Uniformity required by prime patterns

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These author-recorded route comparisons retain source dependence explicitly. Inspected theorem statements are not independent verifications of their proofs.

Growing families and unproved hypotheses

The conditional convergence theorem for Problem 15 uses a quantitative Hardy--Littlewood hypothesis uniform over tuple sizes up to $(\log\log x)^5$ and shifts up to $(\log x)^2$, with a power-saving error. This is Conjecture 1.3 and Theorem 1.4 in Tao's paper; the existing reconstruction records the argument and its qualified review. Fixed-tuple asymptotics do not supply that growing uniform range. The relevant convergence is convergence of partial sums of the alternating series, not absolute summability. The formalization issue on the problem page does not remove this hypothesis.

Similarly, a cluster prime in Problem 17 requires all the specified even differences up to a bound growing with that prime, with both endpoints within the prime's range. The sparse upper bound in Theorem 1 of Elsholtz's paper is compatible with finitely or infinitely many cluster primes. It supplies no growing family of simultaneous representations. The discriminating question is uniformity over that entire growing set of differences, not existence for each fixed difference separately.

Theorem 3 of [[../library/primes/erdos_1949_applications_brun_s_method/_index|Erdős's Brun-sieve paper]] gives, for each fixed separation $c_2>0$, some positive coefficient $c_6(c_2)$ and a run of at least $\lfloor c_6(c_2)\log x\rfloor$ consecutive primes below $x$ whose adjacent gaps exceed $c_2$, for all sufficiently large $x$. The exact question in Problem 238 requires more than $c_1\log x$ consecutive primes for every prescribed $c_1,c_2>0$. Adjacent separation implies pairwise separation within the run, but the coefficient of the number of primes must become arbitrarily large for fixed $c_2$. The theorem only gives one positive coefficient depending on $c_2$. The discriminating test is whether a proposed result supplies every $c_1$, with the threshold allowed to depend on $c_1,c_2$. Stop at a single small coefficient; this is a count of consecutive primes, not an interval-length parameter.

A second moment controls almost all representations

Let $r(n)=\#\{(p,k):n=p+2^k,\ p\text{ prime},\ k\ge1\}$. Display (2.11) of Chen's paper imports the classical sieve estimate $\sum_{n\le X}r(n)^2\ll X$. Its statement and surrounding definition were inspected; the cited second-moment proof was not rechecked. For any fixed $\varepsilon>0$ it implies

$$\#\{X<n\le2X:r(n)>\varepsilon\log n\} \ll_\varepsilon\frac{X}{(\log X)^2}.$$

Allowing $k=0$ adds at most one representation and does not change the limiting issue in Problem 236. The estimate gives a density-zero exceptional set, not the asserted bound $r(n)=o(\log n)$ for every sufficiently large $n$. To see the logical gap, the nonnegative sequence $b(2^j)=j$, $b(n)=0$ otherwise, satisfies $\sum_{n\le X}b(n)^2=O((\log X)^3)=O(X)$ yet $b(2^j)/\log(2^j)=1/\log2$. This is not a representation-count counterexample. A route must rule out sparse spikes using arithmetic information beyond the second moment; stop if Markov's inequality is its only tail control.

Exceptional bases and admissible patterns

The held fourth version of [[../library/primes/ding_2025_two_romanoff_type_problems_erdos/_index|Ding's Romanoff-type paper]] proves positive lower density for almost every real base, as recorded for Problem 244. An almost-everywhere theorem may exclude every member of a countable specified family. It therefore does not settle an arbitrary fixed base. The source card records a golden-ratio result in a later version; that specific-base result is separate from the held version's theorem, and no proof from the later version is imported here. The full target remains every real $C>1$. A useful test names one previously uncovered base and checks its actual hypotheses; stop treating a measure-zero exception as empty.

For Problem 855, the inspected Theorem 1.1 of [[../library/primes/chahal_et_al_2025_second_hardy_littlewood_conjecture/_index|Chahal and collaborators]] assumes a positive nondecreasing $R$ with $|\pi(x)-\operatorname{li}(x)|\le CR(x)$ and $x^{1/2}\le R(x)\le x/(\log x)^3$ for large $x$. Its conclusion holds in the range

$$\frac{3CR(2x)(\log x)^2}{\log\log x}\le y\le x.$$

This is a lower restriction on the smaller variable $y$ in terms of the larger variable $x$. It leaves pairs where both grow but $y$ is below that threshold. The restriction cannot be dropped to obtain the eventual assertion for all pairs.

In the opposite direction, an admissible tuple of diameter $h$ with more than $\pi(h)$ members would yield a violation of the prime-count subadditivity inequality at interval length $h$ if it occurred as a tuple of primes. Admissibility is only the local congruence condition. An unproved prime-tuple conjecture is needed for that occurrence argument. Moreover, one fixed $h$ does not refute a statement restricted to both variables sufficiently large; the construction must allow unbounded $h$. The admissible-set extremal information in Problem 1204 supplies neither of these missing prime-occurrence assertions. The one-step test is to write the actual two parameter ranges and the occurrence hypothesis. Stop at a fixed diameter or an unproved tuple hypothesis rather than declare an unconditional disproof.

A short-interval prime theorem gives only its own exponent

Suppose a set $A\subseteq\mathbb N$ in Problem 428 has arbitrarily large integers $N$ for which every $a\in A$, $0<a<N$, has $N-a$ prime. Fix $0<\theta<1$ and put $x=N^\theta$. Injecting these $0<a\le x$ into primes of $[N-x,N)$ gives

$$A(x)\le\pi(N)-\pi(N-x)+O(1).$$

If a prime-number-theorem asymptotic is available uniformly at that length, then the right side is $(1+o(1))x/\log N$, whereas $\pi(x)\sim x/(\theta\log N)$. Along those $N$ this proves $\liminf A(x)/\pi(x)\le\theta$. A theorem valid only above a fixed positive exponent therefore supplies only a positive upper bound. To answer the existence question negatively by this argument, one needs zero instead of a positive upper bound. That requires the short-interval asymptotic for arbitrarily small fixed $\theta$, or an additional idea.

Local gaps and weighted short intervals

For the midpoint quantity in Problem 454, write $M_n=\min_{1\le i<n}(p_{n+i}+p_{n-i}-2p_n)$. The inner expression is

$$\sum_{j=0}^{i-1}(d_{n+j}-d_{n-j-1}).$$

Its first term being large is necessary, not sufficient, for $M_n$ to be large: every subsequent partial sum must remain large. The data and midpoint discussion on the [[../library/primes/mcnew_2018_convex_hull_prime_number_graph/midpoint_convex_primes_p13|McNew source page]] do not supply the arbitrary-height, all-$i$ conclusion. The discriminating test is control of the minimum of all these partial sums at the same center.

Finally, for Problem 950, set $A_n(t)=\#\{p<n:n-p\le t\}$ for integers $1\le t\le n-1$. Exact summation by parts gives

$$\sum_{p<n}\frac1{n-p} =\frac{A_n(n-1)}{n-1} +\sum_{t=1}^{n-2}\frac{A_n(t)}{t(t+1)}.$$

The largest weights lie on the shortest backward intervals. A global prime number theorem does not give the needed uniform control of these terms for every $n$. No primary proof of the requested bound is supplied here. The next test is a bound for the weighted short-interval sum itself; stop if only a macroscopic prime-count asymptotic is available.