A256598 Irregular triangle where row n contains the odd terms in the Collatz sequence beginning with 2n+1.

1, 3, 5, 1, 5, 1, 7, 11, 17, 13, 5, 1, 9, 7, 11, 17, 13, 5, 1, 11, 17, 13, 5, 1, 13, 5, 1, 15, 23, 35, 53, 5, 1, 17, 13, 5, 1, 19, 29, 11, 17, 13, 5, 1, 21, 1, 23, 35, 53, 5, 1, 25, 19, 29, 11, 17, 13, 5, 1, 27, 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155

Offset: 0

Author

Bob Selcoe, Apr 03 2015

Keywords

• nonn
• tabf

Comments

The Collatz function is an integer-valued function given by n/2 if n is even and 3n+1 if n is odd. We build a Collatz sequence by beginning with a natural number and iterating the function indefinitely. It is conjectured that all such sequences terminate at 1.

In this triangle, row n is made up of the odd terms of the Collatz sequence beginning with 2n+1. Therefore, it is conjectured that this sequence is well-defined, i.e., that all rows terminate at 1.

The last index k in row n gives the number of iterations required for the Collatz sequence to terminate if even terms are omitted.

T(n,k)/T(n,k+1) is of form: ceiling(T(n,k)*3/2^j) for some j>=1. Therefore, the coefficients in each row may be read as a series of iterated ceilings, where j may vary. For example, row 3 has initial term 7, which is followed by ceiling(7*3/2), ceiling(ceiling(7*3/2)*3/2), ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4), ceiling(ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4)*3/8), ceiling(ceiling(ceiling(ceiling(ceiling(7*3/2)*3/2)*3/4)*3/8)*3/16).

The length of row n is A258145(n) (set to 0 if 1 is not reached after a finite number of steps). - Wolfdieter Lang, Aug 11 2021

Examples

			Triangle starts T(0,0):
n\k   0   1   2   3   4    5   6   7   8   9  10 ...
0:    1
1:    3   5   1
2:    5   1
3:    7   11  17  13  5    1
4:    9   7   11  17  13   5   1
5:    11  17  13  5   1
6:    13  5   1
7:    15  23  35  53  5    1
8:    17  13  5   1
9:    19  29  11  17  13   5   1
10:   21  1
11:   23  35  53  5    1
12:   25  19  29  11  17  13   5   1
...
n=13 starts with 27 and takes 41 steps: (27), 41, 31, 47, 71, 107,... 53, 5, 1, (see A372443).
Row 8 is [17, 13, 5, 1] because it is the subsequence of odd terms for the Collatz sequence starting with 17: [17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1].
		

Links

• Eric Weisstein's World of Mathematics, Collatz Problem
• Wikipedia, Collatz conjecture
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006370, A070165, A075677, A258145.

Cf. A372443 (row 13 up to its first 1).

Cf. also array A372283 showing the same terms in different orientation.

Programs

Mathematica

f[n_] := NestWhileList[(3*# + 1)/2^IntegerExponent[3*# + 1, 2] &, 2*n + 1, # > 1 &]; Grid[Table[f[n], {n, 0, 12}]] (* L. Edson Jeffery, Apr 25 2015 *)

PARI

row(n) = {my(oddn = 2*n+1, vl = List(oddn), x); while (oddn != 1, x = 3*oddn+1; oddn = x >> valuation(x, 2); listput(vl, oddn)); Vec(vl);}
tabf(nn) = {for (n=0, nn, my(rown = row(n)); for (k=1, #rown, print1(rown[k], ", ")); print;);} \\ Michel Marcus, Oct 04 2019

Sage

def Collatz(n):
    A = [n]
    b = A[-1]
    while b != 1:
        if is_even(b):
            A.append(b//2)
        else:
            A.append(3*b+1)
    return A
[y for sublist in [[x for x in Collatz(2*n+1) if is_odd(x)] for n in [0..15]] for y in sublist] # Tom Edgar, Apr 04 2015

Formula

T(n,0) = 2n+1 and T(n,k) = A000265(3*T(n,k-1)+1) for k>0. - Tom Edgar, Apr 04 2015