Zeros of a "bad sequence" with the pi function

[Tomasz Ordowski]

Hello Seq Fans!

Let a(n) be the largest 0 <= m < n such that 

2 pi(n) = pi(n-m) + pi(n+m).  Find the zeros. 

Conjecture: for n > 1, a(n) = 0 if and only if

n is a "good prime" of the type A127925.

These are "midpoint convex primes". 

See https://oeis.org/draft/A127925

Is this equivalence provable? 

Requires prior checking. 

Best, 

Tom Ordo 

[Tomasz Ordowski]

PS. Numbers n such that

2 pi(n) > pi(n-m) + pi(n+m)

for all 0 < m < n.  Question:

Are there such composites? 

[M F Hasler]

On Fri, Jun 5, 2026 at 8:57 AM Tomasz Ordowski <tomaszo...@gmail.com> wrote:

PS. Numbers n such that

2 pi(n) > pi(n-m) + pi(n+m)

for all 0 < m < n.  Question:

Are there such composites? 

obviously not -- not even for just m=1, since

2 pi(n) > pi(n-1) + pi(n+1)

<=> pi(n) - pi(n-1) > pi(n+1) - pi(n)

But the LHS > 0 means that n is a prime.

- M.

[Tomasz Ordowski]

Thanks!

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[Tomasz Ordowski]

PS. See https://oeis.org/history/view?seq=A127925&v=36