Also, you can replace

ToExpression@Characters[ToString@x]

with

IntegerDigits[x]

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

Every number eventually returns a 1 or 4 in the ones place. I think what you're trying to do is count how many iterations it takes before you get there. If so, here's the answer:

y = 0; Do[y = x; i = 0; 
 While[y != 1 && y != 4, i++; 
  y = Total[Power[#, 2] & /@ ToExpression@Characters[ToString@y]]];
 Print[{x,i, y}], {x, 99}]

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.

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

Did you mean to use Mod instead of mod. Remember all built in functions in the Wolfram Language are capitalized. When it sees mod and there's nothing in the lookup table, then it returns it as a generic head of your expression.