A palindrome is a number which remains the same when reversing it, for instance 34143. Now pick an arbitrary number, say 26: then 26+62=88 is a palindrome. If the number was 57, then 57+75=132 is not a palindrome; but 132+231=363 is.


In general, iterating $a\_1\ldots a\_n\to a\_1\ldots a\_n+a\_n\ldots a\_1$ always seems to lead to a palindrome. But, the point is that this doesn't work for 196! 


This problem is well-known, and there is a heavy numerical evidence for it, see <http://en.wikipedia.org/wiki/Lychrel_number> and <http://www.p196.org/>. I was wondering, is there any theoretical advance on this subject? Or at least, to which math area does this problem belong to? Many thanks.



Status of the 196 conjecture?
The progress (which may not be recent) appears to be in programming tricks to push the iterations starting from 196 into more and more millions of digits. 


It is not true that iterating almost always seems to lead to a palindrome. Unless iterating leads to a palindrome "fairly quickly," one almost never seems to arrive at a palindrome. For a number with few digits, an eventual palindrome is fairly likely. As the number of digits grows the likelihood seems to decrease rapidly. Let $s(x)$ be the result of applying the reverse and add operation to $x$. There is no proof which rules out the possibility that for all $x$ there is a $j$ with $s^j(x)$ a palindrome, but it seems highly unlikely. If we examine a large range of numbers and classify $x$ as a probably never getting to a palindrome if none of $s(x),s^2(x),..s^{50}(x)$ is a palindrome then we will have some errors (probably under $1\\%$) which can be discovered by pushing out to $s^{500}(x)$. But if we push out that far, or even just to $s^{300}(x)$ then we are likely to never have an error (that we can discover.)


There is (according to Wikipedia) a 19 digit number which arrives at a palindrome after 261 iterations and this is the current record. I tried 500 random 10 digit integers (actually, random integers under $10^{10}.$) Of them, 224 went to 300 iterations without a palindrome. There were 13 cases which did arrive at a palindrome but taking at least 30 iterations. They took 30,30,32,32,32,34,38,41,42,46,49,66 and 88 iterations. 


**Further discussion** Some observations and questions in no special order:


Let $r(x)$ be the reverse of $x$ and $s(x)=x+r(x)$ be the result of applying the reverse and add operation to $x$. 


* If $x$ is a multiple of $11$ so is $r(x)$ and hence $s(x)$ and all future iterates $s^i(x)$
* if $x$ has an even number of digits then $s(x)$ is a multiple of $11$
* A palindrome with an even number of digits is a multiple of 11.
* Of the $9\cdot10^M$ palindromes with $2M+1$ digits, about $\frac{1}{11}$ are multiples of $11.$


Call $x$ *special* if no carries occur in the addition $x+r(x).$ In this case $x+r(x)$ is a palindrome. Call $x$ *exceptional* if $x+r(x)$ is a palindrome but $x$ is not special (i.e. some carries do occur). By definition, $s(x)$ is a palindrome exactly when $x$ is special or exceptional.


If $z=s(y)$ then when we appropriately match up $z$ with $r(z)$, corresponding digits will be equal or differ by $1$ (in case of a carry immediately before one but not the other.) Here appropriately means that when $z$ has one more digit than $y$ we do not match the leading $1$,


So, if $x$ has an even number $2M$ of digits then $s(x)$ is a multiple of $11$ which, if not a palindrome, misses it only by having some positions which should be equal differ by $1$ and perhaps having a leading $2M+1$st digit which is a $1$. If $x$ has an odd number of digits then almost the same is true.


The previous comments shows that integers of the form $s(x)$ are fairly special. What can be said about numbers of the form $s^2(x)$, $s^3(x)$ etc? Certainly by $s^6(x)$ (and usually earlier) we have had an even number of digits at some earlier stage and hence are at a multiple of $11$ from then on.


Clearly there must be many solutions of $s(x)=s(y)$ with $x \ne y$ since the the image set is much sparser. We can see that directly because for $y=\sum\_0^{N-1}a\_i10^i$ we have $s(y)=\sum\_0^{N-1}(a\_i+a\_{N-1-i})10^i.$ This usually presents many chances to increase one of $a\_i,a\_{N-1-i}$ by $1,2,3$ or even more and lower the other by an equal amount to get $x$ with $s(x)=s(y).$


Recall that every $y$ with $s(y)$ a palindrome is special or exceptional (and vice versa). Roughly $(1.8)^{-N/2}$ of the $N$ digit integers $z$ are special and I do not see that knowing also that $z=s^j(x)$ for some $x$ changes that. So for small starting $x$ there is a fair chance of getting to a special number as we repeatedly apply $s$. However this chance gets smaller as the number of digits rises. This suggests that if iterating $s(x)$ gets into enough digits without running into a palindrome, it is unlikely to ever arrive at one.


Here is some justification of the counts: Let $N=2M$ be even. There are $9\cdot10^{N-1}$ integers $x=\sum\_{0}^{N-1}x\_i10^i$ with $N$ digits. The proportions of these which are special and exceptional are easy to give exactly but less than $\frac{1}{1.8^M}$ of them are special and and less than $\frac{1}{10^M}$ are exceptional. This is because there are $100$ ways to pick each of the $M$ pairs $x\_i,x\_{N-i-1}$ (except for $90$ ways to pick $x\_0,X\_N$.) To be special there are only $55$ choices with $x\_i+x\_{N-i-1}\le 9$ (except $45$ for $i=0$). A number is exceptional if each pair $x\_i,x\_{N-1-i}$ has sum $0$ or $11.$ (I think I proved that to my satisfaction.)
For $N=2M+1$ the ratios are about the same taking into account the central digit.


**Final thoughts** Here is an easier related problem which might be called the **49 problem**, I don't know if there is a 49+49=**98 problem** although we have just looked at the 98+98=**196 problem.** 196+196=392 and I've spent time on the **392 problem** but it is unrelated. Here is the **49 problem**: Call an integer $z$ *very even* if all digits are even (equivalently,if $z=d(x)=x+x$ for an $x$ with all digits less than $5$) . 


Q: for which $n$ does $n,d(n),d^2(n),\cdots$ arrive at a very even integer? 


Up to 20001 there is a single starter which gets to a very even number after 47 doublings but no sooner. Curiously 49,98,**196**,392,... is the first example which never seems to arrive at a very even number. I actually included this example because I thought there would be some integers which provably never lead to a very even number. My proposed proof was that $x,d(x),d^2(x),\cdots$ is eventually periodic mod $10^j$ for any $j$, so one would just need to find the appropriate $j$ where every member of the cycle had an odd digit. However now I see that the $\mod 10^j$ orbit has length $4\cdot 5^{j-1}$ so almost surely does have a $j$ digit very even member (which may be the tail end of much much longer integers with plenty of room for odd digits.) However it seems clear that when the number of digits is small there is a reasonable chance of bumping into a very even number but that goes to $0$ exponentially in $d$ for a "random" $d$-digit number. It would be nice to find some invariant for your problem, or for this one, (similar to the proposed reduce mod $10^j$ here) which could rule out a palindrome. Short of something like that it seems hard to imagine a definitive proof (as opposed to probabilistic heuristic) of never arriving at a palindrome.


