A couple other things. These numbers are going to get really big really fast. I didn't look closely at your algorithm, just the syntax, and there are a couple issues.

1. Your For loop won't ever exit if you don't have a test to escape when i gets too large and x doesn't satisfy your conditions.

2. You're not incrementing i in the correct place. Your increment is the third argument, not the 4th.

   Do[For[i = 0, x != 1 && x != 4 && i < 10, i++, 
     x = ((x - Mod[x, 10])/10)^2 + (Mod[x, 10])^2]; 
    Print[{i, x}], {x, 10}]
   

I've decreased the range of i and x to 9 and 10. You'll probably want to skip the print statement if you've hit the upper bound of i that you set i.e. If[i<10,Print[{i,x}]. Even with this very truncated version the numbers get large fast:

{0,1}
{1,4}
{10,485}
{0,4}
{10,29052125}
{10,145}
{10,7341998783795942461}
{10,53905}
{10,2329}
{1,1}

Also, you can replace

ToExpression@Characters[ToString@x]

with

IntegerDigits[x]

Sorry for the delay, hopefully this is still helpful...

Take your number apart by converting it to a string and putting each character (digit) into a list. Look up the documentation for Characters if you're not sure what it does.

Characters[ToString@x]

If the ToString@x notation confuses you it's the same as ToString[x]. Now your list is still a list of strings and you can't add or square strings. So turn the list into an expression

ToExpression@Characters[ToString@x]

You want to square each element so make a Pure Function with the Power function: Power[#,2]&. Then take this function and Map it to each element of the array using /@

Power[#,2]&/@ToExpression@Characters[ToString@x]

That about does it. Just wrap the expression with Total to add up all the elements.

A couple other comments,

1. I switched to While because that's the logical test you're doing. You can use For but it's kind of clumsy.
2. You don't want to alter x. You want x to stay the number between 1 and 99 until you've converged to 1 or 4. So to "keep it safe", set an intermediate variable y equal to x and do your operations on it. Also increment i so you see how many times it takes to satisfy your while loop.

(I've just seen that I've somehow copied your second latest comment before, I don't know what happend but sorry)

Yes, you've exactly done what I've wanted to do, but much more elegant (I've wanted to get there step by step; apparently I have to learn many things).

Can you please tell me how to understand your answer? It's something like: you take the total of every squared digit of "y" and put it back to a string, right?

On your topic I suggest you the book HAZRAT, R., Mathematica: A Problem-Centered Approach, Springer, 2010, Springer Undergraduate Mathematics Series, London, xv+188.

Marian

Thank you so much Eric for your help. I've also just seen that the numbers 3 etc. don't work because they get higher than 100. I've also got another equation, but I'll try yours as well.

Have a good one,

1. I thought, the test is "x != 1 && x != 4": when it is true, it calculates one time, then "x" gets tested again, or otherwise it stops immediately and gives (like the numbers 1 and 4) the result.

2. I didn't want to end the calculations with a certain number of calculations ("i"). The numbers shouldn't get too high, especially because we're just using 1-digit numbers...

I've already said the biggest issue: It doesn't make sense at all that the numbers get really high like the numbers 5 and 7.

Here's an example (this is how the algorithm should work): Number 3: 3^2=9 -> 9^2= 81 -> 8^2+1^2=65 ... Cycle: 3-9-81-65-61-37-58-89-145-42-20-4