Is This A Derivative? I am in a little over my head. This all began with my reading how each level of pascals triangle adds to $2^n$, where n=row# starting with n=0. I then though, "wouldn't it be clever if the rows added to something else--like say $3^n$ instead?" Or even better generalize it for any constant, $a^n$.
All that was needed was to Multiply any given number in pascal's triangle by $a^n/2^n$
$$
\begin{array}{rcccccccccc}
&    &    &    &    &    &  1\\\
&    &    &    &    &  \frac{a}{2} &    &    \frac{a}{2}\\\
&    &    &    &  \frac{a^2}{4} &    &  \frac{a^2}{2} &    &    \frac{a^2}{4}\\\
&    &    &      \frac{a^3}{8} &    &  \frac{3*a^3}{8} &    &  \frac{3*a^3}{8} &    &    \frac{a^3}{8}\\\
&    &      \frac{a^4}{16} &    &  \frac{a^4}{4} &    &  \frac{3*a^4}{8} &    &  \frac{a^4}{4} &    &      \frac{a^4}{16}\\\
&      \frac{a^5}{32} & &   \frac{5*a^5}{32} &    &   \frac{5*a^5}{16} &    &    \frac{5*a^5}{16}&    &  \frac{5*a^5}{32} &    &  \frac{a^5}{32}\\\
&    \frac{a^6}{64}&  &\frac{3*a^6}{32}&   &\frac{15*a^6}{64}& &\frac{5*a^6}{16}& &\frac{15*a^6}{64}& &\frac{3*a^6}{32}& &\frac{a^6}{64}\\\
&    & ... & & & &... & & & & ... & 
\end{array}
$$
Adding any row should give $a^n$. Placing said coefficients in front of a binomial expansion and solving for the binomial expression yields $$(a^n/2^n)*(x+y)^n$$
Letting a=2 makes pascals triangle, but every other value of "a" distorts every value and relation (except n=0 row for obvious reasons). 
This triangle creates some interesting relations that are shared with Pascal's triangle and are immediately obvious: every term in the middle column may be divided by "a" to yield the term above and to the right or left of it--just like Pascals triangle (a=2). 
Next is a really fascinating fluke: 
$$ \frac{\partial \frac {a^4}{4}}{\partial a}=a^3 
$$, which is the sum of the line above it. 
$$
\frac{\partial \frac {a^2}{4}}{\partial a}=\frac {a}{2}
$$, which is found in the line above it. 
I realize that "a" must have a definite value as a coefficient in order to have meaning, and it is not itself a function, but it seems curious that derivative relationship would show up in the relations between the coefficients of this modified triangle.  This serendipitous relation fascinates me to the point of asking "whats up?" here. Is this a derivative? Have derivative relationships popped up organically elsewhere in function theory? 

You may rewrite your $$\frac{a^n}{2^n}(1+1)^n$$ as $$\left(\frac{a}{2} + \frac{a}{2}\right)^n$$
Unfortunately neither of the two examples you mention appear to be specific examples of general patterns.
