Nathan John Eaves has authored 3 sequences.
A351224
Length of record run of consecutive numbers having the same Collatz trajectory length.
Original entry on oeis.org
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
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
-
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)]))
A351104
Numbers that begin a record-length run of consecutive numbers having the same Collatz trajectory length.
Original entry on oeis.org
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
Offset: 1
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.
From _Jon E. Schoenfield_, Feb 01 2022: (Start)
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)
-
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
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(n-c+1)
c=1
lsteps=l[n]
return ciclr
print(", ".join([str(i) for i in find_records(1000000)]))
A282852
37-gonal numbers: a(n) = n*(35*n-33)/2.
Original entry on oeis.org
0, 1, 37, 108, 214, 355, 531, 742, 988, 1269, 1585, 1936, 2322, 2743, 3199, 3690, 4216, 4777, 5373, 6004, 6670, 7371, 8107, 8878, 9684, 10525, 11401, 12312, 13258, 14239, 15255, 16306, 17392, 18513, 19669, 20860, 22086, 23347, 24643, 25974
Offset: 0
-
Table[n(35n-33)/2, {n, 40}]
PolygonalNumber[37,Range[0,40]] (* Requires Mathematica version 10 or later *) (* or *) LinearRecurrence[{3,-3,1},{0,1,37},40] (* Harvey P. Dale, Oct 24 2020 *)
-
for(n=0,100,print1(n*(35*n-33)/2,", ")) \\ Derek Orr, Feb 27 2017
-
for n in range(0,51):
print(n*(35*n-33)//2)
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