# [seqfan] Help with definitions of 3 sequences

Tue Jan 25 14:09:19 CET 2022

```Hi everyone,

Please see the three sequences below. I would really appreciate it if you could help me with the definitions (and finding the terms of the last two).

Best,

Ali

1)
a(1) = 1
a(2) = 2/a(1) = 2/1 = 1
a(3) = (3+4+5)/a(2) = 12/2 = 6
a(4) = 6/a(3)= 6/6 = 1
a(5) = 7/a(4) = 7/1 = 7
a(6) = (8+9+10+11+12+13)/a(5) = 63/7 = 9
a(7) = (14+15+16)/a(6) = 45/9 =5

2)
a(1) =1
a(2) = 2/a(1) = 2/1 = 2
a(3) = 34/a(2) = 34/2 = 17
a(4) = 5678/a(3) = 334
a(4) = 9101112131414…./334 = ?

3)
1, 5, 13, 17, 53, 85, 113, 181, 241, 277, 301, 341, 401, 469, 565, 597, 625, 725, 833, 853, 965, 1109, 1205, 1285, 1477, 1621

The idea is to replace all odd numbers with the terms of this sequence so that when we apply the Collatz algorithm on any of these numbers (>1), we land on a smaller number in the sequence. (I excluded multiples of 3 because no number would land on any of them when we apply the Collatz algorithm)

For example, the Colltaz iterations of 9 are:

9, 7, 11, 17, 13, 5, 1
What we want is to put these numbers in order by replacing them with their 4m+1 iterations, when necessary:

We don’t have problems with 1, 5, 13, or 17.

So, we start by replacing 11 with a number larger than 17.
11*4+1 = 45. But 45 is a multiple of 3, so, we repeat
45*4+1 = 181
181 will replace 11 permanently.  (11*3+1= 34/ 2= 17 and 181*3+1=544/16 = 17)

Now, we want to find an iteration of 7 that is larger than 181
7*4+1 = 29
29*4+1 = 117
117*4+1 = 469 (469 will replace 7 in the sequence.)

Finally, we want to replace 9 with a number larger than 469
4*9+1 = 37
37*4+1 = 149
149*4+1= 597 (a multiple of 3, so we have to continue)
597*4+1 = 2389
Now, the Collatz numbers of 9 are in a fully descending order:
2389, 469, 181, 17, 13, 5, 1

If we, somehow, find a formula for the sequence above, and we prove that all odd numbers have a “representation” in it, would that prove the Collatz?

Best,

Ali

```