Is $\lambda < \kappa \leq 2^\lambda$ special? It is known that if cardinal $\kappa$ is such that there is a cardinal $\lambda$ with $\lambda &lt; \kappa \le 2^\lambda$ then any $\kappa$-complete ultrafilter on $\kappa$ is principal. 

Do such $\kappa$ have a special name and is this property equivalent to some other well known properties? Is this equivalent to being measurable? 

It seems pretty special to me and relates to GCH and critical points of elementary embeddings. 

$\mathfrak{c}$ has the property but $\omega$ doesn't. 

A consequence is that any ultrafilter on $\mathbb{R}$ closed under countable intersections is principle (assuming CH).
As Asaf points out the name of such $\kappa$s is not a strong limit

$\kappa\text{ is not a strong limit}$

$\iff\neg\forall\lambda\left(\left(\lambda&lt;\kappa\right)\to\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\neg\left(\neg\left(\lambda&lt;\kappa\right)\vee\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\left(\left(\lambda&lt;\kappa\right)\wedge\neg\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\left(\lambda&lt;\kappa\le2^{\lambda}\right)$.

When $\kappa$ is not a strong limit then any $\kappa$-complete ultrafilter
$U$ on $\kappa$ is principal (ie.has the form $\left\{ A\subseteq\kappa\mid a\in A\right\} $). This is proved here. 

Therefore in particular no infinite cardinal of the form $2^{\lambda}$
can have a $2^{\lambda}$-complete free ultrafilter, and that includes
$\mathbb{R}$ which has cardinality $2^{\omega}$. 

Note that if $U$ is an ultrafilter on $\mathbb{R}$ and CH is false
then there is a $\omega&lt;\kappa&lt;2^{\omega}$ so the above theorem would
enable us to conclude that if $U$ is $\kappa$-complete then it is
principal.

A measurable cardinal is an uncountable cardinal $\kappa$ with a
non-principle $\kappa$-complete ultrafilter and so is a strong limit.
If the GCH is true then $\lambda^{+}=2^{\lambda}$ for every infinite
cardinal so if measurable cardinals exist they are not successor cardinals.
As Asaf points out the name of such $\kappa$s is not a strong limit

$\kappa\text{ is not a strong limit}$

$\iff\neg\forall\lambda\left(\left(\lambda&lt;\kappa\right)\to\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\neg\left(\neg\left(\lambda&lt;\kappa\right)\vee\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\left(\left(\lambda&lt;\kappa\right)\wedge\neg\left(2^{\lambda}&lt;\kappa\right)\right)$

$\iff\exists\lambda\left(\lambda&lt;\kappa\le2^{\lambda}\right)$.

When $\kappa$ is not a strong limit then any $\kappa$-complete ultrafilter
$U$ on $\kappa$ is principal (ie.has the form $\left\{ A\subseteq\kappa\mid a\in A\right\} $). This is proved here. 

Therefore in particular no infinite cardinal of the form $2^{\lambda}$
can have a $2^{\lambda}$-complete free ultrafilter, and that includes
$\mathbb{R}$ which has cardinality $2^{\omega}$. 

Note that if $U$ is an ultrafilter on $\mathbb{R}$ and CH is false
then there is a $\omega&lt;\kappa&lt;2^{\omega}$ so the above theorem would
enable us to conclude that if $U$ is $\kappa$-complete then it is
principal.

A measurable cardinal is an uncountable cardinal $\kappa$ with a
non-principle $\kappa$-complete ultrafilter and so is a strong limit.
If the GCH is true then $\lambda^{+}=2^{\lambda}$ for every infinite
cardinal so if measurable cardinals exist they are not successor cardinals.