Density, tails and inverse thresholds

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The elementary comparisons here are author-recorded. The linked problem pages own the standing of imported bounds; no independent source-proof audit or new acceptance is claimed.

Multiples and delayed congruences

For Problem 25, normalize $0\le a_i<n_i$ and exclude only $B_i=\{a_i+n_it:t\ge1\}$. Let $A_K=\mathbb N\setminus\bigcup_{i\le K}B_i$, $A=\bigcap_KA_K$, and $\delta_K=d(A_K)\downarrow\delta$. Writing $H_E(X)=\sum_{n\le X,n\in E}1/n$, the exact residual for the proposed value $\delta$ is

$$\lim_{K\to\infty}\limsup_{X\to\infty} \frac{H_{A_K\setminus A}(X)}{\log X}=0.\tag{1}$$

This follows from $H_A=H_{A_K}-H_{A_K\setminus A}$ and the logarithmic density $\delta_K$ of each fixed truncation. The converse follows from the same identity. The existing reduction and proof also explains why universal existence reduces to this candidate-value question. The ambient union $\bigcup_{i>K}B_i$ is the wrong residual: with zero classes for all primes it has density one for every $K$. Conditioning on survival of the prefix is essential. Summable reciprocal moduli and pairwise coprime moduli are already covered by the existing reduction; they do not prove (1) for arbitrary translated classes.

For Problem 691, put $M_D=\bigcup_{d\in D}d\mathbb N$ and $\beta_N=d(M_{D\cap[1,N]})$. The classical multiples theorem identifies the logarithmic density with $\beta=\sup_N\beta_N$; the logarithmic-density proof is also given in the linked reduction. Consequently

$$d(M_D)=1\quad\Longleftrightarrow\quad\sup_N\beta_N=1.\tag{2}$$

For the forward direction natural density one implies logarithmic density one. For the reverse direction $\underline d(M_D)\ge\beta_N$ for every $N$. Each $\beta_N$ can be computed by finite inclusion-exclusion, using $d(d_1\mathbb N\cap\cdots\cap d_r\mathbb N)=1/\operatorname{lcm}(d_1,\ldots,d_r)$. This is a known limiting criterion, not a new structural characterization of the generators. In particular, divergence of $\sum_{d\in D}1/d$ does not suffice: the even generators have divergent reciprocal sum and their multiples have density $1/2$. A proposed improvement must give a new effective or structural condition rather than restate (2).

Covering functions and exact inversion

Let $P(x)=\prod_{p\le x}p$, and let $j(n)$ be the least length such that every interval of that many consecutive integers contains an integer coprime to $n$. The maximal covered run has length $j(n)-1$. A choice of residues $a_p$ covering $1,\ldots,y$ is, by the Chinese remainder theorem, equivalent to an interval $N+1,\ldots,N+y$ with every member sharing a prime factor with $P(x)$: choose $N\equiv-a_p\pmod p$. Conversely any such interval gives these residues. Thus the function of Problem 687 satisfies

$$Y(x)=j(P(x))-1\le h(\pi(x))-1,$$

where $h(r)$ is the maximum Jacobsthal function over integers with at most $r$ distinct prime factors, as in Problem 970. For fixed $k,x$, successful starting points $N$ form a periodic set modulo $P(x)$. Existence is therefore equivalent to positive natural density. This proves the exact inverse description from Problem 929:

$$S(k)=\min\{x:Y(x)\ge k\}.\tag{3}$$

The already recorded bound $Y(x)\ll x^2/(\log x)^2$ gives $S(k)\gg\sqrt{k}\log k$. To see the logarithm without assuming an inverse asymptotic, first $k\ll S(k)^2$ implies $\log S(k)\ge\tfrac12\log k-O(1)$; then $\sqrt{k}\ll S(k)/\log S(k)$. Its exponent remains $1/2$.

The stronger questions $Y(x)\le x^{1+o(1)}$ and $S(k)\ge k^{1-o(1)}$ are equivalent in the usual quantified-exponent sense. If $Y(x)\le x^{1+\varepsilon}$ eventually, (3) gives $S(k)\ge k^{1/(1+\varepsilon)}$ eventually. Conversely, putting $k=Y(x)$ in $S(k)\ge k^{1-\varepsilon}$ gives $Y(x)\le x^{1/(1-\varepsilon)}$ once $k$ is large. Constants in either bound can be absorbed into a slightly larger exponent. No quadratic bound for $h$ yields these stronger exponent-one conclusions. The next useful test is an improvement of the exponent in the covering bound, not another inversion of the existing estimate.

