Recognize this power series? I've been doing some research in ergodic theory (on Perron-Frobenius operators on Banach spaces of analytic functions) and have been led to some power series that may have been studied in other contexts. The power series are
$$
\Phi_k(z)=
\sum_{m\text{ even}}\binom mk z^{m/2},
$$
where the sum is taken over those $m$'s such that $m\ge k$.
I have played around with this and written it as a difference of two rational expressions in $\sqrt z$ (even though, of course $\Phi_k(z)$ is actually an analytic function of $z$ in the unit disk - the odd powers of $\sqrt z$ cancel).

Are these functions known in some existing context?


Are there other well-known examples (possibly series solutions of differential equations?) where one is led to a difference of power series in $\sqrt z$, where the half-integer powers all cancel?

 A: Not an answer, but some data points computed with WA.
We can write $\Phi_k(z)=
\displaystyle\sum_{m=0}^{\infty}\binom {2m}k z^{m}$
because $\displaystyle\binom {2m}k =0 $ for $2m <k$.
Then
$
\Phi_0(z) = -\dfrac{1}{z - 1}
$
$
\Phi_1(z) = \dfrac{2z}{(z - 1)^2}
$
$
\Phi_2(z) = -\dfrac{z (3 z + 1)}{(z - 1)^3}
= -\dfrac{3 z^2 + z}{(z - 1)^3}
$
$
\Phi_3(z) = \dfrac{4 z^2 (z + 1)}{(z - 1)^4}
= \dfrac{4 z^3 + 4 z^2}{(z - 1)^4}
$
$
\Phi_4(z) = -\dfrac{z^2 (5 z^2 + 10 z + 1)}{(z - 1)^5}
= -\dfrac{5 z^4 + 10 z^3 + z^2}{(z - 1)^5}$
$
\Phi_5(z) = \dfrac{2 z^3 (3 z^2 + 10 z + 3)}{(z - 1)^6}
= \dfrac{6 z^5 + 20 z^4 + 6 z^3}{(z - 1)^6}
$
Thanks to @JairTaylor, we can recognize the numerators
from oeis/A109447 and get
$$
\Phi_k(z) = (-1)^{k+1}\dfrac{z^k P_{k+1}(1/z)}{(z - 1)^{k+1}}
= \dfrac{z^k P_{k+1}(1/z)}{(1-z)^{k+1}}
$$
where
$P_1(z)=1, P_2(z)=2, P_{n+1}(z) = 2P_n(z)+(z-1)P_{n-1}(z)$.
This leads to
$$
\Phi_k(z) = \dfrac{Q_k(z)}{(1-z)^{k+1}}
$$
where
$Q_0(z)=1, Q_1(z)=2z, Q_{k+1}(z) = 2zQ_k(z)+z(1-z)Q_{k-1}(z)$, and so
$$
(1-z)\Phi_{k+1}(z) = 2z\Phi_k(z)+z\Phi_{k-1}(z)
$$
