Does light from source one mil. light years away take more than 1 mil. years to reach us? Because of the expansion of space itself, did light from a galaxy one million light years away actually take more than one million years to reach us?
Also, if we are looking at a galaxy one billion light years away, how far away is it today?
Is there a reference chart or equation for this?
I understand (I think) that if an astronomer sees light from 13.8 billion light-years away, it is now 46.6 light years away, a ratio of about 3.375....
But, the ratio for something closer is going to be less, because of Hubble's law/parameter... Right?
 A: The short answer is yes.
As human beings, we (unsurprisingly) tend not to think of velocities in a manner that accounts for the contribution of the expansion of space. Therefore, cosmology has adopted a few terms that make talking about this easier.
Note that I am not an expert, and I defer to any true expert who happens to answer this question. I'm repeating some information off the internet, in the hope of being of some help. Please feel free to comment or edit this answer in case any of my information isn't quite precise enough.
Here are the relevant terms here:


*

*Hubble flow - The motion of distant objects caused by the expansion of the universe

*Recession velocity — The velocity contributed by Hubble flow

*Peculiar velocity — The velocity independent of Hubble flow. In other words, layman velocity

*Redshift velocity — The effective velocity obtained by adding recession and peculiar velocity. Note that this can easily exceed the speed of light for distant objects. Note also that because this fact cannot be used to transmit information faster than light, this is not a violation of the speed of light


If you want to measure how long it would actually take for an object to travel a certain distance through the expanding universe, you'd need to use redshift velocity, which goes to zero for two objects moving toward each other with a velocity exactly equal to their recession velocity, and beyond the speed of light for objects whose recession velocities are superluminal. In the superluminal case, no information will ever be exchanged between the two objects, as even light won't be able to outpace the cosmological redshift.
A: 
Because of the expansion of space itself, did the light from a galaxy one million light-years away actually take more than one million years to reach us?

Yes, that's correct. The proper distance to an object at the time of observation ($d_p(t_0)$) is related to the proper distance when the light was emitted ($d_p(t_e)$) via
$$d_p(t_0) = d_p(t_e) \times (1+z)~~Eqn.(1)$$
For light $ds=0$ and by taking $d\Omega = 0$, the FRWL metric (for flat universe)
$$ds^2 = -c^2dt^2 + a(t)^2[dr^2 + r^2d\Omega^2]$$
takes the form of 
$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)}$$
But you can write this equation in terms of z to make calculations easier. 
So we know that $$1 + z = a(t)^{-1}$$
hence 
$$\frac{dz}{dt} = \frac{dz} {da} \frac{da} {dt}$$
$$\frac{dz}{dt} = -\frac{1}{a^2} \dot{a}$$
$$dz = -\frac{H}{a}dt = -H(1+z)dt$$
$$dt = -\frac{dz}{H(1+z)}$$
Since at $t_e$ corrsponds to $z$ and $t_0 = 0$ we have
$$r = c\int_{t_e}^{t_o}\frac{dt}{a(t)} = -c\int_{z}^{0}\frac{dz}{H}$$
Also we know that $$H = H_0\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}$$
Finally 
$$ r = -\frac{c}{H_0}\int_{z}^{0}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$
or $$ r = \frac{c}{H_0}\int_{0}^{z}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}~~Eqn.(2)$$

Also, if we are looking at a galaxy one billion light-years away, how far away is it today?

If you know the $z$ value you can easily calculate the desired values by using Equation (1) and Equation (2)

I understand (I think) that if an astronomer sees light from 13.8 billion light-years away, it is now 46.6 light-years away, a ratio of about 3.375...

You can easily calculate this factor by using Equation 2. We are looking for the 13.8 billion light-years away so in other terms the beginning of the universe. This implies that in Equation (2) the upper limit of the integral should be infinity. 
$$ r = \frac{c}{H_0}\int_{0}^{\infty}\frac{dz}{\sqrt{\Omega_r(1+z)^4 + \Omega_m(1+z)^3 + \Omega_{\Lambda}}}$$
When you set $\Omega_{r} = 0$, $\Omega_{m} = 0.31$ and $\Omega_{\Lambda} = 0.69$ you ll find that ( you can use this site to do the integral)
$$r = \frac{c}{H_0}\int_{0}^{\infty}\frac{dz}{\sqrt{0.31(1+z)^3 + 0.69}} = 3.26133$$
$$r = 3.2 \times c/H_0 = 3.2 \times 4408.712 Mpc = 14107.88 Mpc = 46.013 Gly$$
Which is the desired result?

But, the ratio for something closer is going to be less, because of Hubble's law/parameter... Right?

It depends on what parameters you are using for $\Omega$'s. But I think you can easily plot a graph by taking the 
$$r = \frac{c}{H_0}\int_{0}^{z}\frac{dz}{\sqrt{0.31(1+z)^3 + 0.69}}$$ and plotting for different z values.
A: I assume you are asking about our local frame here on Earth. And I will assume you are asking about light that was emitted some time ago on our clock.


*

*1 million lightyears is too close, because it is still inside our local gravitational group, where gravity dominates over dark energy, and space expansion is not relevant to your question, so light will exactly arrive in a million years if it was emitted a million lightyears away.

*let's talk a 100 million lightyears away, now we are talking expanding space. In this case, inbetween the emitter and us, space expansion has real effects, dark energy dominates, and light has to travel in expanding space. This will have two effects, that you might now, redshift, and what you are asking about, the photon's delay because everything is getting farther from us, including an effect on the photon that delays its arrival. The photon will not arrive in 100 million years. In fact, the photon might not arrive ever, if space is expanding faster then light inbetween the emitter and us, if it is beyond the event horizon. Certain photons might still reach us, even if the space inbetween is expanding faster then light (particle horizon).
