A351224 Length of record run of consecutive numbers having the same Collatz trajectory length.
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, 120, 127, 130, 136, 152, 174, 176
Offset: 1
Examples
a(10)=17 since the 10th record run of identical consecutive trajectory lengths has a run length of 17.
Used from A351104 by _Jon E. Schoenfield_.
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
Crossrefs
For first term of run see A351104.
Programs
-
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 seq=[] c,lsteps,record=1,0,0 for n in range(1,len(l)): if l[n]==lsteps: c+=1 else: if c>record: record=c seq.append(c) c=1 lsteps=l[n] return seq print(", ".join([str(i) for i in find_records(1000000)]))
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