<s>[INST] $2p-2$ as the sum of consecutive prime numbers Progress:
Let $p$ be a prime such that $p≡1$ (mod 6) then $2p-2$ can be written uniquely (up to the order of addends) as the sum of some consecutive prime numbers.
These are first ten examples:
$$2⋅7-2=12=5+7$$
$$2⋅13-2=24=11+13$$
$$2⋅19-2=36=17+19$$
$$2⋅31-2=60=29+31$$
$$2⋅37-2=72=5+7+11+13+17+19$$
$$2⋅43-2=84=41+43$$
$$2⋅61-2=120=59+61$$
$$2⋅67-2=132=13+17+19+23+29+31$$
$$2⋅73-2=144=71+73$$
$$2⋅79-2=156=17+19+23+29+31+37$$
I don't even have the slightest idea that how and from where to start the proof. Hope you guys can help me. Thanks in advance. 
 [/INST] Let's try the next prime that is $1 \pmod{6}$, i.e. $p = 97$. 
We want either $2$ or $6$ consecutive primes which sum to $2 \cdot 97 - 2 = 192$. 
Since $89,97,101$ are consecutive primes and $89+97 = 186 < 192 < 198 = 97+101$, we see that we cannot write $192$ as the sum of $2$ consecutive primes. 
Since $19,23,29,31,37,41,43$ are consecutive primes and $19+23+29+31+37+41 = 180 < 192 < 204 = 23+29+31+37+41+43$, we see that we cannot write $192$ as the sum of $6$ consecutive primes. 
Therefore, $p = 97$ is a counterexample to your conjecture. 
</s>