Status of the 196 conjecture? 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.
 A: Here are some extensions to Aaron Meyerowitz's comments. (Edit: As Aaron points out in the comments, my primary claim here is actually wrong.)
As Aaron points out, it is clear that if computing the sum $s(x) = x + r(x)$ involves no carries, then $s(x)$ is a palindrome. In this case we call $x$ "special." If computing $s(x)$ does involve carries (i.e., $x$ is not special) but $s(x)$ is nonetheless a palindrome, we call $x$ "exceptional." Aaron asks how common exceptional numbers are.
I claim that exceptional numbers only occur in one very specific situation, and that this gives us a necessary and sufficient condition for $s(x)$ to be a palindrome. Specifically, I claim that a carrying computation of $s(x)$ results in a palindrome if and only if the carry happens in the first and last place of the number, and results in the first two digits and the final two digits all being one. So basically, the rule is that $s(x)$ is a palindrome iff there are no carries in its computation, except in one very specific situation. This applies in all bases.
Given a nonnegative integer $n$ and a base $b \geq 2$, we shall write $\bar{n}$ to denote the number of digits in $n$'s base $b$ representation. We write $n_i$ to denote the $i$th digit from the left, with $n_1$ being the first (least significant) digit, and $n_{\bar{n}}$ being the last (most significant) digit.
We shall write $n_{-i}$ to abbreviate $n_{\bar{n} - i + 1}$. This is the digit "corresponding" to $n_i$ in the reverse of $n$. We have $n_{-i} = r(n)_{i}$.
Given a number $n$, a "carry" is an index $1 \leq i \leq \bar{n}$ such that $n_i + n_{-i} \geq b$. It is a location where a carry happens in computing $s(n)$. If $n$ has no carries, then $s(n)$ is a palindrome.
Define an "inner carry" as a carry $i$ where $1 < i < \bar{n}$. An "outer carry" is a carry $i$ where $i = 1$ or $i = \bar{n}$. An "exceptional outer carry" is an outer carry is an outer carry where, letting $m = s(n)$, we have
$ m_{\bar{m}} = m_{\bar{m}-1} = m_{\bar{n}} = m_2 = m_1 = 1. $
That is, in an exceptional outer carry, the first two digits and the last two digits of $s(n)$ are all $1$.
Proposition. $s(n)$ is a palindrome iff every carry for $n$ is an exceptional outer carry.
(Left to right.) Let $m = s(n)$, and suppose $m$ is a palindrome. Suppose there is an outer carry. Then $m_{\bar{m}} = 1$. Then $m_1 = 1$. Then
$m_{\bar{m}-1} = m_{\bar{n}} = n_1 + n_{\bar{n}} - b = m_1 = 1.$
Then $m_2 = m_{\bar{m}-1} = 1$. So if there is an outer carry, it is exceptional. Now suppose there is an inner carry, and let $i$ be the smallest inner carry. (Observe that $i \leq \lceil \frac{\bar{n}}{2} \rceil$, since if $i$ is a carry, then $-i$ is also a carry.)
To begin, suppose there is no outer carry. Then $i$ is the smallest carry. Then
$m_{i-1} = n_{i-1} + n_{-(i+1)}.$
$-i$ is also a carry, so there is a carry into $-i+1$. But $-i$ is the largest carry, so there is no carry from $-i+1$. So
$m_{-i+1} = m_{-(i-1)} = n_{-(i-1)} + n_{--(i-1)} + 1 = n_{i-1} + n_{-(i-1)} + 1 \neq m_{i-1},$
so $m$ is not a palindrome. So in the case where there is no outer carry, there is no inner carry. Now suppose there is an outer carry. The outer carry is exceptional, and then
$m_{\bar{m}} = m_{\bar{m}-1} = m_{\bar{n}} = m_2 = m_1 = 1.$
If $i \geq 3$, then we can argue as in the case where there is no outer carry, since we have that there is no carry into the $i$th place. $i \neq 1$, since $i$ is inner. Suppose $i = 2$. Then there is a carry from $m_{\bar{n}-1}$ into $m_{\bar{n}}$. $n_1 + n_{\bar{n}} = b+1$,  (i.e., 11 in base $b$), since there is an outer carry and $m_1 = 1$. So
$m_{\bar{n}}  = n_1 + n_{\bar{n}} - b + 1 = b + 1 - b + 1 = 1 + 1 = 2,$
contradicting $m_{\bar{n}} = 1$. So there is no inner carry, and we are done with the left to right case.
(Right to left.) Suppose every carry for $n$ is an exceptional outer carry. If there are no carries, then $m = s(n)$ is a palindrome. Suppose there is an exceptional outer carry. Then $m_i = m_{-i}$ for all $i \in \{1,2,\bar{m},\bar{m}-1\}$ by definition, and $m_{i} = m_{-i}$ for all $2 < i < \bar{m}-1$ by the absence of inner carries.
Comments and criticisms are welcome; I suspect the proof could use refining, and it might actually be wrong!
A: Here are some heuristic calculations. Let $r(x)$ be the reversal of $x$ in base 10, $s(x)=x+r(x)$, and $f(x)=s(x)$ if $x$ is nonzero modulo 10, and $f(x)=s(s(x))$ if $x$ is a multiple of 10.
The palindromes in the sequence $s(x),s^2(x),s^3(x),\dots$ are the same as those in the sequence $f(x),f^2(x),f^3(x),\dots$ [Edit: as per comments this isn't 100% true], and $f$ grows exponentially fast: $$100 x > f(x) > \frac{11}{10} x.$$ The constants here aren't so important: the sequence grows exponentially. Homework: what is $\sup f(x)/x$ ? [Edit: it's still true that $s$ grows exponentially fast on average, and that's enough to justify all below.]
How likely is a random number to be palindromic? Between $a\cdot 10^k$ and $(a+1)\cdot 10^k$ (with $1\leq a \leq 8$), there are $10^{\lfloor k/2 \rfloor}$ palindromes. So a random number of size around $N$ has probability around $1/\sqrt{N}$ of being a palindrome (this is crude, but we're just talking heuristics here). Roughly then, the probability of none of $f(x),f^2(x),f^3(x),\dots$ being palindromes, assuming independence, is something like
  $$p_x:=P(\text{$x$ is Lychrel}) \approx \prod_{i=1}^\infty \left(1-\frac{1}{\sqrt{(11/10)^i x}}\right).$$
Continuing, $$\log(p_x) \approx \sum_{i=1}^\infty \log(1-1/\sqrt{(11/10)^i x})\approx
    -\sum_{i=1}^\infty 1/\sqrt{(11/10)^i x} \approx -\frac{20}{\sqrt{x}},$$
where the $\approx$ symbol means I'm not being precise except in the ways that I think matter. This all comes to $p_x \approx c^{1/\sqrt{x}}$ for some $0 < c < 1$.
Assuming that I can still do calculus, this means that $p_x\to 1$ and so $\sum_{x=1}^\infty p_x$ diverges, and so by the Borel-Cantelli Lemmas there are infinitely many Lychrel numbers. For any particular (random) $x$, it has a nonzero-nonone probability of being Lychrel, and most of that probability is tied up in the first few iterations of $f$. As 196 survives the first so-many iterations, it becomes increasingly unlikely (but not impossible) that a palindrome will be chanced upon.
