Thus \(\sum_pa_pB_p<\infty\) almost surely and \(X>0\).

For any finite set of primes, CRT says that the divisibility indicators of a

uniform integer have limiting joint probabilities

The expected logarithmic contribution from primes \(>y\) is at most

\(\sum_{p>y}1/(p(p-1))\to0\), so Markov's inequality makes the truncation

uniform in the limiting-density calculation. Hence the limiting law is the

law of \(X\). Schoenberg's continuity removes the distinction between \(<\)

and \(\le\). \(\square\)

4. Two lower-bound lemmas

Let

(b) Lemma 2 (elementary endpoint lower bound, modulo Mertens). There are

constants \(c>0\) and \(\eta_0>0\) such that

Proof. Let \(y=\lceil2/\eta\rceil\). Require \(B_p=0\) for every \(p\le y\).

This event has probability

For the independent tail

we have

Markov gives \(\Pr(Z_y\ge\eta)\le1/(y\eta)\le1/2\). On the remaining

event, \(X=e^{-Z_y}>e^{-\eta}>1-\eta\). Independence therefore gives

Mertens's product theorem,

\(M(y)\sim e^{-\gamma}/\log y\), proves the claim. \(\square\)

For a finite prime set \(A\), write

(a) Lemma 3 (cylinder spike). If

\(0<\eta\le\min_{p\in A}1/p\), then

Proof. Let \(Y_A=\prod_{p\notin A}q_p^{B_p}\). If

\(X>1-\eta\), then every \(B_p\), \(p\in A\), must be zero, since otherwise

\(X\le q_p\le1-\eta\). Hence

because the probability that all \(B_p=0\) on \(A\) is \(t_A\).

On the other hand, the event that all \(B_p=1\) on \(A\) and

\(Y_A>1-\eta\) lies inside

\((t_A(1-\eta),t_A]\), and its probability is

This is (1). \(\square\)

(b) Known comparison. Combining (1) with Erdős's sharp endpoint

asymptotic already shows the infinite left derivative at every \(t_A\in E\).

Toulmonde's 2006 theorem gives the matching asymptotic, not merely this lower

bound.

5. An explicit irrational obstruction

Define

(b) Proposition 4. The number \(x_0\) is positive and irrational, and

Proof.

1. Bertrand's postulate gives

\[ Q_k^2

Also \(p_{j+1}>p_j^2\), so \(\sum1/p_j<\infty\) and \(x_0>0\).

2. Put

\[ \eta_k=1-\frac{x_0}{t_k} =1-\prod_{j>k}\left(1-\frac1{p_j}\right). \]

The first omitted factor and the union bound give

\[ \frac1{p_{k+1}}\le\eta_k \le\sum_{j>k}\frac1{p_j}<\frac2{p_{k+1}}. \]

With (2),

\[ \frac1{2Q_k^2}<\eta_k<\frac2{Q_k^2}. \tag{3} \]

3. The rational \(t_k\) has denominator dividing \(Q_k\), while

\(0

\(t_k\ne x_0\) would imply

\[ |t_k-x_0|\ge\frac1{bQ_k}, \]

contradicting the preceding upper bound for large \(k\). Thus \(x_0\)

is irrational.

4. Let \(P_k=\prod_{i\le k}(p_i-1)\). Then \(P_k

\(K_k=1/P_k>1/Q_k\). The upper bound in (3) is at most

\(\min_{i\le k}1/p_i\), so Lemmas 2 and 3 apply:

\[ \frac{f(t_k)-f(x_0)}{t_k-x_0} \ge \frac{cK_k}{t_k\eta_k\log(1/\eta_k)}. \]

From (3), \(t_k\le1\), and \(K_k>1/Q_k\),

\[ \frac{f(t_k)-f(x_0)}{t_k-x_0} \ge \frac{cQ_k}{2\log(2Q_k^2)} \longrightarrow+\infty. \]

Since \(t_k-x_0\downarrow0\), this is the asserted upper right Dini

derivative. \(\square\)

6. Dense continuum version

**(b) Lemma 5 (modulo Mertens, or equivalently Euler's divergence of

prime reciprocals).** The finite products

are dense in \((0,1)\).

