
#1
Jan113, 04:31 PM

P: 2

Hello everyone!
I'm a 1st grade student in a slovenian high school. Recently, I've became very intrested 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 wich 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
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:
I thought this was intresting, so I posted it here. Did anyone find something like this before? 



#2
Jan213, 06:46 AM

P: 688

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/Recurri...Invariant.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. 



#3
Jan213, 07:38 AM

P: 84

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
Jan213, 10:25 AM

P: 220

A Few Numbers With Strange Qualities 



#5
Jan213, 07:28 PM

P: 1,351





#6
Jan213, 09:19 PM

P: 220

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
Jan513, 12:47 PM

P: 2

oay, I have tested every single number to about 13 milion :D




#8
Jan513, 09:29 PM

P: 220





#9
Jan513, 10:11 PM

P: 688

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 4digit 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
Jan613, 07:13 AM

HW Helper
P: 2,866

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
Jan613, 07:20 AM

HW Helper
P: 2,866

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) integerrecursion 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
Jan613, 09:22 PM

P: 220




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