Number of Integer Partitions into Sum of Consecutive Primes

Theorem
Let $n$ be a natural number.

Let $\map f n$ denote the number of integer partitions of $n$ where the parts are consecutive prime numbers.

For example:
 * $\map f {41} = 3$

because:
 * $41 = 11 + 13 + 17 = 2 + 3 + 5 + 7 + 11 + 13$

Then:
 * $\ds \lim_{x \mathop \to \infty} \dfrac 1 x \sum_{n \mathop = 1}^x \map f n = \ln 2$

Proof
Let $\mathbb P$ denote the set of prime numbers.

Every set of consecutive primes whose sum is less than $x$ will contribute $1$ to the sum:
 * $\map f 1 + \map f 2 + \dotsb + \map f x$

The number of such sets of $r$ primes is clearly at most:
 * $\map \pi {x / r}$

and at least:
 * $\map \pi {x / r} - r$

where $\pi$ is the prime-counting function.

Hence:
 * $\ds \sum_{r \mathop = 1}^k \paren {\map \pi {x / r} - r} \le \map f 1 + \map f 2 + \dotsb + \map f x \le \sum_{r \mathop = 1}^k \map \pi {x / r}$

where $k$ is determined by:
 * $(1): \quad p_1 + p_2 + \dotsb + p_k < x < p_1 + p_2 + . . . + p_{k + 1}$

From $(1)$ and the well known $p_r \asymp r \ln r$ we find:
 * $(2): \quad k \asymp \sqrt {\dfrac x {\ln x} }$

Here $f \asymp g$ denotes that $\dfrac f g$ is bounded above and below by positive numbers for large arguments.

This implies:
 * $\ds \sum_{r \mathop = 1}^k r = \map o x$

where $\map o x$ denotes little-o notation.

Hence, if we can show that:
 * $\ds \sum_{r \mathop = 1}^k \map \pi {x / r} \approx \ln 2$

it will follow that:
 * $\map f 1 + \map f 2 + \dotsb + \map f x \approx x \ln 2$

Using $(2)$ and the Prime Number Theorem, the following estimates are justified: