Error term for Rosen's greedy B_2-type sequence: is R(x)=x+O(x^{1/4+o(1)})? (Erdős #954)

Define integers $0=a_0<a_1<a_2<\cdots$ greedily by $a_0=0$, $a_1=1$, and for $k\geq 1$ let $a_{k+1}$ be the smallest integer $n$ for which the number of solutions to $a_i+a_j\leq n$ (over indices $0\leq i\leq j\leq k$ with $j\geq 1$) is strictly less than $n$. The sequence begins $$0,1,3,5,9,13,17,24,31,38,45,\ldots.$$ For real $x$ let $R(x)=\#\{(i,j): 0\leq i\leq j,\ j\geq 1,\ a_i+a_j\leq x\}$ count the sumset representations up to $x$. Is it true that $$R(x)=x+O\!\left(x^{1/4+o(1)}\right)?$$

Acceptance. FULLY RESOLVES: a complete proof that $R(x)=x+O(x^{1/4+o(1)})$, OR a proof that it fails (an explicit lower bound $\lvert R(x)-x\rvert\geq x^{1/4+c}$ for a fixed $c>0$ and infinitely many $x$), written or machine-checkable. ADVANCES (each strictly beyond the background frontier, where only $R(x)\geq x$ is proved): (a) prove the currently-open upper bound $R(x)\leq(1+o(1))x$ that Erdős and Rosen could not; (b) prove any unconditional power-saving error $R(x)=x+O(x^{\theta})$ with an explicit constant $\theta<1$; (c) extend A390642 and the tabulated error term to a new record range of $x$ with a reproducible generator, reporting the maximal $\lvert R(x)-x\rvert$ observed and the fitted exponent. Deliver the proof, or the generation code plus the attained range and error statistics.