A351104 Numbers that begin a record-length run of consecutive numbers having the same Collatz trajectory length.

1, 12, 28, 98, 386, 943, 1494, 1680, 2987, 7083, 57346, 252548, 331778, 524289, 596310, 2886352, 3247146, 3264428, 4585418, 5158596, 5772712, 13019668, 18341744, 24455681, 98041684, 136696632, 271114753, 361486064, 406672385, 481981441, 711611184, 722067240

Offset

1,2

Comments

It appears that this sequence is infinite.

For every record number of consecutive identical terms in A006577, the index of the first of those consecutive terms is a term of this sequence.

This sequence is interesting because when A006577 is graphed on a scatter plot, it is immediately obvious that there are many runs of terms having the same value.

Links

Table of n, a(n) for n=1..32.

Example

a(4)=98 since the length of the Collatz trajectory of each number from 98 through 102 is of length 25 and this is the fourth record length.
From Jon E. Schoenfield, Feb 01 2022: (Start)
      trajectory  numbers in run   run
   n    length    (1st is a(n))   length
  --  ----------  --------------  ------
   1        1        1               1
   2        9       12, 13           2
   3       18       28, 29, 30       3
   4       25       98 ...  102      5
   5      120      386 ...  391      6
   6       36      943 ...  949      7
   7       47     1494 ... 1501      8
   8       42     1680 ... 1688      9
   9       48     2987 ... 3000     14
  10       57     7083 ... 7099     17
(End)

Prog

(Python)
import numpy as np
def find_records(m):
    l=np.array([0]+[-1 for i in range(m-1)])
    for n in range(len(l)):
        path=[n+1]
        while path[-1]>m or l[path[-1]-1]==-1:
            if path[-1]%2==0:
                path.append(path[-1]//2)
            else:
                path.append(path[-1]*3+1)
        path.reverse()
        for i in range(1, len(path)):
            if path[i]<=m:
                l[path[i]-1]=l[path[0]-1]+i
    ciclr=[]
    c=1
    lsteps=0
    record=0
    for n in range(1, len(l)):
        if l[n]==lsteps:
            c+=1
        else:
            if c>record:
                record=c
                ciclr.append(n-c+1)
            c=1
        lsteps=l[n]
    return ciclr
print(", ".join([str(i) for i in find_records(1000000)]))

Crossrefs

For length of run see A351224.

Sequence in context: A043969 A204386 A078441 * A340685 A203026 A189539

Adjacent sequences: A351101 A351102 A351103 * A351105 A351106 A351107

Keyword

nonn

Author

Nathan John Eaves, Jan 31 2022

Status

approved