Reading Eight Erdős Problems Together

Bounded capacity along a subsequence

If $U_N$ does not tend to infinity, there are a constant $B>0$ and arbitrarily large $N$ with $U_N\le B$. Choose one integer $q>B$. Fix a cutoff $K$ after which the allowance and divisibility by $q$ hold, and fix the finite prefix, of value $y_0$. The set $V$ of resulting sums is compact by summability and the finite choices at each remaining position. For every such $N>K$,

Indeed the scaled prefix after $K$ is a multiple of $q$ and the remaining tail is at most $U_N/Q_N$. These grids leave holes inside every fixed open interval for all sufficiently large selected $N$, so $V$ is nowhere dense. For a bounded interval $J$ they also give $|V\cap J|\le (B/q)|J|+2B/Q_N$. Passing to the subsequence and covering $V$ by finitely many intervals with total length at most $|V|+\epsilon$ gives $|V|\le (B/q)(|V|+\epsilon )$, hence $|V|=0$. A countable union over the cutoffs and integer prefixes covers $E_Q(F)$.

One continuation interval for every cumulative residue

Suppose $U_N\to \mathrm{\infty }$ and put $h_N=\underset{k\ge N}{inf}{U}_k$. For sufficiently large $n$, let $M_n$ be the largest factorial at most $min\{n,\sqrt{h_{\lfloor n/2\rfloor }}\}$, and set the finitely many earlier moduli equal to one. These positive integers are nested under divisibility, tend to infinity, and are eventually divisible by each fixed integer. The delayed lower envelope controls the whole future cost of rounding, even when the denominator ratios are unbounded. Indeed, $Q_N/Q_k\le 2^{N-k}$, so for sufficiently large $N$,

The first bound uses $\lfloor k/2\rfloor \le N$ when $k\le 2N$; the second uses $M_k\le k$. Since $U_N\ge h_N\to \mathrm{\infty }$ and $M_N\le \sqrt{h_N}$, this proves $U_N-2C_N\ge M_N$ for every sufficiently large $N$. Start beyond this point and any prescribed cutoff.

Keep only the active positions $S=\{n:F_n\ge 2M_n\}$ after this cutoff. Define two common tail bounds

Both are nonnegative and tend to zero. Discarding an inactive position loses less than $2M_n$; reserving both margins at an active position loses exactly $2M_n$. Thus the preceding estimate gives the decisive overlap inequality

In particular the active set is infinite and each continuation interval has positive width.

Choose a target in $[{\alpha }_{N_0},{\beta }_{N_0}]$ and set all preceding digits to zero. Suppose the current unweighted cumulative sum is $C$ and the remaining target before position $n$ lies in $[{\alpha }_{n-1},{\beta }_{n-1}]$. At an inactive position choose zero. At an active position, the permitted digits in the residue class $e\equiv -C(\mathrm{mod}{M}_n)$, $0\le e\le F_n$, form a nonempty arithmetic progression of step $M_n$. Its least digit is less than $M_n$ and its greatest digit exceeds $F_n-M_n$. By (O), the intervals

for these digits overlap. Their union contains

Choose a digit whose interval contains the remaining target. The next remainder stays between ${\alpha }_n$ and ${\beta }_n$, and these bounds tend to zero, proving exact representation. This is precisely the common-interval feedback construction; no target-independent ordinary block sum is required.

After an active position $n$, the cumulative sum is divisible by $M_n$. Each later active digit is divisible by the previous active modulus, because both adjacent cumulative sums are; the moduli are nested. For a fixed $q$, take one active position with $q\mid M_n$. After that position, all individual digits and all preceding cumulative sums are divisible by $q$. This cutoff is common to all targets in the interval.

From one interval to every nonnegative target

The finite-exception convention in $E_Q(F)$ strengthens the conclusion. Given $y>0$, choose $N$ arbitrarily late with ${\alpha }_N<y$ and put

Then $p\ge 0$, $M_N\mid p$, and (O) gives $y-p/Q_N\in [{\alpha }_N,{\beta }_N]$. Set $e_N=p$ and all earlier digits to zero. Run the same continuation construction after $N$, with initial cumulative sum $C=p$; its interval covering worked for every $C$. All later digits obey their allowances. The zero target uses zero digits.

The congruence cutoffs can also be common to all $y$. For fixed $q$, choose an active position $j$ with $q\mid M_j$. If the target’s exceptional index $N$ precedes $j$, the repair at $j$ gives the required divisibilities after $j$. If $N\ge j$, all preceding digits are zero, $q\mid M_N\mid p$, and all subsequent repairs preserve divisibility by $q$. Thus $j+1$ is a valid cutoff for every target. The allowance-exception index itself may depend on the target. This completes Theorem 3.1.

The growth hypothesis has content. If repeated denominators are allowed, take $Q_n=2^{\lfloor \sqrt{n}\rfloor }$ and $F_n=1$. The allowance series converges, while $U_N\ge \lfloor \sqrt{N}\rfloor$ by counting the next complete denominator block. Yet eventual divisibility by two forces all sufficiently late digits to vanish, so the attainable set is countable. Imposing all cumulative congruences as well leaves only zero. Thus nestedness alone does not justify the criterion.