NPP(n), the Nearest Pythagorean Partner of an integer $n$, I define as the integer nearest to $n$ such that the square of $n$ added to the square of $NPP(n)$ equals a square.
All integers exceeding $2$, have a nearest Pythagorean partner. For instance, $NPP(3)=4$,$NPP(4)=3$, and $NPP(5)=12$. Note that $3$ and $4$ are mutual Nearest Pythagorean Partners. Such mutuality, however, is not always the case. For instance, $NPP(5)=12$, but $NPP(12)=9$.
Now the question arises: what is the density of mutual Nearest Pythagorean Partners? In other words, for all integers $n$ larger than $2$ and smaller than a given maximum $N$, what is the fraction that satisfies $n = NPP(NPP(n))$?
After trying some analytical approach without any success, I have resorted to numerical computation. The following list are the values of $$ 2^{-k}\,\#\bigl\{n\le2^k:n = \operatorname{NPP}(\operatorname{NPP}(n))\bigr\},\quad2\le k\le16. $$
$$ 0.5,\ 0.5,\ 0.375,\ 0.34375,\ 0.375,\ 0.398438,\ 0.402344,\ 0.388672,\\ 0.389648,\ 0.385254,\ 0.377441,\ 0.37146,\ 0.366272,\ 0.363678,\ 0.362701 $$ They seen to stabilize around $0.36\dots$.