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Length of record run of consecutive numbers having the same Collatz trajectory length.
1

%I #8 Mar 09 2022 00:40:35

%S 1,2,3,5,6,7,8,9,14,17,25,27,29,30,40,41,47,52,54,60,65,77,89,96,98,

%T 120,127,130,136,152,174,176

%N Length of record run of consecutive numbers having the same Collatz trajectory length.

%C It appears that this sequence is infinite.

%C For every record number of consecutive identical terms in A006577, the run length is a term of this sequence.

%C 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.

%C This sequence is related to A351104 where instead it returns the run length of the records.

%e a(10)=17 since the 10th record run of identical consecutive trajectory lengths has a run length of 17.

%e Used from A351104 by _Jon E. Schoenfield_.

%e trajectory numbers in run run

%e n length (1st is a(n)) length

%e -- ---------- -------------- ------

%e 1 1 1 1

%e 2 9 12, 13 2

%e 3 18 28, 29, 30 3

%e 4 25 98 ... 102 5

%e 5 120 386 ... 391 6

%e 6 36 943 ... 949 7

%e 7 47 1494 ... 1501 8

%e 8 42 1680 ... 1688 9

%e 9 48 2987 ... 3000 14

%e 10 57 7083 ... 7099 17

%o (Python)

%o import numpy as np

%o def find_records(m):

%o l=np.array([0]+[-1 for i in range(m-1)])

%o for n in range(len(l)):

%o path=[n+1]

%o while path[-1]>m or l[path[-1]-1]==-1:

%o if path[-1]%2==0:

%o path.append(path[-1]//2)

%o else:

%o path.append(path[-1]*3+1)

%o path.reverse()

%o for i in range(1,len(path)):

%o if path[i]<=m:

%o l[path[i]-1]=l[path[0]-1]+i

%o seq=[]

%o c,lsteps,record=1,0,0

%o for n in range(1,len(l)):

%o if l[n]==lsteps:

%o c+=1

%o else:

%o if c>record:

%o record=c

%o seq.append(c)

%o c=1

%o lsteps=l[n]

%o return seq

%o print(", ".join([str(i) for i in find_records(1000000)]))

%Y For first term of run see A351104.

%K nonn,more

%O 1,2

%A _Nathan John Eaves_, Feb 04 2022