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A351224 Length of record run of consecutive numbers having the same Collatz trajectory length.

%I A351224 #8 Mar 09 2022 00:40:35
%S A351224 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 A351224 120,127,130,136,152,174,176
%N A351224 Length of record run of consecutive numbers having the same Collatz trajectory length.
%C A351224 It appears that this sequence is infinite.
%C A351224 For every record number of consecutive identical terms in A006577, the run length is a term of this sequence.
%C A351224 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 A351224 This sequence is related to A351104 where instead it returns the run length of the records.
%e A351224 a(10)=17 since the 10th record run of identical consecutive trajectory lengths has a run length of 17.
%e A351224 Used from A351104 by _Jon E. Schoenfield_.
%e A351224       trajectory  numbers in run   run
%e A351224    n    length    (1st is a(n))   length
%e A351224   --  ----------  --------------  ------
%e A351224    1        1        1               1
%e A351224    2        9       12, 13           2
%e A351224    3       18       28, 29, 30       3
%e A351224    4       25       98 ...  102      5
%e A351224    5      120      386 ...  391      6
%e A351224    6       36      943 ...  949      7
%e A351224    7       47     1494 ... 1501      8
%e A351224    8       42     1680 ... 1688      9
%e A351224    9       48     2987 ... 3000     14
%e A351224   10       57     7083 ... 7099     17
%o A351224 (Python)
%o A351224 import numpy as np
%o A351224 def find_records(m):
%o A351224     l=np.array([0]+[-1 for i in range(m-1)])
%o A351224     for n in range(len(l)):
%o A351224         path=[n+1]
%o A351224         while path[-1]>m or l[path[-1]-1]==-1:
%o A351224             if path[-1]%2==0:
%o A351224                 path.append(path[-1]//2)
%o A351224             else:
%o A351224                 path.append(path[-1]*3+1)
%o A351224         path.reverse()
%o A351224         for i in range(1,len(path)):
%o A351224             if path[i]<=m:
%o A351224                 l[path[i]-1]=l[path[0]-1]+i
%o A351224     seq=[]
%o A351224     c,lsteps,record=1,0,0
%o A351224     for n in range(1,len(l)):
%o A351224         if l[n]==lsteps:
%o A351224             c+=1
%o A351224         else:
%o A351224             if c>record:
%o A351224                 record=c
%o A351224                 seq.append(c)
%o A351224             c=1
%o A351224         lsteps=l[n]
%o A351224     return seq
%o A351224 print(", ".join([str(i) for i in find_records(1000000)]))
%Y A351224 For first term of run see A351104.
%K A351224 nonn,more
%O A351224 1,2
%A A351224 _Nathan John Eaves_, Feb 04 2022