Is there an online calculator in which you can type a number and have it tell you if it could be a Lychrel number or not?

Say you type $$7326$$ into it, it runs a few calculations and tells you it reaches $$99099$$ in three iterations. But if you type in a number like $$887$$, it runs a reasonable number of iterations (say, twenty) and tells you it doesn't reach a palindrome in that number of iterations.

In response to another question, I tried to determine if $$9988$$ is a Lychrel number. After ten iterations on a general purpose calculator, I could not find a palindrome, but I could have made a mistake somewhere. I also tried asking Google "Is $$9988$$ a Lychrel number?" The results were unenlightening.

• It's unknown, as well as all the other terms in oeis.org/A023108 and many more. Dec 21, 2014 at 3:01
• But surely there is some tool that can reliably tell me that it doesn't reach a palindrome in some small number of iterations, some tool that's not going to inadvertently mix up the digits (e.g., type 68779 instead of 86779). Dec 21, 2014 at 3:14
• One can write a short program to do this in any high-level language, e.g. Python. Dec 21, 2014 at 3:15
• If I really wanted to, I could do it in Javascript. But wouldn't I be reinventing the wheel? Hasn't anyone done this and made it available online? Dec 21, 2014 at 3:18
• @vadim123 See Code Review question 44272 for a Python implementation. Dec 28, 2014 at 2:34

I tried to make the code semi-readable, just set $$N$$ to the number of iterations you wish to go before rejecting; by default I have it set to $$25$$. The value of the iterations rises very quickly. As a result, too high of iterations are not supported by javascript. Without a moderate amount of additional work (e.g. custom classes) this is roughly the best javascript can do. If there are any obvious improvements I can make let me know.
Math.log(1000)/Math.log(10) Returns 2.9999999999999996 rather than 3.
This correction will not affect any "ordinary" inputs. I added $$10^{-14}$$ to the logarithm before taking the floor of the input.