6
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Let $\mathbb{P}$ be the set of all primes, $p_n$ the $n^{\text{th}}$ prime and $S_i$ a subset of $\mathbb{P}$, consisting of the smallest amount of consecutive primes for which $\sum S_i = \text{prime}$.

Rule 1: if the last prime of $S_i$ is $p_n$, then the first prime of $S_{i+1}$ must be $p_{n+1}$.

Rule 2: $|S_i| \ge 2 $

If we start with the first primes we get $$S_1 = \{ 2,3 \} \quad \to \quad \sum S_1 = 5$$

followed by

$$S_2 = \{ 5,7,11 \} \quad \to \quad \sum S_2 = 23$$ $$S_3 = \{ 13,17,19, 23, 29 \} \quad \to \quad \sum S_3 = 101$$ $$S_4 = \{ 31,37,41 \} \quad \to \quad \sum S_4 = 109$$ $$\dots$$

After calculating the subsets for the first $10^4$ primes, I'm inclined to say that there are infinitely many of these subsets. Is there any way we can prove this?

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  • $\begingroup$ A rigorous proof will be difficult, but because the partial sums grow slowly and there is no reason that the partial sums must have small prime factors, your claim should be true. $\endgroup$ – Peter Sep 26 '15 at 10:59
  • $\begingroup$ I checked it upto the prime $10^9$. The last sum occuring is : $10,999,998,463$. $\endgroup$ – Peter Sep 26 '15 at 11:02
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This PARI/GP-program shows the sums having more summands than every earlier sum.

? maxi=0;s=0;p=1;while(p<10^9,s=0;p=nextprime(p+1);s=s+p;p=nextprime(p+1);s=s+p;
l=2;while(isprime(s)==0,p=nextprime(p+1);s=s+p;l=l+1);if(l>maxi,maxi=l;print(p,"
  ",s,"  ",l)))
3  5  2
11  23  3
29  101  5
127  757  7
173  1367  9
983  44843  57
15641  936697  61
77069  5747459  75
102329  12913867  127
848273  119470049  141
2090071  323789369  155
2501357  422521873  169
4053473  765829241  189
9145921  1801447657  197
17298467  3615013973  209
30961577  6594416419  213
39716671  9015231847  227
45057613  11849608241  263
72835471  24981493099  343
491344573  178356736889  363
?

The last sum has only $11$ summands.

? print(p," ",s,"  ",l)
1000000009 10999998463  11
?

The following conjectur would imply your conjecture.

The sequence of partial sums of consecutive primes, starting with an arbitary prime $p$, contains a prime greater than $p$ itself.

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