# Complete cyclic-frame estimates

Let `m(k)` and `R(k)` have the meanings in `solution_draft.md`. This note gives all quantitative details omitted from the formal frame/CRT lemmas F-034--F-040.

I use these standard PNT consequences (all follow by partial summation): if `p_j` is the j-th prime, then

\[
p_j=(1+o(1))j\log j,
\quad \sum_{i\le j}(p_i-1)=(1/2+o(1))j^2\log j,
\quad \sum_{p\le y}\frac{\log p}{p}=\log y+O(1).
\tag{A}
\]

In particular, for all sufficiently large j one may use the explicit coarse inequalities `p_j≤2j log j` and
`(1/8)j²log j≤sum_{i≤j}(p_i-1)≤j²log j`.

## 1. Exact coordinate weights

List the odd primes as `r_1=3,r_2=5,...` and put `a_j=(r_j-1)/2`. The sequence `(a_j)` is complete (every nonnegative integer is a sum of distinct terms).

Proof. A standard induction says a nondecreasing positive sequence is complete if `a_1=1` and

\[
a_{j+1}\le1+\sum_{i\le j}a_i.\tag{1}
\]

The first three cases are direct. Set `S_j=sum_{i≤j}a_i`; `S_3=6=r_3-1`. If `S_j≥r_j-1`, Bertrand gives `r_{j+1}<2r_j`, hence `a_{j+1}≤r_j-1≤S_j`; it also gives `S_{j+1}≥2a_{j+1}=r_{j+1}-1`. This proves (1) and the induction. ∎

Fix large k and N=k−1. Start with one binary digit if N is odd and two binary digits if N is even; let b∈{1,2} be their total weight. Thus N−b is nonnegative and even, and the first odd layer has at least two distinct 2-adic prefix tags.

Choose

\[
h=\left\lceil A(k\log k)^{1/4}\right\rceil
\]

where A is a sufficiently large absolute constant. Force the first h odd-prime coordinates. Their total half-weight `S_h=O(h²log h)=O(sqrt(k) log(k)^{3/2})=o(k)`. Put

\[
T=(N-b)/2-S_h\ge0.
\]

By completeness, choose a finite set J with `T=sum_{j∈J}a_j`. Use one coordinate of size r_j for each j∈J. If j≤h, this is the second base-r_j digit; otherwise it is the first. Together with the forced block and binary prefix, the coordinate weight is exactly

\[
b+2S_h+2T=N=k-1.\tag{2}
\]

No odd prime has exponent above two.

To bound coordinate size, choose the complete-sequence representation inside the first t terms, where t is minimal with S_t≥T. Since S_{t-1}<T≤k and (A),
`t=O(sqrt(k/log k))`; hence every coordinate size is at most

\[
x:=r_t=O(\sqrt{k\log k}).\tag{3}
\]

Order the binary prefix first, then forced/represented first digits in increasing prime order, then any second digits after their first digit. This is an arithmetic digit order.

## 2. Enough small distinct supports

Let y=r_h. By (A),

\[
y=O(k^{1/4}(\log k)^{5/4}).\tag{4}
\]

The h forced first-digit coordinates have `choose(h,2)` unordered pairs. From (3), choose A once and for all so large that

\[
\binom h2-h>x\tag{5}
\]

for every sufficiently large k. Every pair has support-product at most y².

For each completion coordinate u of size q≤x, its q−1 non-axis values receive q−1 distinct supports: the mandatory same-prime prefix through u, united with distinct seed pairs. If u is a second digit of a seed prime, discard pairs containing its first digit; (5) still leaves more than q−1 choices. Reserve one support for the chain condition in Section 3 and use pairs for the other q−2.

For forced seed layer i, only earlier coordinates may be tags. For all sufficiently large i,

\[
1+(i-1)+\binom{i-1}{2}\ge r_i-1
\]

because `r_i=O(i log i)`; hence subsets of at most two predecessors suffice. The finitely many smaller i are checked directly (the initial one/two binary digits supply 2-adic prefix choices). Their contribution is absorbed in the constant.

