I ain't no Fortunate sum
Haskell, 130 128 bytes

a n=or[f x&amp;&amp;f(n-x)|x&lt;-[2..n],2*x&lt;n]
f m=or[and[p(x+y)/=(y&lt;m)|y&lt;-[2..m]]|x&lt;-scanl1(*)$filter p[2..m]]
p n=all((&gt;0).mod n)[2..n-1]

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This solution is based on the following observation.
Fact. An integer \$m&gt;1\$ is Fortunate if and only if, for some prime number \$p_n&lt;m\$, \$m\$ is the smallest integer \$&gt;1\$ such that
\$
p_1 p_2\cdots p_n+m
\$
is prime.
Proof. The claim follows easily from the fact that if \$p_n\ge m\$ then there is some \$p_i\$ with \$i\le n\$ such that \$p_i\mid m\$, and therefore \$p_i\mid p_1 p_2\cdots p_n+m\$.
At this point, checking if a number \$m\$ is Fortunate is trivial: we just have to check the condition for all the prime numbers up to \$m\$.
Explanation of the code
p n=all((&gt;0).mod n)[2..n-1]

The standard prime-checking function. Only works for n&gt;1, but that's ok.

f m=or[and[p(x+y)/=(y&lt;m)|y&lt;-[2..m]]|x&lt;-scanl1(*)$filter p[2..m]]

A function to check whether a number m is Fortunate. As explained above, it calculates the primorials \$\texttt{x}=p_1p_2\cdots p_n\$ until \$p_n\le\texttt{m}\$ and tests whether m is the smallest integer \$\texttt{y}&gt;1\$ such that \$\texttt{x}+\texttt{y}\$ is prime.

a n=or[f x&amp;&amp;f(n-x)|x&lt;-[2..n],2*x&lt;n]

The final function, to check whether n is a Fortunate sum. Pretty straightforward, the only thing to be careful of is that x and n-x must be different: this is the reason why we only iterate over values of x such that 2*x&lt;n.