Is this Prime Sequence the longest?

Question

So you are interested in Prime Numbers and puzzles thereof. You saw the following on PSE and gave it a try and got it long after the correct answer was posted by @hexomino.

My Eight Cousins

But then you think that you could design another puzzle.

So you start with 2 prime numbers say A and B with B > A

Then you put another Prime C (C>B) in there.

Turns out B is the average of the three numbers! That is nice.

Then you put another Prime D (D>C) with them. You guessed it.

Now C is the average of A,B,C and D!

Then you put another Prime number E (E>D) with them.

Sure enough now D is the average of A,B,C,D and E!

Now you are excited. You think you have discovered something great. Shake up the Math world??

So you continue with another prime F>E. And yes same thing! E is the average of A to F.

You continue these steps until your bubble bursts when you get the number 5917. Oh no. It is not a Prime.

So you wonder. Is this the longest sequence you can do or is there a sequence longer than one you have discovered? How do you explain this anyway? Is there a math equation for this?

So how many total numbers were in this sequence?

Answer 1

The sequence you tried is

3,5,7,13,27,157,877,5917

So how many total numbers were in this sequence?

7, the 8th number failed

is there a sequence longer than one you have discovered?

13 numbers long: 6947,9173,11399,18077,44789,178349,979709,6589229,51465389,455350829,4494205229 48921603629,582050384429. Haven't proven if this is the longest sequence or not...

Is there a math equation for this?

Start with your first prime, A. Then B is A+X where X is a multiple of 2. From this, we know that C=A+2X, D=A+5X, E=A+17X... where the multiplier on X is https://oeis.org/A014288

Answer 2

arbitrahj and Paul Panzer have already provided great analysis of this problem.

I would like to demonstrate that

It is possible to find arbitrarily long sequences of primes which satisfy this property (there is no longest sequence). To show this, I need a small lemma and a big theorem.

Lemma

For any positive integers $a_0$ and $d$, I can generate a sequence of positive integers $a_0, a_1, a_2, \ldots$ such that $a_1 = a_0+d$ and for $n \geq 1$, $a_n$ is the average of $a_0, a_1, \ldots a_{n+1}$.
Furthermore for each $n$, $$a_n = a_0 + b_n d$$ where $b_n$ is an integer.

This has already essentially been shown by arbitrahj and Paul Panzer in their answers but to convince yourself, we can use Paul's difference of two consecutive equations to get $$a_{n+1} = (n+2)a_n - (n+1)a_{n-1}$$ and it's a very easy induction argument from there.

Theorem

The very big theorem I need is the Green-Tao Theorem which states that the sequence of prime numbers contains arbitrarily long arithmetic progressions.

Putting these together

Suppose we wish to find a sequence of prime numbers of length $N$, with elements $a_0, a_1, \ldots, a_{N-1}$, which satisfies the property that $a_n$ is the average of the numbers $a_0, \ldots, a_{n+1}$ for each $1 \leq n\leq N-2$. (henceforth termed the averaging property)

By the lemma, any sequence of the form $a_0, \,\,a_0 + b_1d, \,\,a_0+b_2d,\ldots, a_0+b_{N-1}d$ will satisfy the averaging property for some specific integer values $b_1, b_2, \ldots, b_{N-1}$ and arbitrary $d$.

Now let $b_{\min} = \min\{0,b_1,\ldots,b_{N-1}\}$ and $b_{\max} = \max\{0,b_1,\ldots,b_{N-1}\}$.
Then by the theorem, there exists integers $A_0$ and $D$ such that each of the elements in the arithmetic progression $$A_0 + b_{\min}D, A_0 + (b_{\min}+1)D, A_0 + (b_{\min}+2)D, \ldots , A_0 + (b_{\max}) D$$ is prime.

Finally, setting $a_0 = A_0$ and $d=D$, we see that we have constructed a sequence $a_0, a_1, \ldots a_{N-1}$ of prime numbers which satisfy the averaging property.

Answer 3

Partial answer, heuristic argument that it is plausible to get reasonably long sequences like this 10-prime example:

89 101 113 149 293 1,013 5,333 35,573 277,493 2,454,773

Let us rename the sequence elements for convenience: p_1,p_2,p_3 etc. Then (n+1) x p_n = p_1 + p_2 + ... + p_{n+1} for n>=2. Subtracting two consecutive versions of this identity we get (n+1) x p_n - n x p_{n-1} = p_{n+1}
Now define q_n = p_{n+1} - p_n and rewrite n x q_{n-1} = q_n. This holds from n>=3; We can directly check that q_2 = q_1, so in the end we get q_n = q_1 / 2 x n! (note that q_1 / 2 will be an integer, otherwise the sequence stops right at the beginning).
Now recall that q_n was the difference of consecutive elements of our original sequence: therefore, if p_n happens to be a prime > n then we know already that p_{n+1} is not divided by any number up to n, which nicely improves the odds for it to be prime.