Question 1: Let $a$ be a real number with a base-10 decimal representation $a_1a_2\ldots a_n \ldots$ Denote the number of ways to write $a_n$ as the sum of positive integers as $p(a_n)$ - also called the partition of $a_n$. I write $A(n)=\sum_{k=1}^n p(a_k)$ and $B(n)= \sum_{k=1}^n a_k$ and set $$\beta(a)=\lim_{n\to \infty}{A(n)\above 1.5 pt B(n)}$$ If $a$ is a base-10 simply normal number is it true that $\beta(a)={97 \above 1.5 pt 45}$ ?
Attempted Solution: Let $a$ be a base-10 simply normal number with decimal representation $a_1a_2\ldots a_n \ldots$. Explicitly we can write \begin{equation}\lim_{n\to \infty}{A(n)\above 1.5 pt B(n)} =\lim_{n\to \infty}{p(a_1)+p(a_2)+\ldots +p(a_n)\above 1.5pt a_1+a_2\ldots +a_n }\end{equation} Surely $a_k \in [0,9]$ for every $1\leq k \leq n$. Now let $\nu_n(a_k)$ count occurrences of the digit $a_k$ up to and including the n-th decimal place of $a$. Observe the following: the numerator in the above equation can be written explicitly as:
$$ \underbrace{(p(0)+\ldots +p(0))}_{\nu_n(0)\text{ times}}+\ldots+\underbrace{(p(9)+\ldots + p(9))}_{\nu_n(9)\text{ times}}$$ and similarly the denominator may be written as
$$\underbrace{(0+\ldots +0)}_{\nu_n(0)\text{ times}}+\ldots+\underbrace{(9+\ldots + 9)}_{\nu_n(9)\text{ times}}$$ Combining those two observation together we can rewrite the RHS of our first equation as
$$ \lim_{n \to \infty}{\nu_n(0)\times p(0)+\nu_n(1)\times p(1)+\ldots + \nu_n(9)\times p(9)\above 1.5pt \nu_n(0)\times 0+ \nu_n(1)\times1+\ldots + \nu_n(9)\times9} $$
Recall that every digit of $a$ occurs with equal frequency. In particular since $a$ is simply normal we have that $\lim_{n \to \infty}\nu_n(a_j)=\lim_{n \to \infty}\nu_n(a_k)$ for every $j$ and $k$. With out loss to generality we can substitute $\nu_n(j)$ with $\nu_n(1)$ in the numerators and denominators of the equation above and evaluate the limit,
$$\lim_{n \to \infty}{\nu_n(1)\times p(1)+\ldots + \nu_n(1)\times p(9) \above 1.5pt \nu_n(1)\times1+\ldots + \nu_n(1)\times9}$$
We can then pull out the term $ \nu_n(1)$, noticing those terms cancel out in the limit, and make a final simplification to show that $$\lim_{n \to \infty}{\nu_n(1) \above 1.5pt \nu_n(1)}{p(1) +\ldots +p(9)\above 1.5 pt 1+ \ldots +9}={1+1+1+2+3+5+7+11+15+22+30 \above 1.5pt 0+1+2+3+4+5+6+7+8+9}={ 97 \above 1.5pt 45}.$$
I believe the above solution to Question 1 is correct? Now my understanding is that we have the following implication: base-10 normal $\implies$ base-10 simply normal. So the follow up question seems natural.
Question 2: If $a$ is a base-10 normal is it true that $\beta(a)={97 \above 1.5 pt 45}$ ?
The answer to Question 2 should be yes by implication alone. Moreover it should be true for both algebraic and transcendental base-10 normal numbers. Since I believe I have shown every base-10 simply normal number converges to ${97 \above 1.5 pt 45}$ under $\beta$ and every base-10 normal number is simply normal then so to will every base-10 normal number converge to ${97 \above 1.5 pt 45}$ under $\beta$. It should be clear that the converse two Question 2 is false !
If we believe Questions 1 and 2 are true then we would expect several well known provably base-10 normal constants to converge to ${97 \above 1.5pt 45}$ under $\beta$. So I ran an experiment using data from Sloan's Database on the Erdos-Copeland constants which is proven to be normal in base-10. Using the first $2\text{ }000$ digits I plotted the values of $\beta(\text{Erdos-Copeland})$. The graph below illustrates the convergence. The blue line is the value of ${97 \above 1.5 pt 45}$ and the green graph are the values of the Erdos-Copeland constant under $\beta$. Up to the $2\text{ }000$-th digit $\beta(\text{Erdos-Copeland})=2.12999037999\ldots$
I believe given enough terms one can plot a similar graph for the Champernowne constant and one should expect the convergence to be significantly slower.
Finally it is not known if $\pi$ is normal in base-10. Out of curiosity I computed $\beta(\pi)$ up to the first $500\text{ }000$ digits of $\pi$. See the graph below. Again the blue line is the value of ${97 \above 1.5 pt 45}$ and the orange graph are the values of the $\pi$ under $\beta$. Up to the $500\text{ }000$-th digit $\beta(\pi)=2.153781\ldots$. $\pi$ appears to converge faster under $\beta$ than any other mathematical constant I have looked into. 
I have numerous plots for zeta values and square roots and they all appear to approach ${97 \above 1.5 pt 45}$ which can be written numerically as $2.15555\ldots$. I have not attempted to plot numbers that are known to be abnormal in base-10.
For simply normal numbers in base $b$ we have $\displaystyle \lim_{n\rightarrow\infty}\sum_{k=1}^n f(a_k) = \frac{n}{b}\sum_{d=0}^{b-1} f(d),$ as for simply normal numbers the digits in the limit all have frequency $\frac{1}{b}$. With that we have
$$\displaystyle \lim_{n\rightarrow\infty}\frac{A(n)}{B(n)} = \frac{\sum_{d=0}^{b-1}p(d)}{\sum_{d=0}^{b-1}d},$$
which for base 10 does result in $\frac{97}{45}$.
Since base-10 normal implies base-10 simply normal, yes, this also holds for base-10 normal numbers.