Conjecture: 5 is the only prime that is the sum of all preceding primes.

Question

ganeshie8 · Answer

Interesting... The difference between two consecutive prime numbers appears to mostly increase for large numbers. So there might exist a large prime that is sum of all the preceding primes.

ganeshie8 · Answer

2 + 3 = 5
2 + 3 + 5 = 10 **
2 + 3 + 5 + 7 = 17
2 + 3 + 5 + 7 + 11 = 28 **

$p_{2n}$ cannot be the sum of its preceding primes as the sum is even. So 50% of the primes are in favor of your conjecture ! :)

kainui · Answer

Haha nice good thinking!

ganeshie8 · Answer

How strongly do you believe your conjecture is true ? 
(I'm looking at the lower/upper bounds for the primes - might give us something to mess with...)

ganeshie8 · Answer

Nvm, twin primes kill that idea of mine

kainui · Answer

Yeah this is sorta related to prime gaps so I don't know if it's different enough or not to be provable. I think it might but not sure, I think perhaps there existing at least one prime between powers of 2 could help?

ganeshie8 · Answer

Definitely, bertrand's postulate right ? https://en.wikipedia.org/wiki/Bertrand%27s_postulate

ganeshie8 · Answer

$p_{n+1} \le 2^{2^{n}}$

ganeshie8 · Answer

$$S_n = \sum\limits_{k=1}^n p_k \le  \sum\limits_{k=1}^n 2^{2^{k-1}}$$

ganeshie8 · Answer

I think we're done if we could show that sum is less than $p_{n+1}$

kainui · Answer

Hmm that seems difficult I'm not sure if this will end up working out.

ganeshie8 · Answer

Yeah, I don't really know. I'm just typing whatever comes in my head haha

kainui · Answer

I'm feeling that I thought this was gonna work out too but it's sorta like bounded on the wrong side of the fence.

I am looking at this now:

$$p_{n+1} < \sum_{k=1}^n p_k$$

I think this slightly stronger statement is true for $n>3$.

ganeshie8 · Answer

Ahh right, the sum might exceed our fancy prime; so we should be looking at the lower bound

imqwerty · Answer

Maybe we can use that formula for nth prime which ganeshie once posted.. That won't be easy tho

kainui · Answer

I think I might have almost proved it!
Bertrand's postulate:
$$p_{k+1} < 2 p_k$$$$p_{k+1}-p_k < p_k$$$$\Delta p_k < p_k$$$$\sum_{k=1}^n \Delta p_k < \sum_{k=1}^n p_k$$$$p_{n+1}-2 < \sum_{k=1}^n p_k$$
which is almost 
$$p_{n+1} < \sum_{k=1}^n p_k $$

kainui · Answer

Ah ok I have found the last piece of the puzzle, it's actually really amazing how perfectly it fits, @wio gave me the last info I needed!

Instead of summing from 1, I start at some variable.

$$\sum_{k=a}^n \Delta p_k < \sum_{k=a}^n p_k$$$$p_{n+1}-p_a < \sum_{k=a}^n p_k$$$$p_{n+1}< p_a+ \sum_{k=a}^n p_k$$
Now since I already know $5=3+2$ or in other words: $p_3=p_2+p_1$ I plug in $a=3$ and get:
$$p_{n+1}< \sum_{k=1}^2p_k+ \sum_{k=3}^n p_k$$
$$p_{n+1}< \sum_{k=1}^n p_k$$
Uncannily exactly what I was seeking to prove... Spooky how well that worked out!

ganeshie8 · Answer

Very clever use of the telescoping sum !