Why does this pattern with consecutive numbers in the Collatz Conjecture work?

Question

I have been getting into the Collatz Conjecture and I have noticed something interesting. Consecutive numbers often take the same amount of numbers to get to 1. For example, $84$ goes to $42$ to $21$ to $64$ to $32$ to $16$ to $8$ to $4$ to $2$ to $1$. That's ten numbers. $85$ goes to $256$ to $128$ to $64$ to $32$ to $16$ to $8$ to $4$ to $2$ to $1$. That's also ten numbers. There are many more examples, like $60$ and $61, 76$ and $77,$ and $92$ and $93.$ As the numbers get larger, the number of consecutive integers that take the same amount of numbers to get to one grows. $386, 387, 388, 389, 370,$ and $371$ all take $121$ numbers to get back to $1$! I have thought a lot about this, but I can't figure out why this pattern works. I also cannot find a pattern of when this pattern occurs. It seems to be random, but I might be wrong. Can someone help with these questions?

Answer 1

There are indeed patterns of this form.

If, say, you start with a number of the form $8n+4$ the chain begins

$$8n+4\mapsto 4n+2\mapsto 2n+1\mapsto 6n+4$$

While if you add $1$ to get $8n+5$ you get $$8n+5\mapsto 24n+16\mapsto 12n+8\mapsto 6n+4$$

Thus the consecutive numbers $8n+4,8n+5$ always have the same Collatz length. That explains your pairs $(84,85)$, $(60,61)$, $(76,77)$, $(92,93)$. I expect there are other patterns as well.

Answer 2

There are many patterns where consecutive numbers sync up in their Collatz progressions after a certain number of steps. (As of this writing, I’ve found over 8,600, most of which I have to confirm through proofs such as below.) So far, all that I’ve found are based on multiples of powers of 2, just as you see with 8n+4 (2^3+4) and 16n+2 (2^4+2). As you prove each one of these equations, you’ll notice more patterns. For purposes of this analysis, if one of the two numbers reaches 1 before the other, just keep applying the equations until they either sync (usually at 4) or don’t. The extra steps past reaching 1 count toward the number of steps until they sync.

[NOTE: This analysis has nothing to do with predicting how many steps until they reach 1, only with how many steps before consecutive numbers sync up.]

(8n+4,8n+5) or (2^3+4,2^3+5)
  8n+  4→
  4n+  2→  2n+  1→  6n+4

  8n+  5→
 24n+ 16→ 12n+  8→  6n+4


(16n+2,16n+3) or (2^4+2,2^4+3)
 16n+  2→  8n+  1→ 24n+  4→ (This is the same as 8(3n)+4)
 12n+  2→  6n+  1→ 18n+  4

 16n+  3→ 48n+ 10→ 24n+  5→ (This is the same as 8(3n)+5)
 72n+ 16→ 36n+  8→ 18n+  4


(32n+22,32n+23) or (2^5+22,2^+23)
 32n+ 22→ 16n+ 11→ 48n+ 34→ (This is the same as 16(3n+2)+2)
          24n+ 17→ 72n+ 52→ (This is the same as  8(9n+6)+4)
 36n+ 26→ 18n+ 13→ 54n+ 40

 32n+ 23→ 96n+ 70→ 48n+ 35→ (This is the same as 16(3n+2)+3)
         144n+106→ 72n+ 53→ (This is the same as  8(9n+6)+5)
216n+160→108n+ 80→ 54n+ 40


(32n+5,32n+6) or (2^5+5,2^5+6)
 32n+ 5→ 96n+16→
         48n+ 8→ 24n+ 4→ (This is the same as 8(3n)+4)
 12n+ 2→  6n+ 1→ 18n+ 4

 32n+ 6→ 16n+ 3→
         48n+10→ 24n+ 5→ (This is the same as 8(3n)+5)
 72n+16→ 36n+ 8→ 18n+ 4


