a(1) = 1, ...

[Tomasz Ordowski]

Hello Math Fans! 

Let a(n) be the largest m >= n such that 

prime((m+n)/2) = (prime(m)+prime(n))/2,

where m and n must have the same parity.

How far should we stop searching for such m? 

Without this knowledge there is no reliable data. 

However, someone can calculate something. 

Refutable conjecture: a(n) > n for n > 1.   

Best, 

Tom Ordo 

[Allan Wechsler]

I'm not even convinced that it can be proven that there is such an m. It's true that the general slope of prime(n) increases, but it increases so slowly that the increase over a span of 2*k is not that much more than twice the increase over a span of k, so a slightly-unusually large value of prime((m+n)/2) offers an opportunity.

It might be worthwhile asking a similar question for an easier but similarly-growing function like floor(n ln n) ( oeis.com/A050504 ) or the closely related sum n/[1..n] ( oeis.com/A006218 ) .

-- Allan

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[Tomasz Ordowski]

There is always m = n, which gives the trivial a(n) = n.
But my bold conjecture is that a(n) > n for every n > 1.

We have a(n) >= 1, 8, 7, 8, 15, 18, 13, ... Right? 

Maybe someone will find more and better data. 

Need an estimation of the type a(n) << n + f(n).
Could it be related to the k-tuple conjecture?
Just like this A289827 (with the pi function).

It is known that pi(2x) < 2 pi(x) for x >= 11.

See https://oeis.org/A289827 

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[Tomasz Ordowski]

PS. We considered arithmetic progressions of triples of primes with indices also in arithmetic progressions. 

I conjecture that every odd prime number initiates such a progression with a positive difference.

In general, there should be arbitrarily long arithmetic sequences of prime numbers with indices that also form arithmetic progressions.