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
Keywords
Examples
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)
Crossrefs
For length of run see A351224.
Programs
-
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)]))
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