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library/integer_sequences/chojecki_2026_truncated_congruence_sieves_erdos_problem_25


Przemyslaw Chojecki, Truncated Congruence Sieves and Erdős Problem 25, unpublished preprint, 19 March 2026. Source: https://www.ulam.ai/research/erdos25.pdf. Local artifacts: the retained PDF, held on the source card, and its canonical conversion.

The note treats the singleton-residue truncated sieve in Erdős problem 25. For 1≤n1<n2<⋯1\leq n_1<n_2<\cdots and one chosen class ai(modni)a_i\pmod {n_i} per modulus, put

Bi={n∈N:n≥ni, n≡ai(modni)},A=N∖⋃iBi,B_i=\{n\in\mathbb N:n\geq n_i,\ n\equiv a_i\pmod {n_i}\},\qquad A=\mathbb N\setminus\bigcup_iB_i,

and let A(k)=N∖⋃i≤kBiA^{(k)}=\mathbb N\setminus\bigcup_{i\leq k}B_i. The paper proves two positive cases and gives a conditional local-to-global reduction for the general case. It does not claim a solution: it is a GPT-assisted, unrefereed preprint, and its arguments remain unreviewed here.

Read status: claims checked. The full paper in Markdown was read end to end, and the hypotheses and conclusions listed below were checked clause by clause. Their proofs have not been independently verified.

Finite truncations and positive cases

  • Lemma 2.1 (p. 3). For every fixed k≥1k\geq1, A(k)A^{(k)} is eventually periodic, with period dividing Lk=lcm⁡(n1,…,nk)L_k=\operatorname{lcm}(n_1,\ldots,n_k). Consequently its natural and logarithmic densities exist and agree. Writing their common value as δk\delta_k, the inclusions A(k+1)⊆A(k)A^{(k+1)}\subseteq A^{(k)} give a decreasing limit δ=lim⁡k→∞δk∈[0,1]\delta=\lim_{k\to\infty}\delta_k\in[0,1].
  • Theorem 3.1 (pp. 3--4). If ∑i1/ni<∞\sum_i1/n_i<\infty, then AA has natural density, hence logarithmic density, and its value is δ\delta. The tail union bound d‾(A(k)∖A)≤∑i>k1/ni\overline d(A^{(k)}\setminus A)\leq\sum_{i>k}1/n_i squeezes the upper and lower densities.
  • Theorem 3.2 (p. 4). If the nin_i are pairwise coprime, then AA has natural density in all cases. Its finite-truncation densities are ∏i≤k(1−1/ni)\prod_{i\leq k}(1-1/n_i); convergence of ∑i1/ni\sum_i1/n_i invokes Theorem 3.1, while divergence sends this product to zero as k→∞k\to\infty, and therefore d(A)=0d(A)=0.

First kills and quotient sieves

  • Proposition 4.1 (pp. 4--5). The first-kill sets Ei=A(i−1)∩BiE_i=A^{(i-1)}\cap B_i are pairwise disjoint and N∖A=⨆i≥1Ei\mathbb N\setminus A=\bigsqcup_{i\geq1}E_i. Each EiE_i is eventually periodic; if ei=ld⁡(Ei)e_i=\operatorname{ld}(E_i), then ei=δi−1−δie_i=\delta_{i-1}-\delta_i and ∑iei=1−δ\sum_i e_i=1-\delta.
  • Proposition 4.2 (pp. 5--6). For j<ij<i, set gij=(ni,nj)g_{ij}=(n_i,n_j). Incompatible classes ai≢aj(modgij)a_i\not\equiv a_j\pmod {g_{ij}} impose no condition. A compatible class induces one forbidden quotient residue bij(modqij)b_{ij}\pmod {q_{ij}}, where qij=nj/gijq_{ij}=n_j/g_{ij} and ai+nit≡aj(modnj)a_i+n_it\equiv a_j\pmod {n_j} exactly when t≡bij(modqij)t\equiv b_{ij}\pmod {q_{ij}}. Thus the finite periodic quotient sieve $S_i={t\in\mathbb N:t\not\equiv b_{ij}\pmod {q_{ij}} \text{ for every compatible }j<i}$ satisfies Ei={ai+nit:t∈Si}E_i=\{a_i+n_it:t\in S_i\}. Its density did_i exists and obeys di=nieid_i=n_i e_i.