(64n+14,64n+15) or (2^6+14,2^6+15)
 64n+ 14→ 32n+  7→ 96n+ 22→ (This is the same as 32( 3n  )+22)
          48n+ 11→144n+ 34→ (This is the same as 16( 9n+2)+ 2)
          72n+ 17→216n+ 52→ (This is the same as  8(27n+6)+ 4)
108n+ 26→ 54n+ 13→162n+ 40

 64n+ 15→192n+ 46→ 96n+ 23→ (This is the same as 32( 3n  )+23)
         288n+ 70→144n+ 35→ (This is the same as 16( 9n+2)+ 3)
         432n+106→216n+ 53→ (This is the same as  8(27n+6)+ 5)
648n+160→324n+ 80→162n+ 40


(64n+45,64n+46) or (2^6+45,2^6+46)
          64n+ 45→192n+136→
          96n+ 68→ 48n+ 34→ (This is the same as 16(3n+2)+2)
          24n+ 17→ 72n+ 52→ (This is the same as  8(9n+6)+4)
 36n+ 26→ 18n+ 13→ 54n+ 40

          64n+ 46→ 32n+ 23→
          96n+ 70→ 48n+ 35→ (This is the same as 16(3n+2)+3)
         144n+106→ 72n+ 53→ (This is the same as  8(9n+6)+5)
216n+160→108n+ 80→ 54n+ 40


(128n+29,128n+30) or (2^7+29,2^7+30)
128n+ 29→384n+ 88→
   1      192n+ 44→ 96n+ 22→ (This is the same as 32( 3n  )+22)
          48n+ 11→144n+ 34→ (This is the same as 16( 9n+2)+ 2)
          72n+ 17→216n+ 52→ (This is the same as  8(27n+6)+ 4)
108n+ 26→ 54n+ 13→162n+ 40

128n+ 30→ 64n+ 15→
         192n+ 46→ 96n+ 23→ (This is the same as 32( 3n  )+23)
         288n+ 70→144n+ 35→ (This is the same as 16( 9n+2)+ 2)
         432n+106→216n+ 53→ (This is the same as  8(27n+6)+ 4)
648n+160→324n+ 80→162n+ 40


(128n+45,128n+46) or (2^7+45,2^7+46)
128n+ 45→384n+136→
         192n+ 68→ 96n+ 34→ (This is the same as 16( 6n+2)+ 2)
          48n+ 17→144n+ 52→ (This is the same as  8(18n+6)+ 4)
 72n+ 26→ 36n+ 13→108n+ 40

128n+ 46→ 64n+ 23→
         192n+ 70→ 96n+ 35→ (This is the same as 16( 6n+2)+ 2)
         288n+106→144n+ 53→ (This is the same as  8(18n+6)+ 5)
432n+160→216n+ 80→108n+ 40
 

(128n+94,128n+95) or (2^7+94,2^7+95)
 128n+  94→  64n+  47→ 192n+ 142→ (This is the same as 64(3n+2)+14)
             96n+  71→ 288n+ 214→ (This is the same as 32(9n+6)+22)
            144n+ 107→ 432n+ 322→ (This is the same as 16(27n+20)+2)
            216n+ 161→ 648n+ 484→ (This is the same as  8(81n+60)+4)
 324n+ 242→ 162n+ 121→ 486n+ 364

 128n+  95→ 384n+ 286→ 192n+ 143→ (This is the same as 64(3n+2)+15)
            576n+ 430→ 288n+ 215→ (This is the same as 32(9n+6)+23)
            864n+ 646→ 432n+ 323→ (This is the same as 16(27n+20)+3)
           1296n+ 970→ 648n+ 485→ (This is the same as  8(81n+60)+5)
1944n+1456→ 972n+ 728→ 486n+ 364

As I said, there are thousands of such patterns. Where it starts to get interesting is when three consecutive numbers sync up on the same step (and I’ve notice as many as five consecutive numbers syncing up.) The above pairs sync up in 11 steps or fewer, and there are a couple of others that sync up that quickly. I have not yet determined the patterns where three or five consecutive numbers sync up on the same step, but I’m sure there is one. I may post those when I find them.