Massive Even Length Palindromes

We can notice a pattern if we look at the first few palindromes which gives intuition for the answer immediately: (quotes ' added for emphasis, they mean nothing) $1'1, 2'2, 3'3, 4'4, 5'5, 6'6, 7'7, 8'8, 9'9, 10'01, 11'11, 12'21, 13'31, 14'41, ...$

Hypothesis: the $k^{\text{th}}$ even length palindrome is the number k written forwards concatenated with it written backwards. Call this map (which takes an integer and returns the first even length palindrome starting with that integer) $\phi$ , it is injective since it sends no two elements to the same place.

To prove this, note that any even length palindrome is a number written forwards concatenated with itself written backwards, by definition of a palindrome, so $\phi$ is surjective. As it's also injective, $\phi$ is a bijection. Notice that for any two natural numbers a and b, $\phi(a) > \phi(b)$ if and only if a is larger than b. Thus the map $\phi$ also preserves order, and is a isomorphism, and so the $k^{\text{th}}$ palindrome is given by $\phi(k)$ .

In particular, $\phi(3,141,592,653) = 3141592653'3562951413.$