This representation does the essential localization: the infinite complement is partitioned by the first congruence that kills each integer, while the interaction with all earlier congruences becomes a finite sieve on the quotient variable tt.

Conditional reduction

  • Conjecture 5.1 (p. 6). With αi=ai/ni\alpha_i=a_i/n_i, there should be nonnegative charges τi\tau_i such that, for every ii and every Y≥1Y\geq1, with an absolute implied constant, $\sum_{\substack{t\leq Y\t\in S_i}}\frac1{t+\alpha_i} =d_i\log Y+O\left(d_i\log\frac2{d_i}+\tau_i\right)$, where the entropy term is zero when di=0d_i=0, and such that $\sum_{n_i\leq X}\tau_i/n_i=o(\log X)$.
  • Theorem 5.4 (pp. 7--9). Assuming Conjecture 5.1, the logarithmic density of AA exists and equals δ=lim⁡kδk\delta=\lim_k\delta_k. The proof sums the first-kill harmonic masses. Lemma 5.2 and Corollary 5.3 make the total entropy error only O(log⁡log⁡X)O(\log\log X), so the genuinely missing input is the sublogarithmic global charge.

Obstructions and the remaining gap

  • Proposition 6.1 (p. 10). There is no universal bound μX(⋃i>kBi)≤Φk+o(1)\mu_X(\bigcup_{i>k}B_i)\leq\Phi_k+o(1) with Φk→0\Phi_k\to0. Taking the nin_i to be the primes and ai=0a_i=0 makes every tail union have logarithmic density one. Any viable estimate must instead control the conditioned tail A(k)∩⋃i>kBiA^{(k)}\cap\bigcup_{i>k}B_i.
  • Proposition 6.3 (pp. 10--11). For (u,6)=1(u,6)=1, r≥1r\geq1, and mj=u2r−j3jm_j=u2^{r-j}3^j, the stated choice of residues ar=0a_r=0 and aj≡u3r(1+2r−j−1)(modmj)a_j\equiv u3^r(1+2^{r-j-1})\pmod {m_j} for j<rj<r leaves, at the top modulus mr=u3rm_r=u3^r, exactly Emr={u3r(1+2rt):t≥0}E_{m_r}=\{u3^r(1+2^rt):t\geq0\} and Smr={t∈N:t≡1(mod2r)}S_{m_r}=\{t\in\mathbb N:t\equiv1\pmod {2^r}\}. Hence emr=1/(u6r)e_{m_r}=1/(u6^r), but its first survivor already contributes 1/(u3r)1/(u3^r), far larger than the entropy scale emrlog⁡(1/emr)≍r/(u6r)e_{m_r}\log(1/e_{m_r})\asymp r/(u6^r).
  • Conjecture 7.1 (p. 12). Every first-kill quotient sieve should split as Si=Sitr⊔SitwS_i=S_i^{\mathrm{tr}}\sqcup S_i^{\mathrm{tw}} into a transverse part and a prime-power-tower part, with a nonnegative τi\tau_i satisfying exactly the uniform harmonic estimate and global o(log⁡X)o(\log X) charge condition of Conjecture 5.1. Theorem 5.4 would then answer problem 25 positively.

The exact remaining singleton-residue gap is therefore to prove that the early survivors created by infinitely many such tower compressions cannot synchronize strongly enough to make the global logarithmic mass oscillate. Quantitatively, one must obtain Conjecture 5.1's uniform estimate for every finite quotient sieve while charging all non-entropy tower spikes by τi\tau_i with ∑ni≤Xτi/ni=o(log⁡X)\sum_{n_i\leq X}\tau_i/n_i=o(\log X). Proposition 6.3 shows why the charge cannot simply be omitted; the paper supplies neither this charging theorem nor an unconditional replacement.

Bears on

  • E0025: supplies two positive cases and a conditional reduction, but leaves the full singleton-residue problem open.