# Is the fundamental partition associated to $n$ the partition of $r_{0}(n)$ in $k_{0}(n)$ parts that maximizes entropy?

As usual, under Goldbach's conjecture, let's define for a large enough composite integer $$n$$ the quantities $$r_{0}(n):=\inf\{r\geq 0,(n-r,n+r)\in\mathbb{P}^{2}\}$$ and $$k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$$.

The sequence of prime gaps in the interval $$[n-r_{0}(n),n+r_{0}(n)]$$ sorted in decreasing order can be listed as $$g_{1}(n),\cdots,g_{k_{0}(n)}(n)$$ and forms a partition of $$2r_{0}(n)$$. As such, the sequence $$(h_{i}(n))_{1\leq i\leq k_{0}(n)}$$ defined by $$h_{i}(n)=\frac{g_{i}(n)}{2}$$ is a partition of $$r_{0}(n)$$ in $$k_{0}(n)$$ parts. Let's call this partition the fundamental partition associated to $$n$$.

Among all partitions of $$r_{0}(n)$$ in $$k_{0}(n)$$ parts, is the fundamental partition associated to $$n$$ the one that maximizes entropy in Shannon's sense?

• I suspect not. You might tabulate all fundamental partitions (as n varies) for a given r and see what allowed partitions arise. For r larger than 3 you will find the count of allowed partitions drops off relative to the number of all partitions of n, and congruence restrictions explain some of this relative drop. Then compute entropies and report back for each r. Because of the number theory involved my guess is the answer to your question is no. Gerhard "Entropy Is A Group Concept" Paseman, 2019.03.08. – Gerhard Paseman Mar 8 at 18:41
• Hello Gerhard "enjoys Starbucks coffee" Paseman. What do you mean by "entropy is a group concept"? To me it is a measure of closeness to a uniform distribution. Sylvain "Boltzmann was a genius" Julien, 2019.03.08 :-) – Sylvain JULIEN Mar 8 at 21:59
• When you talk entropy of the fundamental partition associated to n, it suggests to me a value computed using r and k and n. I think instead it should be a value computed just using r and k, while taken in the context of all possible partitions arising as n varies. Gerhard "Not That I Understand Entropy" Paseman, 2019.03.08. – Gerhard Paseman Mar 8 at 23:23
• Actually, when I say "that maximizes entropy", I mean implicitly "given the constraints weighing on n due to congruence/prime factorization related issues". – Sylvain JULIEN Mar 8 at 23:34
• Otherwise we could as well get rid of the necessity for the $h_i$ to be integers as halves of prime gaps! – Sylvain JULIEN Mar 8 at 23:36