Reading the preprint Bounded gaps between primes in short intervals by Ryan Alweiss and Sammy Luo (https://arxiv.org/abs/1707.05437), I came up with the following question:
Can the difference $\Delta$ between prime powers $p^a$ and $q^b$ not exceeding $x$ always be written as a finite linear combination $\sum_{i=1}^{k'} c_{i}J_{i}(x)$ with non negative integral coefficients $c_{i}$ of the jumping champions up to $x$ $J_{i}(x)$ with $i<j\Longrightarrow J_{i}(x)>J_{j}(x)$ and $\lVert\Delta\rVert:=\sum_{i=1}^{k'}c_{i}$ minimal? I'm especially interested in the case $ab\leq 2$.
Edit: assuming an affirmative answer to the previous question can one get $\lVert\Delta\rVert=O(\log^{O(1)}x)$?
Here comes a vague idea: assuming the jumping champions are the primorials $P_{m}:=\prod_{i=1}^{m}p_{i}$, from $P_{m}=m^{m(1+o(1))}$ and writing $w_{x}$ for the "power square root" of $x$ i.e. $w_{x}^{w_{x}}=x$, from $\log^{\log x}x=(e^{\log\log x})^{\log x}=e^{\log x.\log\log x}=x^{\log\log x}$ one gets $w_{x}=o(\log x)$. Now from Bertrand's postulate one has $P_{m+1}\leq 2p_{m}P_{m}\lesssim 2m\log m P_{m}$ so one should have $\lVert\Delta\rVert\leq 2k'w_{x}\log w_{x}$. But $k'\leq w_{x}$ hence $\lVert\Delta\rVert\leq 2w_{x}^{2}\log w_{x}=o(\log^{2}(x)\log\log x)=O(\log^{O(1)}x)$.
Edit April 24th 2023: by the way, can one prove that $H_{m}:=\lim\inf_{n\to\infty}p_{n+m}-p_{n}$ is reached for $\lVert\Delta\rVert=m$ with the finite sequence $(c_{i})_{1\leq i\leq k'}$ non decreasing? This may lead to a combinatorial prediction of diameters of prime constellations. Also we may make an analogy between $\lVert\Delta\rVert$ and the dimension of a vector space and between the number of non vanishing coefficients $c_{i}$ and the rank of some endomorphism defined thereon.
Edit April 27th 2023: I've just seen that someone voted to close because this question "needs details or clarity", so I refer to https://mathoverflow.net/questions/61842/about-goldbachs-conjecture and especially to the definition of natural configuration therein so as to provide an insight of why primorials and thus conjecturally jumping champions are expected to preserve the primality of a number to which it's added or substracted (and thus why the difference of prime powers should be a linear combination with non negative integral coefficients of jumping champions provided they are primorials). From the point of view of configurations the sequence of primorials "converges to $0$" as the configuration of order $u$ of both a primorial greater or equal to $p_{u}\sharp$ and of $0$ only consists of a sequence of $0$. Taking $J_{1}(x)$ as a value for $d$ in the preprint of Alweiss and Luo may allow to pass from $2n=p+q^2$ with $p$ and $q$ prime to $2n=p'+q'$ with $p'$ and $q'$ prime with $O(\log^{O(1)}n)$ shifts of the integer $r$ such that both $n-r$ and $n+r$ are prime powers.
The general idea would be to first strengthen Chen's theorem to "every large enough even integer is the sum of a prime and the square of a prime or the sum of two primes" and then reduce the first case to the second one.
Edit April 29th 2023: say an interval $I$ (respectively its length $g=\lambda(I)$) of the form $[P^{a},Q^{b}]$ with $P$ and $Q$ prime and $Q-P$ minimal is an $m$-jumping prime power interval (respectively gap) of level $l=ab$ if $m$ is the infimum of the set of integers $t$ such that $g$ is the sum of $t$ jumping champions up to $Q^{b}$. As an example $g_{30}=p_{31}-p_{30}=127-113=14=6+6+2$ is a $3$-jumping prime power gap of level $1$.
Now denote by $\gamma_{m,l}(J)$ the number of $m$-jumping prime powers intervals of level $l$ contained in $J$ such that the total number of jumping prime power intervals of level at most $l_{0}$ contained in $J$ is $N_{l_{0}}(J)=\sum_{m\geq 1,l\leq l_{0}}\gamma_{m,l}(J)$. The idea is to express the lengths of the intervals $I_{1}:=[\min(p',q'),\max(p',q')]$ and $I_{2}:=[\min(p,q^{2}),\max(p,q^{2})]$ with $2n=p'+q'=p+q^{2}$ with $p, q, p'$ and $q'$ prime as sums of jumping prime power gaps of level at most $2$, each being of the form $\Delta$ and show that the number of $m$-jumping prime power intervals of level at most $2$ contained in $I_{j}$ with $m>1$ and $j\in\{1,2\}$ is an $o(N_{2}(I_{j}))$.
