One can see that all values of $c$ in the table of the parent entry are hypotenuses in a right triangle with integer sides. E.g., 41 is the contraharmonic mean of 5 and 45; $9^2\!+\!40^2 \;=\; 41^2$ .

Theorem. Any integer contraharmonic mean of two different positive integers is the hypotenuse of a Pythagorean triple. Conversely, any hypotenuse of a Pythagorean triple is contraharmonic mean of two different positive integers.

Proof. $1^\circ.$ Let the integer $c$ be the contraharmonic mean $$c \;=\; \frac{u^2\!+\!v^2}{u\!+\!v}$$ of the positive integers $u$ and $v$ with $u > v$ . Then $u\!+\!v \,\mid\, u^2\!+\!v^2 \,=\, (u\!+\!v)^2-2uv$ , whence $$u\!+\!v \,\mid\, 2uv,$$ and we have the positive integers $$a \;:=\; u\!-\!v \;=\; \frac{u^2\!-\!v^2}{u\!+\!v}, \quad b \;:=\; \frac{2uv}{u\!+\!v}$$ satisfying $$a^2\!+\!b^2 \;=\; \frac{(u^2\!-\!v^2)^2\!+\!(2uv)^2}{(u\!+\!v)^2} = \frac{u^4\!-\!2u^2v^2+v^4\!+\!4u^2v^2}{(u\!+\!v)^2} = \frac{u^4\!+\!2u^2v^2\!+\!v^4}{(u\!+\!v)^2} = \frac{(u^2\!+\!v^2)^2}{(u\!+\!v)^2} \;=\; c^2.\\$$

$2^\circ.$ Suppose that $c$ is the hypotenuse of the Pythagorean triple $(a,\,b,\,c)$ , whence $c^2 = a^2\!+\!b^2$ . Let us consider the rational numbers $$ u := \frac{c\!+\!b\!+\!a}{2}, \quad v := \frac{c\!+\!b\!-\!a}{2}. $$ If the triple is primitive, then two of the integers $a,\,b,\,c$ are odd and one of them is even; if not, then similarly or all of $a,\,b,\,c$ are even. Therefore, $c\!+\!b\!\pm\!a$ are always even and accordingly $u$ and $v$ positive integers. We see also that $u\!+\!v = c\!+\!b$ . Now we obtain