Reverse, Add, and 196

Write a positive whole number in base 10, reverse its digits and add the reversed value to the original. If the result reads the same forwards and backwards, it is a palindrome and this laboratory stops. Otherwise repeat with the new total.

For 59, the calculations are 59 + 95 = 154, then 154 + 451 = 605, then 605 + 506 = 1111. The third addition reaches a palindrome. Leading zeroes in a reversal do not change its value: reversing 120 gives 021, which is 21.

The sweep also counts a start that is already palindromic as finished with zero additions. With its sixty-step limit, 9,754 starts from 1 to 10,000 are already palindromic or reach a palindrome within the limit.

The start 196 does not reach a palindrome in the laboratory's 400 additions. Larger published searches have also failed to find one, but a finite unsuccessful search cannot establish that a palindrome will never appear.

Look at what the addition actually does. You are adding a digit string to its own reverse, so column $i$ from the left gets the same pair of digits as column $i$ from the right - just the other way round. Addition does not care about order, so without carries the sum is a palindrome automatically.

Try it: 1234 + 4321 = 5555. Every column adds to 5, nothing carries, done in one step.

So the question is never "will it be a palindrome" in general. It is only ever "will a column overflow". A carry pushes a 1 into the neighbouring column and breaks the symmetry, and then you go round again with a number whose digits are no longer mirror-friendly.

Most starts in the finite ranges shown here run out of carries quickly. The step limit becomes interesting when a value keeps producing digits large enough to carry, step after step - which is what every completed search from 196 has observed, and observation is as far as anyone has got.

Here is the sentence this page exists to make impossible to misread.

196 is a candidate, not a proved failure. Not one base-10 starting value has ever been proved never to reach a palindrome.

The word for such a number is Lychrel, and it is defined as "never reaches a palindrome". Nobody has ever shown that any base-10 number satisfies that definition. The published lists - 196, 295, 394, 493, 592, 689 … - are lists of numbers that searches have not finished.

You can watch this happen. In the sweep view, set the step limit to 10 and count how many starts below 300 look unresolved: ten of them. Set it to 120 and there are two. The list is a fact about the search.

There is also a genuinely settled case this page can show, which is the sharpest way to see what kind of question this is. In base 2, the number written 10110 - that is 22 - has been proved never to reach a palindrome. Reversing digits is a question about notation, so the answer is allowed to depend on the base. What that settles is base 2: the argument uses the binary form, and no result in one base transfers a conclusion to another. The other bases in that view - 3, 4, 8, 16

• are there to be tried, not to be pronounced on; this page cites a proof for base 2 and makes no claim either way about the rest.

Fix a base $b\ge 2$. For $n$ written in base $b$, let ${\rho }_b(n)$ be the value obtained by reversing its digits, and define

Call $n$ $b$-Lychrel if no iterate $R_b^k(n)$, $k\ge 1$, is a palindrome in base $b$.

1. $R_b(n)\ge n$ for all $n\ge 0$, with equality only at $n=0$. Proved.
2. There exists a 2-Lychrel number: $10110_2=22$. Proved.
3. Whether any 10-Lychrel number exists is open. In particular, whether 196 is one is open.

Statement 1 is what makes this different from every convergent digit routine on this site: the orbit is strictly increasing, so there is no finite region to trap it in and no space to exhaust.

The orbit never decreases. ${\rho }_b(n)\ge 0$ for every $n$, so $R_b(n)=n+{\rho }_b(n)\ge n$. Equality requires ${\rho }_b(n)=0$, which requires every digit of $n$ to be 0 - that is, $n=0$. So for $n\ge 1$ the sequence is strictly increasing, and its digit count is non-decreasing. ∎

A carry-free step lands immediately. Let $n$ have base-$b$ digits $d_1{d}_2\dots d_k$. Then ${\rho }_b(n)$ has digits $d_k\dots d_1$, and column $i$ of the sum receives $d_i+d_{k+1-i}$ - which is the same quantity as column $k+1-i$ receives. If no column reaches $b$, no carry occurs, each column stands alone, and the resulting digit string is its own reverse. ∎

Consequently, every step that fails to produce a palindrome carried. The contrapositive of the above. This is what the marked columns in the laboratory are showing. ∎

What is not proved, and why the gap is not technical. Nothing above constrains how long the carries can keep happening. To show 196 never lands you would have to rule out a palindrome at every one of infinitely many steps. Across the 400-step window drawn here the values grow by roughly 0.44 digits each time, so even this small computation illustrates how the number of digit patterns to exclude keeps growing. The base-2 proof does not provide a corresponding argument for 196 in base 10; it works by an argument specific to how the value looks in binary.

What this laboratory computes. Four hundred steps per run, in exact arbitrary-precision arithmetic. That is not a shortcut: 196 leaves the range where ordinary whole numbers are exact at step 33, and every value printed after that in floating point would be a rounded fiction that looks exactly like arithmetic. The published search has gone more than a billion digits farther than this page ever will, and it has settled nothing either.

Candidate lists depend on both the range and the step limit. Hold the range fixed and a shorter step limit produces more candidates; raising the limit can only remove them. This page scans fewer starting values than published lists, so it may still print fewer candidates overall for that different reason. Two honest searches can disagree about their lists without disagreeing about any number, because each list describes its search.

Read this next to the 6174 attractor. They are the same kind of object - a digit routine, iterated, on ordinary decimal numbers - and one is completely settled while the other is completely open. The difference is not sophistication. It is that Kaprekar's routine maps four-digit numbers to four-digit numbers, so the whole space is ten thousand states and can be exhausted; and reverse-and-add only ever grows, so there is no space to exhaust and nothing for a finite computation to close.

That is the most useful thing this page can leave you with. From the outside - a rule you can explain in one line, orbits you can watch, a pattern that holds every time you try it - a settled problem and an open one are indistinguishable. What tells them apart is whether the thing being iterated has anywhere to escape to.