If the average consecutive difference of a sequence of numbers is asymptotically the same as $f(n)$. Then what can be said about numbers in the sequence, asymptotically as $n \to \infty$. Let $\Delta a_k = a_k - a_{k-1}$. Can it be said that $\dfrac{\# \{\Delta a_k :\Delta a_k \lt f(n), \ k= 2 \dots n\}}{\# \{ \Delta a_n: \Delta a_n \gt f(n), \ k =2 \dots n\}} \to 1$ or just some constant or not even a constant?
Also, given this asymptotic information, can we say something about odd prime gaps? Since if $a_k$ were the average consecutive odd prime gap, then aymptotically there must be a large enough number of prime gaps less than $f(n) = \log n$ to compensate for the denominator in the limit above.
Is there not enough information, such as bounds on $\Delta a_n$?
The average of the consecutive gaps is just $\dfrac{a_n - a_2}{n-2} \sim \dfrac{a_n - a_2}{n} = \dfrac{1}{n}\sum (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \dots + (a_3 - a_2)$
The average of the differences $\Delta a_k = a_k - a_{k-1}$ for $k = 1 \ldots N$ is just $(a_N - a_0)/N$. This is asymptotic to $f(N)$ iff $a_N \sim N f(N)$ as $N \to \infty$. Let $b_n = a_n/n$. For $n > 1$, the gap $\Delta a_n < b_n$ iff $b_n < b_{n-1}$. The asymptotic density of $n$ for which this is true could be anywhere from $0$ to $1$, inclusive, or undefined.