Mathematics of apportionment of representation in a legislative body It seems to be eclipsed by coronavirus, but today is the U.S. Census day for this coming decade.
If you are American, where you are living today, where your children are living today, is where they are enumerated in the Census. This will determine, first hand, how many U.S. Representatives your state will be apportioned. It will also affect how redistricting will occur in your state and, perhaps, in your city if it is divided politically into wards or districts.
Okay, this is about once the census populations of each state are finalized, about distributing (or "apportioning") a fixed number of representatives among the states.  I know about the Huntington-Hill method and this will eventually be about that.  But I want to get more fundamental about the problem.
This is what is given:


*

*$N$: The number of U.S. states.  Currently 50.

*$H$: The number of representatives in the House of Representatives.  Currently 435.

*$P_n$: The census population of state $n$.


All of these parameters are positive integers $\in \mathbb{N}$.
What we need to determine:


*

*$R_n$: The number of representatives apportioned to state $n$.


Also all of these $R_n \in \mathbb{N}$.
We know that the total population of the 50 states (this leaves out the District of Columbia, Puerto Rico, Guam, American Samoa, and other territories whose residents are American, but they get no voting representation in Congress) is the sum of the populations of all states:
$$ P_\mathrm{total} = \sum\limits_{n=1}^{N} P_n $$
And we know also that the total number of representatives is the sum of apportioned representatives of all states:
$$ H = \sum\limits_{n=1}^{N} R_n $$
Now, the simplest meaning of the concept of apportionment is that the number of representatives a state has is directly proportional to the population of that state.  That would say that there exists a constant (w.r.t. all of the states) of proportionality, $\alpha$, such that:
$$ R_n = \alpha \cdot P_n $$
Now, if there were no problems regarding fractional representatives, we know that:
$$\begin{align}
 \alpha \cdot P_\mathrm{total} &= \sum\limits_{n=1}^{N} \alpha \cdot P_n \\
\\
&= \sum\limits_{n=1}^{N} R_n \\
\\ 
&= H \\
\end{align} $$
So we can solve that for $\alpha$ and have an idea what it might be, but we cannot have a fraction of a representative, $R_n$ must be a positive integer, so then quantization or rounding to an adjacent integer is necessary.  We know that if rounding down or even rounding to nearest may cause the rounded value of $R_n$ for the least populous states to possibly be zero, and the U.S. Constitution does not allow for that.  It seems to me that the only consistent simple rule of rounding would be to always round up:
$$ R_n = \big\lceil \alpha \, P_n \big\rceil $$
where $\lceil x \rceil$ is the ceiling function which is the smallest integer $m$ such that $m-1 < x \le m$, or 
$$ \lceil x \rceil \in \mathbb{Z} \\
\\
  x \le \lceil x \rceil < x+1 $$
So, it seems to me that the simplest consistent rule to guarantee that each state gets a whole number of representatives and at least one representative is to find the constant of proportionality, $\alpha$ such that:
$$\begin{align}
H &= \sum\limits_{n=1}^{N} R_n \\
\\ 
&= \sum\limits_{n=1}^{N} \big\lceil \alpha \, P_n \big\rceil \\
\end{align} $$
Now couldn't we simply define an increasing function $h(\alpha)$ as
$$ h(\alpha) \triangleq \sum\limits_{n=1}^{N} \big\lceil \alpha \, P_n \big\rceil $$
and, starting at $\alpha=0$ (and we know that $h(0)=0$), then increase that value $\alpha$ until $h(\alpha)=H$?  Then we know the number of representatives for all of the states $R_n = \big\lceil \alpha \, P_n \big\rceil$ for all $n$.
Is this consistent with the Huntington-Hill method?
If needed, I will explain the Huntington-Hill method here, but I need to figure out a good set of symbols that is consistent with the symbols I use above.  Give me a couple hours to do that.
 A: Your method of apportionment is the one that John Quincy Adams proposed in 1832. It has never been used to apportion seats in the U.S. House of Representatives.
Most methods of apportionment in widespread use can be described in one of two equivalent ways.
One way is the one you use. A common divisor is defined for all states (or parties, when apportioning seats after an election), and each state’s population (or each party’s vote tally, in case of an election) is divided by this divisor and then rounded in some prescribed manner. The divisor is selected such that the desired total of seats results. The three obvious choices of rounding down, up or to the nearest integer are typically referred to by different names in Europe and the United States:
\begin{array}{c|c}