Proof. The numbers \(a_p=-\log(1-1/p)\) tend to zero and have divergent

sum because Mertens's product theorem implies

\(\prod_{p\le y}(1-1/p)\to0\). Given \(L>0\) and an error \(\delta>0\),

start beyond a prime for which \(a_p<\delta\), and add consecutive \(a_p\)'s

until their sum first exceeds \(L\). The overshoot is \(<\delta\). Exponentiate

with a minus sign. \(\square\)

(b) Proposition 6. The points in Proposition 4 may be chosen to form a

dense continuum \(\mathcal S\) of irrationals.

Proof. Start with any finite seed set \(A\), its product \(t_0\), and a

large integer \(M\) exceeding all seed primes. For each infinite bit string

\(\omega=(\omega_1,\omega_2,\ldots)\), choose:

• \(p_1\) as the least prime \(>M\) if \(\omega_1=0\), or the least prime

\(>4M\) if \(\omega_1=1\);

• after \(p_1,\ldots,p_k\), put

\(Q_k=(\prod_{q\in A}q)(\prod_{i\le k}p_i)\), and choose \(p_{k+1}\) as

the least prime \(>Q_k^2\) for bit 0, or the least prime \(>4Q_k^2\)

for bit 1.

Bertrand places the two choices respectively in

The proof of Proposition 4, with

works unchanged and makes every limit irrational with infinite upper right

Dini derivative.

The bit-string map is injective. At its first split after a common \(Q\), the

bit-0 tail deficit is \(>1/(2Q^2)\), whereas the bit-1 tail deficit is

\(<2/(4Q^2)=1/(2Q^2)\). At the initial split, the identical argument uses

\(M\) in place of \(Q^2\): the two deficits lie respectively above and below

\(1/(2M)\). Hence there are continuum many limits.

Finally, given an open interval \(I\subset(0,1)\), use Lemma 5 to choose

\(t_0\in E\cap I\), and then choose \(M\) so large that the total future

decrement \(<2/M\) keeps every resulting limit inside \(I\). Thus the union

of these continuum families is dense. \(\square\)

7. Independent checker and exact output

Standalone verifier:

runs/erdos50_wave7b_verify.py.

Run command:

python runs/erdos50_wave7b_verify.py

Key lines from the observed terminal output (the executable also prints the

two exact rational enclosure endpoints):

explicit all-zero path
 k | p_k        | Q_k            | t_k=prod(1-1/p)       | K(t_k)
 1 | 3          | 3              | 2/3                    | 1/2
 2 | 11         | 33             | 20/33                  | 1/20
 3 | 1091       | 36003          | 21800/36003            | 1/21800
 4 | 1296216011 | 46667665044033 | 28257509018000/46667665044033 | 1/28257509018000

exact four-stage enclosure for x_0
lower decimal = 0.6055050963303562347352374259555010341886043447949594679254273662051248
upper decimal = 0.6055050963303562347352374263725402715451782744597739002740422247180453
width = 4.170392373565739296648144323486148585129204514754755901586056880084122E-28

finite Euler product: 1024 distinct subset products; exact probabilities sum to 1

first two binary branch levels
bits | previous Q | chosen prime
   0 |          3 | 11
   1 |          3 | 37
  00 |         33 | 1091
  01 |         33 | 4357
  10 |        111 | 12323
  11 |        111 | 49297

cost estimate: pi(10000) = 1229 ; balanced tables have 2^614 and 2^615 states; smaller table > 10^174 core-hours at 10^7 states/core-second

VERIFIED: all exact arithmetic and finite structural checks passed.

(d) What the script certifies. It uses standard-library trial division,

not SymPy or a prime table; exactly checks the four displayed least primes,

the rational identity

all \(2^{10}=1024\) distinct products and probabilities for the first ten

primes, the exact enclosure, and the first binary splits. It does not pretend

to computationally prove Mertens or the infinite limiting step.

The compact mathematical core is below; the standalone file additionally

formats and asserts every displayed rational value and cost estimate.

