The claimed new quantum factoring record is $n=a(a+b)$ with $a=1048589=2^{20}+13$ and $b=12$. It could well be that the form $n=a(a+b)$ with $0\le b\le12$ (or other small upper bound for $b$) defines all the sizable numbers the new record-claiming method can factor (with the indicated $3$ qubits on the hardware at hand and the ∼30% success rate, based on the picture in that answer). Such $n$ are scarce: their proportion near $n$ thins out as $6.5/\sqrt n+\mathcal o(1/\sqrt n)$ ^{[*]}. The method lacks generality.

Further, the factor $a$ of such $n$ is effortlessly found using Fermat's factorization method (exposed in a fragment of letter likely to Mersenne circa 1643). If $n=a(a+b)$ with $0\le b\le2\sqrt a$ (a much larger class of integers), only the first step of Fermat's factoring is required (for all except very small $n$): compute $r=\bigl\lceil\sqrt{4n}\,\bigr\rceil$, then $a=(r-\sqrt{r^2-4n})/2$. That's enough to factor **1099551473989** by hand.

The picture and that other answer refer to Eric R. Anschuetz, Jonathan P. Olson, Alán Aspuru-Guzik, Yudong Cao's *Variational Quantum Factoring* (arXiv:808.08927, Aug 2018). That reduces factorization to a combinatorial problem with number of unknowns proportional to the bit length of the factors. I find nothing suggesting the preprocessing makes that sublinear in the general case, and conclude that Variational Quantum Factoring is exponential in the bit length of $n$, thus essentially pointless since we have sub-exponential algorithms for factorization on classical computers.

Similar stunts of stretching to artificially large numbers what an experimental setup allows have already been pulled. Take the record of **143**=11×13 by Nanyang Xu, Jing Zhu, Dawei Lu, Xianyi Zhou, Xinhua Peng, and Jiangfeng Du in *Quantum Factorization of 143 on a Dipolar-Coupling Nuclear Magnetic Resonance System* (Physical Review Letters, 2012). Their experimental technique factors an integer product of two odd exactly-4-bit integers (thus with two unknown bits per factor). The scarcity of primes in range [8,15] makes 143 the only product of two distinct primes that the technique can factor. Their experimental setup iteratively minimizes a function with a 2-bit input. This achievement (I'm *not* joking up to this point) has been stretched, without new experiment AFAIK, to:

**56153**=233×241 by Nikesh S. Dattani and Nathaniel Bryans, *Quantum factorization of 56153 with only 4 qubits* (arXiv:1411.6758, 2014)
**4088459**=2017×2027, by Avinash Dash, Deepankar Sarmah, Bikash K. Behera, and Prasanta K. Panigrahi, *Exact search algorithm to factorize large biprimes and a triprime on IBM quantum computer* (arXiv:1805.10478, May 2018)
**383123885216472214589586724601136274484797633168671371**=618970019642690137449562081×618970019642690137449562091 by myself (crypto.SE, June 2018), using the technique of the above paper.

^{[*]} Proof: for each $a$, there are $13$ values of $b$ leading to $13$ values of $n$ of the form $n=a(a+b)$ with $0\le b\le12$. For large enough $a$ there are no two $(a,b)$ leading to the same $n$^{[#]}. When we increase $n$ by $1$, $n^2$ increases by $2n+1$. Taking the inverse, the density of squares near large $n$ is $\displaystyle\frac1{2\sqrt n}+\mathcal o(\sqrt n)$. Thus the density of $n$ of the form $a(a+b)$ with $0\le b\le12$ is $\displaystyle\frac{13}{2\sqrt n}+\mathcal o(1/\sqrt n)$.

^{[#]} Auxiliary proof: Assume $a(a+b)=a'(a'+b')$ with $0\le b\le12$ and $0\le b'\le12$. Let $c=2a-b$ and $c'=2a'-b'$. It comes $(c-b)(c+b)/4=(c'-b')(c'+b')/4$, thus $c^2-b^2=c'^2-b'^2$, thus $c^2-c'^2=b^2-b'^2$, thus $|(c-c')(c+c')|\le144$, thus for $c\ge72$ and $c'\ge72$ the only solution is $c=c'$, hence $b=b'$. Thus for $a\ge42$ and $a'\ge42$ the only solution is $a=a'$ and $b=b'$. The bound on $a$ can be further lowered.

1Could you add a reference for the scarcity of $n=(a-b)(a+b)$ with $|b|\leq 6$ – kelalaka – 2019-12-15T15:40:05.490

3@kelelaka: I found no reference. Therefore I attempted a derivation, found a mistake in my statement, fixed that, and while at it allowed for about twice as many $n$. – fgrieu – 2019-12-15T17:49:26.913

1@kelalaka: the scarcity of $n = (a-b)(a+b)$, $|b| \le 6$ is easy to derive; $n$ is of this form iff one of $n, n+1, n+4, n+9, n+16, n+25, n+36$ is a perfect square;the probability of a random value circa $n$ being a perfect square is circa $1/(2\sqrt{n})$, and hence the probability that one of the above will be a square (assuming $n$ large) is $7/(2\sqrt{n})$ – poncho – 2019-12-17T21:44:23.537

1@poncho: yes. But I moved to the form $a(a+b)$ with $0\le b\le12$ and that simple and nice reasoning became the current mess at the bottom of the question :-( – fgrieu – 2019-12-17T21:46:40.470

1I would like to point out that the value of $a$ is 100000000000000001101 in binary, and b=100000000000000011001. Therefore the number of qubits needed to factor this using the method used for the previous record of 291311, might turn out to be 2, not 3. However such pre-processing has not been done successfully yet for this combination of a(a+b), as far as I know. – user1271772 – 2020-05-10T18:22:43.670