\text{rounding}&\text{European name}&\text{United States name}&\text{divisor offset}\\\hline
\text{up}&\text{Adams}&\text{Adams}&0\\
\text{nearest}&\text{Sainte-Laguë}&\text{Webster}&\frac12\\
\text{down}&\text{D'Hondt}&\text{Jefferson}&1\\
\end{array}
Equivalently, the seats can be apportioned one by one, with each state or party being assigned a divisor of its own that is determined by the seats already allocated to it. An offset $\Delta$ is added to this seat count, that is, the quotients
$$
\frac{\text{population or #votes}}{\text{#seats}+\Delta}
$$
are calculated, and the state or party with the largest current quotient gets the next seat. The divisor offsets corresponding to the three methods above are shown in the table.
The equivalence of the two ways of describing these methods can be shown by noting that in the first approach, when the global divisor is adjusted to achieve the desired total, a state gets an additional seat whenever the divisor crosses its rounding threshold, and this is precisely when its population is $\text{#seats}+\Delta$ times the global divisor.
The Huntington–Hill method currently in use in the U.S. House of Representatives is easier to describe in the second way. Here the divisor is the geometric mean of $\text{#seats}$ and $\text{#seats}+1$, that is, of the values that correspond to rounding up and down, respectively. Thus, it is somewhat similar to the Sainte-Laguë / Webster method, which uses the arithmetic mean $\text{#seats}+\frac12$ instead of the geometric mean. For large seat counts, the two are very similar, and even when a single seat has been apportioned, the geometric mean of $1$ and $2$ is $\sqrt2\approx1.414$, not all that different from $\frac32=1.5$; but when zero seats have been apportioned, the divisor is $0$ rather than $\frac12$, thus ensuring that all states get at least one seat. So in a sense Huntington–Hill is Webster with a smooth transition to the guaranteed minimum seat provided by Adams.
Here’s Java code that implements all four methods of apportionment. Applying them to the $2010$ United States census data yields the following differences relative to the Huntington–Hill method (which I’m using as a reference since it’s the one actually being used):
\begin{array}{l|r|c}
\text{state}&\text{rank}&\text{Adams}&\text{Webster}&\text{Jefferson}\\\hline
\text{California}&1&-3&&+2\\
\text{Texas}&2&-2&&+1\\
\text{New York}&3&-1&&+1\\
\text{Florida}&4&-1&&+1\\
\text{Illinois}&5&&&+1\\
\text{Pennsylvania}&6&-1&&&\\
\text{Ohio}&7&&&+1\\
\text{Georgia}&9&-1&&&\\
\text{North Carolina}&10&&+1&+1\\
\text{New Jersey}&11&&&+1\\
\text{Missouri}&18&+1&&&\\
\text{Minnesota}&21&&&-1\\
\text{South Carolina}&24&&&-1\\
\text{Louisiana}&25&+1&&&\\
\text{Oregon}&27&+1&&&\\
\text{Oklahoma}&28&+1&&&\\
\text{Iowa}&30&+1&&&\\
\text{West Virginia}&37&&&-1\\
\text{Nebraska}&38&&&-1\\
\text{Idaho}&39&+1&&\\
\text{Maine}&41&&&-1\\
\text{New Hampshire}&42&&&-1\\
\text{Rhode Island}&43&&-1&-1\\
\text{Montana}&44&+1&&\\
\text{Delaware}&45&+1&&\\
\text{South Dakota}&46&+1&&\\
\text{Vermont}&49&&&-1\\
\text{Wyoming}&50&&&-1\\
\end{array}
The remaining states get the same number of seats under all four methods.
The similarity between Huntington–Hill and Webster is evident, with the expected very slight advantage for smaller states under Huntington–Hill. The two “extreme” methods that always round up or down instead of taking one of the two means show a clear relative advantage for smaller and larger states, respectively. There would be quite significant changes if the Adams–Bristow-Johnson method were used instead of Huntington–Hill; so the answer to your question is “no”.
An interesting statistic that illustrates a sense in which the two methods that take a mean are more fair than the two others is the variance in the weight of voters in the House. Ideally, each voter should have exactly the same weight there, so this variance is a measure of the unfairness of the apportionment. The mean weight is the same for all methods; it’s just the number of representatives divided by the total population. In 2010 this was about $1.41\cdot10^{-6}$. The standard deviations under Huntington–Hill and Webster are quite similar; they are $6.54\cdot10^{-8}$ and $6.49\cdot10^{-8}$, respectively. The standard deviations under Adams and Jefferson are about twice as large, $1.14\cdot10^{-7}$ and $1.32\cdot10^{-7}$, respectively. So under these methods, the weight of voters in the House varies by about $±10\%$; whereas under the methods that use a mean, the variation is only about $±5\%$.
A: Perhaps consult one of the hundred or so books written on the mathematics of the topic.  For example
El-Helaly, Sherif, The mathematics of voting and apportionment. An introduction. Compact Textbooks in Mathematics. Birkhäuser/Springer, 2019
