2.3 Why every \(T_n\) is in the live set \(A\)

(a) First rule out a prime plus one power. For any exponent \(e\geq0\), choose a covering class \(e\equiv a_i\pmod {m_i}\). Then

If this positive number equalled \(p_i\), reducing

\(T_n=p_i+2^e\) modulo 8191 would give

which the checked choice \(c=1\) excludes. Thus \(T_n-2^e\) is never prime.

(a) Now suppose \(a>b\geq1\), and put \(a-b=2^ru\) with \(u\) odd. If

\(T_n-2^a-2^b>1\), inequality (1) gives \(a<2^n\), hence \(r\leq n-1\). Since

the factor

divides both \(Q_n\) and \(2^a+2^b\). Therefore

\(B_r\mid T_n-2^a-2^b\).

This divisor is proper. If equality held, then

\(T_n=2^a+2^b+B_r\). For \(a\geq4\), a direct reduction modulo 16 using

\(B_0\equiv3\), \(B_1\equiv5\), and \(B_r\equiv1\) for \(r\geq2\)

never gives \(T_n\equiv-1\pmod {16}\). For \(a\leq3\), the three possible

pairs give a right side at most 15, whereas \(T_n>15\). Hence the prime

candidate has a nontrivial divisor.

(a) Equal exponents satisfy

\(2^a+2^a=2^{a+1}\), already excluded by the one-power argument. This includes \(a=0\), which is the live statement's \(1+1\) case.

(a) If exactly one exponent is zero and the other is \(b\geq1\), then

\(T_n-1-2^b\) is even. It could be prime only if it equalled 2, i.e.

\(T_n=3+2^b\). But \(3\mid Q_n\mid T_n\), while \(3\nmid3+2^b\), a contradiction.

Therefore \(T_n\in A\) for all \(n>10\).

(a) The family is strictly increasing: \(T_{n+1}\) is divisible by

\(F_n=2^{2^n}+1\), whereas \(T_n<2^{2^n}-1\). It supplies one exception per \(n\), hence only the known \(\asymp\log\log x\) scale; it does not address positive density.

2.4 A concrete 420-digit exception

For \(n=11\), the least CRT solution is

114692399417541627831390678344111455559634149945994060064370176174761757107094384161208131810508010278443673581829982252271454885733150869511794121524724926719189042703019347118118212960843353505774370088928506106578466372689421320193947115016412660586173360894622173451791309400443984363146771213896025262579711774536791930091548873856297462778888609711986475144538417671136309198210638114258170393070241496204972122655

(d) Its decimal SHA-256 is

67b68d67710b3e0f9c0d70e30a32968fc7f6406fecc6944ce284682f919bac5d.

The checker tests all 1,393 relevant one-power candidates and all 968,132 unequal positive-exponent pairs, producing a proper divisor in every case. It also checks all mixed-zero parity cases.

3. Exact computation through \(10^9\)

Define

(a) The compressed indexing used by the checker is exact. Write an odd prime as \(p=2i+1\) and an odd target as \(n=2j+1\). For \(a,b\geq1\),

Thus each power-pair sum is a Boolean shift of an odd-prime sieve. The case \(a=b=0\) is shift 1. If exactly one exponent is zero, parity forces \(p=2\), so the only such targets are \(3+2^b\), which are marked separately.

(d) A from-scratch odd Eratosthenes sieve and these exact shifts gave:

| exponent cap \(K\) | \(U_K(10^9)\) |

|---:|---:|

| 0 | 449,152,467 |

| 4 | 141,873,646 |

| 8 | 8,212,948 |

| 12 | 225,829 |

| 16 | 10,572 |

| 20 | 495 |

| 23 | 69 |

| 24 | 38 |

| 25 | 13 |

| 26 | 8 |

| 27 | 3 |

| 28 | 2 |

At cap 27 the exact survivors are

At cap 28,

and the checker independently verifies \(19586773\) prime by trial division. It also directly checks that 288026325 has no representation with both exponents at most 27.

Therefore:

(d) Every odd integer \(5\leq n\leq10^9\) is \(p+2^a+2^b\).

(d) More sharply, every such \(n\) has a representation with

\(\max(a,b)\leq28\), and 28 is the least uniform exponent cap for this interval.

The exact array audit values are:

• odd primes through \(10^9\): 50,847,533 (excluding the prime 2);
• odd-prime Boolean-array SHA-256:

74d176c4598da9365337436398bedd09312587810e7355be4038cebb449d804f;

• final represented-array SHA-256:

3c8e4dc51e9093f0a102844ee4f19e738486fa2086ffb479e06cca484c5a089d.

(d) As an indexing/parity cross-check, a separate ordinary enumeration of primes and exponent pairs agrees with the vectorized array on every odd target through 100,000.

4. A local reduction and a four-prime obstruction theorem

This addresses precisely the covering-system issue mentioned on the live page.

4.1 The exact local obstruction

Let \(\mathcal P\) be a finite set of odd primes and

\(Q=\prod_{p\in\mathcal P}p\).

