# On the sum of consecutive primes and product of first and last

Lets consider the sequence of consecutive prime numbers $p_1=2 , p_2=3 ,p_4=5 , \cdots$

. $(p_n,p_{j})$ is to be called good prime pair if $$\sum_{i =n }^{j}p_i= p_n p_{j}$$ Meaning the sum of set of consecutive term of the sequence of prime number equals to the product first and the last number of the set.

For example we have $$2+3+5=10= 2 \times 5$$Thus $(2,5)$ are good prime pairs . Similarly $$3+5+7+11+13=39=3 \times 13$$thus $(3,13)$ are good prime pairs And similarly $(5 ,31)$ is other such example . On this I have two questions

1) Are there infinitely many good prime pairs? and also 2) For every prime $p$ does there exist another prime $p'$ such that those form a good prime pair $(p , p')$

• 2) is pretty much unlikely and, if false, has a good chance to be settled with a brute force computer search (for large $p'$, we certainly have LHS>RHS, so it is a finite problem for every single $p$). Have you tried it? Needless to say, there is no known reason for the sum of all primes between two large numbers to be or not to be any particular number that is within the limits dictated by the PNT asymptotics, so my prediction is that Question 1 will outlive any of us. I'll be happy to be proved wrong, of course. – fedja Jun 8 '14 at 15:42
• @fedja I'll be happy it it will be settled – Shivam Patel Jun 8 '14 at 16:58
• $(7,53)$ is the next such pair, but there does not seem to be a prime $p'$ such that $(11,p')$ is such a pair. In general, when $p$ is large, any prime $p'$ such that $(p,p')$ is a good prime pair must be around $p'\approx (1+\sqrt2)p$ in size. – Greg Martin Jun 8 '14 at 19:29
• @GregMartin: Surely $p^{\prime} \sim 2p \log p$? To Shivam Patel: Since you're 15 perhaps you won't be offended by this unsolicited advice (or feel free to ignore it)? I think you should take fedja's comment seriously, and use this as an opportunity to learn about the prime number theorem (for example, you can find a copy of Apostol's book on Analytic Number Theory inexpensively in India), and maybe also do a little coding so you can experiment with your questions yourself. After all, the fun in math is in learning and doing it, and not simply getting answers from others. All the best! – Lucia Jun 8 '14 at 21:49
• @Lucia: you're right, I lost the $\log p$. – Greg Martin Jun 8 '14 at 23:21