Thus every layer has exactly q−1 distinct arithmetic supports, all using only earlier coordinates besides their owner.

## 3. Cover, distinctness, and strong irreducibility

For each coordinate after the first, designate one owned support that also contains the entire arithmetic prefix of the immediately preceding coordinate. This is possible by the reserved choice above. Let E be the same-prime prefix through the last coordinate. If the final chain support would equal E, add one unused seed tag to it.

All supports are distinct:

1. within a layer their tags differ;
2. supports from different layers differ at the later owner coordinate, which cannot occur in an earlier support;
3. E differs from the final owned support by construction.

Arithmetic initial-prefix supports have different prime-exponent vectors, so their support products are distinct integer moduli.

Fix every non-owner support coordinate to zero and assign the q−1 nonzero owner values bijectively to its q−1 supports. Use E, fixed to zero, as anchor. Every nonzero digit vector is covered at its first nonzero coordinate; the zero vector lies in E. Thus the k hyperplanes cover.

For a nonempty missing-coordinate set L, either L meets E, or its largest coordinate i is not last. In the latter case i+1 lies outside L and its designated owned support meets L. F-037 proves exactly this boundary condition in Lean; F-034/F-035 prove that no proper support subfamily covers with arbitrary new values. F-040 proves that arithmetic prime-power boundary frames transport to prompt-level strong irreducible modulus sets. Thus (2) constructs a size-k irreducible modulus set.

## 4. Largest modulus

The largest seed prime is y and every completion first digit is at most x.

- A seed-layer support has at most owner plus two predecessor factors, so is `O(y³)` (finitely many initial exceptions are constant).
- A first-digit completion support contributes at most x from its owner and y² from its pair tag, hence `O(xy²)=O(k(log k)^3)` by (3),(4).
- A second seed digit contributes at most y² from its mandatory prefix and y² from its tag, hence `O(y^4)=O(k(log k)^5)`.
- A chain support is the union of two prime-power prefixes. It is at most `O(x²)`, `O(xy²)`, or `O(y^4)`, according as zero, one, or two prefixes are repeated seed primes.
- E is one prefix and is smaller.

Therefore a constant C exists such that, for all sufficiently large k,

\[
m(k)\le Ck(\log k)^5.\tag{6}
\]

The finitely many smaller k≥5 enlarge C. Since k distinct integers in `[2,m(k)]` require `m(k)≥k+1` (F-031),

\[
\boxed{k+1\le m(k)\le O(k(\log k)^5),\qquad m(k)=k^{1+o(1)}.}
\]

## 5. Reciprocal lower bound

For this separate extremal construction, force a longer initial odd-prime block whose total weight is at most `k/log k` and maximal with that property. This is still o(k), so Section 1 completes the remaining weight exactly; the tag pool is even larger than needed.

At forced prime p, include among its p−1 tags every squarefree integer `d≤p−1`, represented by its set of prime factors. All those factors are earlier forced primes. There are at most p−1 such d, and the remaining support slots are filled as above. Consequently the reciprocal sum is at least

\[
\sum_{p\le z}\frac1p\sum_{d\le p-1}\frac{\mu(d)^2}{d},\tag{7}
\]

where z is the largest forced prime. The standard squarefree harmonic estimate

\[
\sum_{d\le y}\frac{\mu(d)^2}{d}=\frac6{\pi^2}\log y+O(1)\tag{8}
\]

follows, for example, from `mu²(n)=sum_{a²|n}mu(a)` and harmonic summation. Equations (A),(7),(8) give

\[
(7)=\left(\frac6{\pi^2}+o(1)\right)\log z.
\]

The maximality and (A) for a block of weight `k/log k` imply
`z²/(2log z)=(1+o(1))k/log k`, hence `log z=(1/2+o(1))log k`. Thus

\[
R(k)\ge\left(\frac3{\pi^2}+o(1)\right)\log k.\tag{9}
\]

F-030 proves for every admissible k-set

\[
R(k)\le\sum_{j=2}^{k+1}\frac1j=(1+o(1))\log k.
\]

Therefore

\[
\boxed{R(k)=\Theta(\log k).}
\]
