A351122 Irregular triangle read by rows in which row n lists the number of divisions by 2 after tripling steps in the Collatz 3x+1 trajectory of 2n+1 until it reaches 1.
1, 4, 4, 1, 1, 2, 3, 4, 2, 1, 1, 2, 3, 4, 1, 2, 3, 4, 3, 4, 1, 1, 1, 5, 4, 2, 3, 4, 1, 3, 1, 2, 3, 4, 6, 1, 1, 5, 4, 2, 1, 3, 1, 2, 3, 4, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 4, 2, 2, 4, 3, 1, 1, 5, 4, 3, 1, 2, 3, 4
Offset: 1
Examples
Triangle starts at T(1,0):
n\k 0 1 2 3 4 5 6 7 8 ...
1: 1 4
2: 4
3: 1 1 2 3 4
4: 2 1 1 2 3 4
5: 1 2 3 4
6: 3 4
7: 1 1 1 5 4
8: 2 3 4
9: 1 3 1 2 3 4
10: 6
11: 1 1 5 4
12: 2 1 3 1 2 3 4
13: 1 2 1 1 1 1 2 2 1 2 1 1 2 ... (see A372362)
...
For n=6, the trajectory of 2*n+1 = 13 is as follows. The tripling steps ("=>") are followed by runs of 3 and then 4 halvings ("->"), so row n=6 is 3, 4.
13 => 40 -> 20 -> 10 -> 5 => 16 -> 8 -> 4 -> 2 -> 1
triple \------------/ triple \---------------/
3 halvings 4 halvings
Runs of halvings are divisions by 2^T(n,k). Row n=11 is 1, 1, 5, 4 and its steps starting from 2*n+1 = 23 reach 1 by a nested expression
(((((((23*3+1)/2^1)*3+1)/2^1)*3+1)/2^5)*3+1)/2^4 = 1.
Crossrefs
Programs
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PARI
row(n) = my(m=2*n+1, list=List()); while (m != 1, if (m%2, m = 3*m+1, my(nb = valuation(m,2)); m/=2^nb; listput(list, nb));); Vec(list); \\ Michel Marcus, Jul 18 2022
Extensions
Corrected by Michel Marcus, Jul 18 2022