A Few Numbers With Strange Qualities

In summary, a 4-digit number can only become smaller when adding the squared digits, and no number of 5,6,7... digits can loop back to themselves.
  • #1
suzi9spal
2
0
Hello everyone!
I'm a 1st grade student in a slovenian high school. Recently, I've became very interested in numbers and math itsself.
I was just playing around with a few numbers, when I came across a few that are unique. I was wondering which numbers can loop back
to themselfes after this procedure;
n is a digit, m is a digit, o is an integer, p is another integer
Code:
o = mn
p = m^2 + n^2
o = p
...repeat...

For instance, 16 is one of these numbers;
1^2 + 6^2 = 1+36 = 37 ==> 3^2 + 7^2 = 9+49 = 58 ==> 5^2 + 8^2 = 25+64 = 89 ==>
8^2 + 9^2 = 64+81 = 145 ==> 1^2 + 4^2 + 5^2 = 1+16+25 = 42 ==>
4^2 + 2^2 = 16+4 = 20 ==> 2^2 + 0^2 = 4+0 = 4 ==> 4^2 = 16

I then created a program to find every single number that has this property. It turns out, there are only a limited few:
  • 1
  • 4
  • 16
  • 20
  • 37
  • 42
  • 58
  • 89
  • 145

I thought this was interesting, so I posted it here.
Did anyone find something like this before?
 
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  • #2
Hello, suzi9spal, and happy new year!

Personally I've never seen this before, but I found something after googling a bit. The following link describes it, and also contains a reference to a book by J. Madachy (which I haven't read) that explores this and other similar number games.

http://mathworld.wolfram.com/RecurringDigitalInvariant.html

Notice that, apart from 1 where the property is trivially true, the other numbers that you discover form a cycle, and you can start from any of them and obtain the same collection of numbers. These are the "cycles" that the link above refers to.

And keep these kinds of hobbies! They will greatly pay off when you go to the university.
 
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  • #3
here's a helpful link. If you sequence is new, you will not find it here.

http://oeis.org/

if it is known, then you will find it.
 
  • #4
suzi9spal said:
It turns out, there are only a limited few
And you have proved that?
 
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  • #5
oay said:
And you have proved that?

It's not very hard. If you start with an n-digit number, the largest number that could follow it is 81n. A 4-digit number would have at most 324 following it. All the numbers following it will have 3-digits or less, so a 4-digit or larger number can't be part of a loop, so there are only 1000 numbers you have to check.
 
  • #6
willem2 said:
It's not very hard. If you start with an n-digit number, the largest number that could follow it is 81n. A 4-digit number would have at most 324 following it. All the numbers following it will have 3-digits or less, so a 4-digit or larger number can't be part of a loop, so there are only 1000 numbers you have to check.
I'm sure you're correct but I don't follow what you've said. (You're probably right and I just need a lie down...)

However, it is intriguing that, other than the trivial 0 and 1, the other numbers in this list all sum up to 16 in their digits.
 
  • #7
oay, I have tested every single number to about 13 milion :D
 
  • #8
willem2 said:
It's not very hard. If you start with an n-digit number, the largest number that could follow it is 81n. A 4-digit number would have at most 324 following it. All the numbers following it will have 3-digits or less, so a 4-digit or larger number can't be part of a loop, so there are only 1000 numbers you have to check.
I don't get how you've gone from an n-digit to a 4-digit with no explanation.

suzi9spal said:
oay, I have tested every single number to about 13 milion :D
I'm sure you have, and I don't disbelieve you, I just don't see the proof!
 
  • #9
oay said:
... I don't follow what you've said.

What willem2 was saying is that, for all possible n-digit numbers, the sum of squared digits is at its largest when the n-digit number is 9999... (n nines), and that sum is 9^2 . n = 81n.

So, for different values of n (number of digits), you can see that
n=1: 81n = 81 (so, the sum of squared digits has at most 2 digits)
n=2: 81n = 162 (so, the sum of squared digits has at most 3 digits)
n=3: 81n = 243 (so, the sum of squared digits has at most 3 digits)
n=4: 81n = 324 (so, the sum of squared digits has at most 3 digits)

You can see that, when applying the process of adding the squared digits, a 4-digit number can only become smaller (since the result won't be larger than 324).

For n>4, you have that 81n < 10^n (just like the case n=4), so numbers of 5,6,7... digits also become smaller when adding the squared digits.

Therefore, for numbers of 4,5,6,7,... digits, the process of adding the squared digits has no chance of returning to the original number. So you only need to test numbers up to 3 digits.
 
  • #10
This has already been described on Sloane's database: http://oeis.org/A039943

It's tagged "fini" for finite - confirming that this is the exhaustive listing.

I love that resource! I've got a few sequences accepted long ago.
 
  • #11
I like the OP's interest in this sort of thing, because it reminds me of myself at that age. OP, if you're interested in a genuinely unsolved (and maddening) integer-recursion problem, look up the Collatz conjecture. And before you write a program for that one, try starting out with 27 using just pen and paper (and a calculator, if you wish).
 
  • #12
Dodo said:
What willem2 was saying is that, for all possible n-digit numbers, the sum of squared digits is at its largest when the n-digit number is 9999... (n nines), and that sum is 9^2 . n = 81n.

So, for different values of n (number of digits), you can see that
n=1: 81n = 81 (so, the sum of squared digits has at most 2 digits)
n=2: 81n = 162 (so, the sum of squared digits has at most 3 digits)
n=3: 81n = 243 (so, the sum of squared digits has at most 3 digits)
n=4: 81n = 324 (so, the sum of squared digits has at most 3 digits)

You can see that, when applying the process of adding the squared digits, a 4-digit number can only become smaller (since the result won't be larger than 324).

For n>4, you have that 81n < 10^n (just like the case n=4), so numbers of 5,6,7... digits also become smaller when adding the squared digits.

Therefore, for numbers of 4,5,6,7,... digits, the process of adding the squared digits has no chance of returning to the original number. So you only need to test numbers up to 3 digits.
Nope, you need to explain it more clearly!
 

What is the concept behind "A Few Numbers With Strange Qualities"?

The concept behind "A Few Numbers With Strange Qualities" is to explore and understand the unique and often unexpected properties of certain numbers, such as their relationships to other numbers and their significance in different mathematical and scientific contexts.

What are some examples of numbers with strange qualities?

Some examples of numbers with strange qualities include the golden ratio, pi, e, and the imaginary number i. These numbers have unique properties and are often found in nature and used in various mathematical equations.

Why are these numbers considered significant?

These numbers are considered significant because they often arise in unexpected ways and have important implications in various fields, such as mathematics, physics, and biology. They also have fascinating patterns and relationships that have captured the curiosity of scientists and mathematicians for centuries.

How do scientists study and analyze these numbers?

Scientists study and analyze these numbers using various mathematical tools and techniques, such as equations, graphs, and models. They also conduct experiments and observations to understand how these numbers appear in nature and how they can be applied in different contexts.

What is the importance of understanding these numbers with strange qualities?

Understanding these numbers with strange qualities can lead to new discoveries and advancements in various fields, from developing more efficient algorithms to understanding the laws of nature. It also helps us appreciate the beauty and complexity of mathematics and its role in our world.

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