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A351104
Numbers that begin a record-length run of consecutive numbers having the same Collatz trajectory length.
1
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.
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
KEYWORD
nonn
AUTHOR
Nathan John Eaves, Jan 31 2022
STATUS
approved