Mixing
Another mathematical number sequence
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
2
Can you work out the mathematical rule for this number sequence?
354, 72, 345, 132, 6, 216, 9, 729, 354, 72, 345........
Hint 1
this is a mathematical puzzle
From time to time hints will be updated until solved.
Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences
number-sequence
asked Aug 24 at 12:34
tom
1,499121
 |Â
show 1 more comment
up vote
3
down vote
favorite
2
Can you work out the mathematical rule for this number sequence?
354, 72, 345, 132, 6, 216, 9, 729, 354, 72, 345........
Hint 1
this is a mathematical puzzle
From time to time hints will be updated until solved.
Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences
number-sequence
asked Aug 24 at 12:34
tom
1,499121
This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
1
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
1
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09
 |Â
show 1 more comment
up vote
3
down vote
favorite
2
up vote
3
down vote
favorite
2
2
Can you work out the mathematical rule for this number sequence?
354, 72, 345, 132, 6, 216, 9, 729, 354, 72, 345........
Hint 1
this is a mathematical puzzle
From time to time hints will be updated until solved.
Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences
number-sequence
asked Aug 24 at 12:34
tom
1,499121
Can you work out the mathematical rule for this number sequence?
354, 72, 345, 132, 6, 216, 9, 729, 354, 72, 345........
Hint 1
this is a mathematical puzzle
From time to time hints will be updated until solved.
Note: this sequence cannot be found in The On-Line Encyclopedia of Integer Sequences
number-sequence
asked Aug 24 at 12:34
tom
1,499121
asked Aug 24 at 12:34
tom
1,499121
asked Aug 24 at 12:34
tom
1,499121
asked Aug 24 at 12:34
tom
1,499121
1,499121
This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
1
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
1
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09
 |Â
show 1 more comment
This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
1
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
1
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09
This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
1
1
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
1
1
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Take
the $n$-th digit, where $n$ is the first digit (if $n$ is more than the number of the digits then "modulo" it -- wrap around)
and
cube it
then
sum with other digits
so you will get
$354$ -> $3+5+4^3$ (3rd digit) = $72$
$72$ -> $7^3+2$ (7th digit) = $345$
$345$ -> $3+4+5^3$ (3rd digit) = $132$
$132$ -> $1^3+3+2$ (1st digit) = $6$
$6$ -> $6^3$ (6th digit) = $216$
$216$ -> $2+1^3+6$ (2nd digit) = $9$
$9$ -> $9^3$ (9th digit) = $729$
$729$ -> $7^3+2+9$ (7th digit) = $354$
...
answered Aug 24 at 12:59
athin
6,08412257
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Take
the $n$-th digit, where $n$ is the first digit (if $n$ is more than the number of the digits then "modulo" it -- wrap around)
and
cube it
then
sum with other digits
so you will get
$354$ -> $3+5+4^3$ (3rd digit) = $72$
$72$ -> $7^3+2$ (7th digit) = $345$
$345$ -> $3+4+5^3$ (3rd digit) = $132$
$132$ -> $1^3+3+2$ (1st digit) = $6$
$6$ -> $6^3$ (6th digit) = $216$
$216$ -> $2+1^3+6$ (2nd digit) = $9$
$9$ -> $9^3$ (9th digit) = $729$
$729$ -> $7^3+2+9$ (7th digit) = $354$
...
answered Aug 24 at 12:59
athin
6,08412257
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
 |Â
show 4 more comments
up vote
6
down vote
accepted
Take
the $n$-th digit, where $n$ is the first digit (if $n$ is more than the number of the digits then "modulo" it -- wrap around)
and
cube it
then
sum with other digits
so you will get
$354$ -> $3+5+4^3$ (3rd digit) = $72$
$72$ -> $7^3+2$ (7th digit) = $345$
$345$ -> $3+4+5^3$ (3rd digit) = $132$
$132$ -> $1^3+3+2$ (1st digit) = $6$
$6$ -> $6^3$ (6th digit) = $216$
$216$ -> $2+1^3+6$ (2nd digit) = $9$
$9$ -> $9^3$ (9th digit) = $729$
$729$ -> $7^3+2+9$ (7th digit) = $354$
...
