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%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