I ain't no Fortunate sum The primorial \$p_n\#\$ is the product of the first \$n\$ primes. The sequence begins \$2, 6, 30, 210, 2310\$.
A Fortunate number, \$F_n\$, is the smallest integer \$m &gt; 1\$ such that \$p_n\# + m\$ is prime. For example \$F_7 = 19\$ as:
$$p_7\# = 2\times3\times5\times7\times11\times13\times17 = 510510$$
Adding each number between \$2\$ and \$18\$ to \$510510\$ all yield composite numbers. However, \$510510 + 19 = 510529\$ which is prime.
The Fortunate numbers below \$200\$ are
$$3, 5, 7, 13, 17, 19, 23, 37, 47, 59, 61, 67, 71, 79, 89, 101, 103, 107, 109, 127, 151, 157, 163, 167, 191, 197, 199$$
We'll say an integer \$n\$ is a &quot;Fortunate sum&quot; if it can be expressed as the sum of two distinct Fortunate numbers. For example, \$22 = 3 + 19 = 5 + 17\$, so \$22\$ can be expressed as the sum of two Fortunate numbers, and so is a &quot;Fortunate sum&quot;
You are to take an integer \$n\$ as input and output a truthy value if \$n\$ is a Fortunate sum and a falsey value otherwise. You may swap the order (falsey indicates it is a Fortunate sum) if you wish. You may take input and output in any convenient format.
This is code-golf so the shortest code in bytes wins
Test cases
The first line is the Fortunate sums less than 100 (truthy values) and the second are the integers less than or equal to 100 that aren't Fortunate sums
8 10 12 16 18 20 22 24 26 28 30 32 36 40 42 44 50 52 54 56 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98
1 2 3 4 5 6 7 9 11 13 14 15 17 19 21 23 25 27 29 31 33 34 35 37 38 39 41 43 45 46 47 48 49 51 53 55 57 58 59 61 63 65 67 69 71 73 75 77 79 81 83 85 87 89 91 93 95 97 99 100
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]

Try it online!
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.