#!/usr/bin/env python3
from decimal import Decimal, getcontext
from fractions import Fraction
from itertools import product
from math import isqrt

def is_prime(n):
    if n < 2:
        return False
    if n % 2 == 0:
        return n == 2
    d = 3
    while d <= isqrt(n):
        if n % d == 0:
            return False
        d += 2
    return True

def next_prime(n):
    q = n + 1
    if q <= 2:
        return 2
    if q % 2 == 0:
        q += 1
    while not is_prime(q):
        q += 2
    return q

def dec(x, digits=70):
    getcontext().prec = digits
    return str(Decimal(x.numerator) / Decimal(x.denominator))

def explicit_path():
    ps = [3]
    while len(ps) < 4:
        Q = 1
        for p in ps:
            Q *= p
        ps.append(next_prime(Q * Q))
    assert ps == [3, 11, 1091, 1296216011]

    Q = P = 1
    t = Fraction(1)
    for p in ps:
        Q *= p
        P *= p - 1
        t *= Fraction(p - 1, p)
        assert t == Fraction(P, Q)
        assert Fraction(1, Q) / t == Fraction(1, P)

    lo = t * (1 - Fraction(2, Q * Q))
    hi = t * (1 - Fraction(1, 2 * Q * Q))
    assert Q == 46667665044033
    assert P == 28257509018000
    assert hi - lo < Fraction(1, 10**27)
    print(ps)
    print(dec(lo))
    print(dec(hi))
    return lo, hi

def finite_model():
    ps = [2, 3, 5, 7, 11, 13, 17, 19, 23, 29]
    law = {}
    for bits in product((0, 1), repeat=len(ps)):
        x = w = Fraction(1)
        for p, bit in zip(ps, bits):
            if bit:
                x *= Fraction(p - 1, p)
                w *= Fraction(1, p)
            else:
                w *= Fraction(p - 1, p)
        assert x not in law
        law[x] = w
    assert len(law) == 1024
    assert sum(law.values(), Fraction(0)) == 1

def branch(Q, bit):
    base = Q * Q * (1 if bit == 0 else 4)
    p = next_prime(base)
    assert base < p < 2 * base
    return p

def binary_prefixes():
    level = [("", 3)]
    for _ in range(2):
        nxt = []
        for word, Q in level:
            A = Q * Q
            p0, p1 = branch(Q, 0), branch(Q, 1)
            assert Fraction(1, p0) > Fraction(1, 2 * A)
            assert Fraction(2, p1) < Fraction(1, 2 * A)
            nxt += [(word + "0", Q * p0), (word + "1", Q * p1)]
        level = nxt

if __name__ == "__main__":
    explicit_path()
    finite_model()
    binary_prefixes()
    print("VERIFIED")

8. Exact remaining wall

(c) Logical wall. The cylinder inequality (1) turns an approximation

\(t_A\downarrow x\) into an infinite right Dini derivative when

The construction forces (4) by making each next prime much larger than the

product of all previous primes. For an arbitrary candidate \(x\), no theorem

found in the audited literature supplies appropriately sided approximants

\(t_A\) with the needed simultaneous control of distance and

\(K_A=\prod_{p\in A}(p-1)^{-1}\). This weighted subset-product approximation

is the exact missing arithmetic lemma for extending this particular method.

It cannot hold for every \(x\), because pure singularity gives derivative zero

almost everywhere; a full solution would have to derive a contradiction only

under the hypothesis \(0

(c) Analytic wall. Conditioning on any finite prime set \(A\) decomposes

the law into positive scaled copies of the tail law. A finite positive

derivative would require these copies to have mutually compatible local

linear increments at all relevant scaled points. The known global bound

\(\sup_t(f(t)-f((1-\varepsilon)t))\ll1/\log(1/\varepsilon)\) is far too large

after division by \(\varepsilon\), and the available right-increment estimates

do not provide the cancellation or uniform pointwise upper bound needed to

rule out that compatibility.

(d/c) Computation wall and cost. A straightforward certified truncation at

primes \(p\le y\) has \(2^{\pi(y)}\) subset-product cylinders (meet-in-the-middle

still needs about \(2^{\pi(y)/2}\)). The elementary tail certificate used

above is

To make this error at most \(h\), it requires \(y\ge h^{-2}\). Already at

\(h=10^{-2}\), \(y=10^4\) and \(\pi(y)=1229\), so a balanced

meet-in-the-middle split has tables of \(2^{614}\) and \(2^{615}\) states; the

smaller alone has \(2^{614}\approx6.8\times10^{184}\) states. Even a

generous throughput of \(10^7\) certified states per core-second would require

about \(1.9\times10^{174}\) core-hours for that smaller table, before accounting

for memory or sorting. I did not run such a computation; progress must use an

analytic or structural lemma rather than finer brute force.

PARTIAL: Proved, modulo Schoenberg/Mertens/Bertrand, a dense continuum-sized family of irrational points generated by an explicit recursion with infinite upper-right Dini derivative; the possibility of a finite positive derivative at all remaining exceptional points stays open.