ketju
As a bonus conjecture, I would go out on a limb and say that any number multiplied by 3 will result in the same or more 1s when represented in binary than the original number. 3 just has some neat properties in binary.
Lataaminen näyttää kestävän hetken.
Twitter saattaa olla ruuhkautunut tai ongelma on muuten hetkellinen. Yritä uudelleen tai käy Twitterin tilasivulla saadaksesi lisätietoja.