Fixed sieve density does not locate a first hit

For fixed $k$, the integers $n$ for which no $n+i$, $1\le i\le k$, has a prime divisor in $(k,2k)$ have the exact positive density

$$\prod_{k<p<2k}\left(1-\frac{k}{p}\right).$$

Each prime excludes $k$ distinct residues and the Chinese remainder theorem multiplies their surviving proportions. This addresses the periodic density in Problem 451, not its least positive successful $n$. A periodic set of positive density can concentrate its residues in part of a huge period. A first-hit bound requires control of incomplete periods uniform in $k$.

Likewise the density of a fixed sifted set in Problem 688 does not guarantee a survivor in the particular interval $[1,n]$ when the prime cutoff and excluded residues depend on $n$. The elementary covering count supplies a necessary condition, not a sufficient construction or uniform survivor theorem. For both problems the discriminating test is the size and uniformity of the error in the proposed incomplete-period estimate. Stop using a fixed-parameter density formula as a first-hit argument if that error dominates the main term.

Moment and prime-product comparisons

Write $s_i$ for the squarefree integers and $\Delta_i=s_{i+1}-s_i$. If the moment assertion in Problem 145 held for every $\alpha>0$ with finite limiting constants, its nonnegative summands would give $\Delta_i^\alpha\ll_\alpha s_i$. For any $\varepsilon>0$, choose $\alpha>1/\varepsilon$ to obtain $\Delta_i=o(s_i^\varepsilon)$. This would answer the subpower part of Problem 208, but not a logarithmic gap estimate. Moments in a fixed bounded range of $\alpha$ give only fixed powers. A claimed distributional limit must also control the tails of these unbounded powers before this implication is usable.

For Problem 1181, let $k=\lfloor\log n\rfloor$ and let $q$ be the smallest prime dividing none of $n+1,\ldots,n+k$. Every prime below $q$ divides their product, hence

$$\vartheta(q^-)\le\log\prod_{i=1}^k(n+i)\le k\log(n+k).$$

The prime number theorem gives only $q\le(1+o(1))(\log n)^2$. Obtaining a fixed coefficient smaller than one requires additional uniform information beyond this product comparison. The Jacobsthal covering problem does not supply it: there one asks for a small prime in each integer, whereas here every small prime must occur somewhere in the block.

An upper bound $q=o((\log n)^2)$ for all sufficiently large $n$ would suffice, for example by taking the required fixed saving to be $c=1/2$. This differs from improving the coefficient-one bound by only a subleading amount: $q\le(\log n)^2-r(n)$ with $r(n)=o((\log n)^2)$ need not give any fixed positive fractional saving.

For the sums of two squares in Problem 222, an elementary rounding argument gives only a quarter-power gap. If $a=\lfloor\sqrt X\rfloor$ and $b=\lceil\sqrt{X-a^2}\rceil$, then $0\le a^2+b^2-X\le2\sqrt{X-a^2}+1=O(X^{1/4})$. This produces a sum of two squares just above any $X$, but does not close the gap with the much smaller lower-bound scales on the problem page.

Finally, normalize an admissible $k$-tuple by $a_1=0$. Every initial $j$-tuple remains admissible, so the diameter function in Problem 1204 gives $a_j\ge A(j)$. Since that problem defines $B(k)$ as a mean, not a sum,

$$B(k)\ge\frac1k\sum_{j\le k}A(j).$$

Thus $A(j)\sim j\log j$ would imply $B(k)\ge(1/2+o(1))k\log k$, matching the upper asymptotic bound recorded on the problem page. This is a conditional transfer between the two questions, not a proof of the diameter asymptotic. The hard step is its leading constant; fixed-size admissibility tests cannot determine it.

A negative answer has a different sufficient interface: for some fixed $\eta>0$, admissible $k$-tuples with diameter at most $(1-\eta)k\log k$ for unbounded $k$ would refute $A(k)\sim k\log k$. Finite tables and second-order refinements do not supply that family. The problem page's Siegel-zero obstruction is conditional; refuting its hypothesis merely disables that route and proves neither asymptotic. An unconditional counterexample family must remain a possible disproof route.