answered Aug 24 at 12:59
athin
6,08412257
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
 |Â
show 4 more comments
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Take
the $n$-th digit, where $n$ is the first digit (if $n$ is more than the number of the digits then "modulo" it -- wrap around)
and
cube it
then
sum with other digits
so you will get
$354$ -> $3+5+4^3$ (3rd digit) = $72$
$72$ -> $7^3+2$ (7th digit) = $345$
$345$ -> $3+4+5^3$ (3rd digit) = $132$
$132$ -> $1^3+3+2$ (1st digit) = $6$
$6$ -> $6^3$ (6th digit) = $216$
$216$ -> $2+1^3+6$ (2nd digit) = $9$
$9$ -> $9^3$ (9th digit) = $729$
$729$ -> $7^3+2+9$ (7th digit) = $354$
...
answered Aug 24 at 12:59
athin
6,08412257
Take
the $n$-th digit, where $n$ is the first digit (if $n$ is more than the number of the digits then "modulo" it -- wrap around)
and
cube it
then
sum with other digits
so you will get
$354$ -> $3+5+4^3$ (3rd digit) = $72$
$72$ -> $7^3+2$ (7th digit) = $345$
$345$ -> $3+4+5^3$ (3rd digit) = $132$
$132$ -> $1^3+3+2$ (1st digit) = $6$
$6$ -> $6^3$ (6th digit) = $216$
$216$ -> $2+1^3+6$ (2nd digit) = $9$
$9$ -> $9^3$ (9th digit) = $729$
$729$ -> $7^3+2+9$ (7th digit) = $354$
...
answered Aug 24 at 12:59
athin
6,08412257
answered Aug 24 at 12:59
athin
6,08412257
answered Aug 24 at 12:59
athin
6,08412257
answered Aug 24 at 12:59
athin
6,08412257
6,08412257
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
 |Â
show 4 more comments
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
Hhhmm.... if you followed this pattern starting from a random number with at least $3$ digits.... will you always reach $345$??
– user477343
Aug 24 at 13:27
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@user477343 - no - take 512 for example. see my comment on the question for an excel formula to try for yourself.
– Collett89
Aug 24 at 13:35
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@Collett89 You need to teach me.... it even looks cool in Rot13! (Oh, and by the way, $512=2^9$)
– user477343
Aug 24 at 13:36
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
@user477343 if only I had the time - it is also 8 cubed. I'd like there to be a simple mathematical explanation as to why numbers x loop like this and numbers y dont - but I can't see one (does not mean one does not exist)
– Collett89
Aug 24 at 13:42
1
1
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
@Collett89 $2^9=2^3times 3=(2^3)^3=8^3$ and also goes to $8$... weird, but interesting! Experimenting just a little further, I also found out that $17^2=289$ just goes to $10$ which of course goes to $1$ and does not reach $345$. But $18^2=324$ works :)
– user477343
Aug 24 at 13:44
 |Â
show 4 more comments
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This is an interesting looking puzzle! Neat that you've created a loop with 354 as a starting point....I wonder if loops are possible with every starting point.... Great puzzle! :D
– El-Guest
Aug 24 at 12:38
1
@El-Guest If you want to do the work I have got a working excel formula - Pop this in field A2 and copy on downwards - then change the number at the top>> ROT13(=CBJRE(ZVQ(N1,VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1))),1),3)+VS(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>1,AHZOREINYHR(ZVQ(N1,1,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>2,YRA(N1)>1),AHZOREINYHR(ZVQ(N1,2,1)))+VS(NAQ(VS(ZBQ(ZVQ(N1,1,1),YRA(N1))=0,YRA(N1), ZBQ(ZVQ(N1,1,1),YRA(N1)))<>3,YRA(N1)>2),AHZOREINYHR(ZVQ(N1,3,1))))
– Collett89
Aug 24 at 13:30
(and yes most of the ones I have looked at do hit a loop pretty quickly)
– Collett89
Aug 24 at 13:31
Wow! Thanks for doing all of that work, @Collett89! I'm definitely finding the same thing, although I'm trying to derive a mathematical reason for why that is the case...
– El-Guest
Aug 24 at 14:04
1
@tom I think what might be happening (what I'd be curious in proving or disproving) is whether there's a stable loop (ie. the long-term behaviour of the sequence converges to a cyclical pattern) for all sequences. There might always be numbers that start out of a loop, but might converge to one. 0 and 1 are self-contained, but after that it looks like it could be possible!
– El-Guest
Aug 24 at 14:09