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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?
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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 $$ (1-z)\Phi_{k+1}(z) = 2z\Phi_k(z)+z\Phi_{k-1}(z) $$

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    $\begingroup$ The numerators appear to be oeis.org/A109447 $\endgroup$ Apr 26, 2018 at 1:04
  • $\begingroup$ @JairTaylor, good find! Specially this: $P_1(x)=1, P_2(x)=2, P_{n+1}(x) = 2P_n(x)+(x-1)P_{n-1}(x)$. $\endgroup$
    – lhf
    Apr 26, 2018 at 1:10
  • $\begingroup$ That’s great. Thanks! $\endgroup$ Apr 26, 2018 at 13:13

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