Are there results in the literature which may help to reach this goal? Note that this partly boils down to showing that most coefficients $c_{i}$ in a prime gap of the form $\Delta$ equal $0$ or $1$.
Also, assuming an affirmative answer to the first question, we may set $I_{j}$ with $j\in\{1,2\}$ such that $\Delta:=\lambda(I_{j})$ is a $\lVert\Delta\rVert$-jumping prime power gap of level $j$. Denoting by $r_{0}(n)$ the smallest non negative integer $r'$ such that $(n-r',n+r')\in\mathbb{P}^{2}$ and by $k_{0}(n):=\pi(n+r_{0}(n))-\pi(n-r_{0}(n))$ we get $2r_{0}(n)=:\Delta\Longrightarrow k_{0}(n)\leq\lVert\Delta\rVert$ and the conjecture in the current edit implies $k_{0}(n)\lesssim\lVert\Delta\rVert$. Writing $2r_{0}(n)=k_{0}(n)\log^{1+\varepsilon_{k_{0}(n)}}n=\lVert\Delta\rVert\log^{1+\varepsilon'_{\lvert\Delta\rVert}}n$ this and the PNT imply that $\lim_{k_{0}(n)\to\infty}\varepsilon_{k_{0}(n)}=\lim_{k_{0}(n)\to\infty}\varepsilon'_{\lVert\Delta\rVert}=0$.
From the reasoning leading to $\lVert\Delta\rVert=o(\log^{2}n.\log\log n)$ we get $k_{0}(n)=o(\log^{2}n.\log\log n)$ and thus $2r_{0}(n)=o(\log^{3+h}n.\log\log n)$ where $h=\sup_{k_{0}(m),m\leq n}\varepsilon_{k_{0}(m)}$, giving further evidence to the hypothesis that one may pass from $2n=p+q^{2}$ to $2n=p'+q'$ in $O(\log^{O(1)}n)$ shifts of the integer $r$ such that both $n-r$ and $n+r$ are prime powers.
Edit April 30th 2023: a way to do so would be to use the rank reducing shift $S_{n}(r):=\frac{(\Omega(n+r)+\Omega(n-r)-2)+\vert\Omega(n+r)-\Omega(n-r)\vert}{2}g_{min}(n)$ with $g_{min}(n)=2.3^{1_{3\not\mid n}}$ iteratively to get a finite sequence $(r'_{l}(n))$ with $r'_{0}(n):=r$ and $r'_{l'+1}(n):=r'_{l'}(n)+S_{n}(r'_{l'}(n))$ to get a primality radius $r_{0}(n)$, i.e get $(n-r_{0}(n),n+r_{0}(n))\in\mathbb{P}^{2}$.
As $\Omega(m)\leq\frac{\log m}{\log 2}$ for any positive integer $m$, one may be able to prove that this algorithm with initial value $r'_{0}(n):=0$ terminates after $O(\log^{1+o(1)} n)$ steps and thus that $r_{0}(n)=O(\log^{2+o(1)}n)$.
Edited on May 5th 2023 to try to resolve conflict of notations.
Edit May 7th 2023: in the special case $ab=1$ $\Delta_{1}=q-p$ is obviously a linear combination of prime gaps with positive integral coefficients. Now consider the case $\Delta_{2}=q^{2}-p=q^{2}-q+q-p=q(q-1)+q-p$. As $q-1=q-3+2$ and $2$ being both a prime gap and a jumping champion, assuming $\Delta_{1}=\sum_{i}c_{i}J_{i}(x)$ one gets $\Delta_{2}=\Delta_{1}+q\Delta'_{1}$ where $\Delta'_{1}:=\sum_{i}c'_{i}J_{i}(x)$ hence $\Delta_{2}=\sum_{i}c''_{i}J_{i}(x)$.
Edit January 10th 2024: the dual question, namely viewing jumping champions as linear combinations with non negative coefficients of prime gaps, may help reduce the known upper bound of $246$ of Polymath8b to $30=5\times 6$ replacing thrice $6$ by $2+4$ to get an admissible $k$-tuple with $k=9$ of diameter $30$, hence a prime constellation. In that case one would get $\lVert\Delta\rVert=3+3+2=8=k-1$. The fact that the $8$-simplex is convex and self-dual may be useful in the variational method used in the aforementioned Polymath project.