

A351104


Numbers that begin a recordlength 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
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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



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.
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(m1)])
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(nc+1)
c=1
lsteps=l[n]
return ciclr
print(", ".join([str(i) for i in find_records(1000000)]))


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



