If you set $\alpha=1$ and impose the condition that $P_{2i}=P_{2i+1}$, then you get the sequence $P_0,P_2,P_4,... = 1,1/2,0,1/4,-1/4,0,1/4,1/8,-3/8,-1/8,1/8,0,...$. Define $Q_{2i}=P_{2i}-P_{2i-2}$. The pattern of these differences is simpler: $Q_2,Q_4,Q_6,...=-1/2,-1/2,1/4,-1/2,1/4,-1/8,-1/2,1/4,...,(-1/2)^{b(i)}$ where $b(i)$ is the number of $1$s in the binary expansion of $i$. To prove this, one can check that $Q_{2k}=Q_{4k}=-2Q_{4k+2}$ using $P_{2k}=2P_{4k+2}$.
Since there are arbitrarily large numbers of low binary weight, the differences $Q_{2i}$ do not converge to $0$ so the terms $P_i$ do not converge to $0$, so the sum does not converge and can't be normalized to $1$. Nevertheless, for other values of $\alpha$ I expect that you can produce a convergent series this way involving the binary weight function $b$.
For $\alpha \ne 1$, we can still impose that $P_{2i}=P_{2i+1}$ for $i\gt 0$. It appears that there is the following formula for $Q_{2i}$ where $2i =2^j + \epsilon 2^{j-1} + k$, $\epsilon \in \{0,1\}, k \lt 2^{j-1}$,$2i \gt 0$:
$$Q_{2i} = 2^{-j} \alpha^2 (\alpha-3)(\alpha+1)^{j-2} \left( -\frac{\alpha-2}{\alpha-3} \right)^\epsilon \left(-\frac{\alpha}{\alpha+1} \right)^{b(k)}$$
This means $Q_{2i}$ can be written as a product over the binary digits of $2i$ where the first two digits are special, or we can express $Q_{2i}$ recursively.
$Q_4 = \frac{1}{4} \alpha^2 (\alpha-3)$
$Q_6 = -\frac{1}{4} \alpha^2 (\alpha-2)$
And then for $k \gt 1$,
$Q_{4k} = \frac{1+\alpha}{2} Q_{2k}$
$Q_{4k+2} = -\frac{\alpha}{2} Q_{2k}$
I expect that the proofs for the case $\alpha=1$ don't need to be modified significantly to prove these.
This question is a moving target, so this answer no longer satisfies all of the conditions. I think I'm done.