(a) If some odd residue \(r\bmod Q\) satisfies

then all but \(O_{\mathcal P}((\log x)^2)\) integers

\(n\leq x\) in the odd class \(r\bmod 2Q\) are in \(A\). Indeed, in a representation the prime \(n-2^a-2^b\) would have to equal one of the finitely many primes in \(\mathcal P\); there are only \(O((\log x)^2)\) corresponding power pairs. The zero-exponent cases add only \(O(\log x)\) possibilities.

Thus (3) would prove positive lower density, not merely positive upper density. This is the exact finite local lemma a covering-system attack would need.

4.2 A union bound for exponent pairs

For an odd prime \(p\), put \(d_p=\operatorname{ord}_p(2)\), and define

(a) For each fixed \(a\), at most one \(b\bmod d_p\) solves the equation, so

The exact exceptional small-order values are

and every other odd prime has \(\rho(p)\leq1/12\).

(d) The checker proves the last classification from scratch by factoring \(2^d-1\) for \(2\leq d\leq11\), checking exact orders, and enumerating each small subgroup. The remaining primes have \(d_p\geq12\), when (4) applies.

4.3 At most four primes can never give (3)

(a) If four distinct primes do not contain all of \(3,5,7\), the sum of their four largest possible bad-pair proportions is at most

The union bound therefore leaves an exponent pair avoiding every prime.

It remains to handle \(\{3,5,7,p\}\).

(d) For every residue \(r\bmod105\), exact enumeration of the \(12^2\) ordered exponent pairs modulo 12 leaves at least 28 pairs for which

is divisible by none of \(3,5,7\). The minimum 28 first occurs at \(r=45\bmod105\).

(a) If \(d_p\nmid12\), fix one of those pairs \((a_0,b_0)\) and vary

\(a=a_0+12t\). The element \(2^{12}\bmod p\) then has order

\(d_p/\gcd(d_p,12)>1\). For fixed \(b_0\), the unwanted equation modulo \(p\) holds for at most one value in this cycle, so another \(t\) avoids \(p\) while preserving avoidance of \(3,5,7\).

(a) If \(d_p\mid12\), then

\(p\mid2^{12}-1=3^2\cdot5\cdot7\cdot13\). Since \(p\) is distinct from \(3,5,7\), it is \(13\). A residue modulo 13 excludes at most 12 of the 144 exponent pairs, fewer than the 28 already left.

(d) A direct joint cross-check over all \(r\bmod1365\) actually leaves at least 25 pairs avoiding \(3,5,7,13\).

Combining these cases gives:

(d) For every integer \(r\) and every set \(\mathcal P\) of at most four odd primes, there are positive exponents \(a,b\) such that

Residue representatives equal to zero can be increased by a common exponent period, so the exponents may indeed be taken positive. The infinite reasoning is elementary; the label is (d) because it depends on the exact 105-residue finite lemma.

(b) By Dirichlet's theorem, every infinite arithmetic progression of odd integers whose modulus has at most four distinct odd prime divisors contains infinitely many integers \(p+2^a+2^b\). Choose \(a,b\) by (5); the residue for \(p\) is coprime to the modulus, and Dirichlet supplies infinitely many primes in it.

Consequently, any fixed-modulus local obstruction of form (3) needs at least five distinct odd primes. This does not assert that an obstruction with five primes exists.

5. What remains and why the standard machinery stalls

(b) The Ding–Sun–Zhao construction chooses one residue class

\(\beta\bmod W\) with

It proves that a positive fraction inside that class is exceptional, yielding about \(x/W=x^{1-o(1)}\) exceptions.

(a) Since \(W\to\infty\), one class has ambient density \(1/W\to0\). To reach positive density by this route, one needs a genuinely new uniform lemma of one of the following forms:

1. keep \(W=O(1)\), which is exactly the fixed local obstruction Erdős doubted and which Section 4 rules out when its odd radical has at most four primes; or

2. construct \(\gg W\) compatible residue classes modulo the growing \(W\), so their total proportion is bounded below; or

3. replace the CRT thinning by a lower-tail theorem proving that the representation count is zero for a positive proportion of \(n\).

No cited sieve theorem supplies any of these. The current proofs control primes after first thinning to one very sparse CRT class.

(a) The finite interval computation cannot bridge this uniformity gap. It proves a sharp statement for \(n\leq10^9\), but upper density is a limsup at unbounded scales; no finite cutoff decides it.

(c) A dense extension to \(10^{10}\) with the present algorithm is estimated from the measured \(10^9\) run to require roughly 15 GiB of working memory and 12–18 minutes of memory-bound work (about \(0.2\)–\(0.3\) core-hours). It was not run because it cannot settle the density question and exceeds the useful few-minute scope.

6. Reproduction

The standalone checker is runs/verify_erdos9_wave5d.py. Its SHA-256 is

32b9ffe5f4c2be218e88ff9c1b7ba40a4d4ab767d57781d8feb5f8c934f2a2ab.

Run all mathematical checks, including the \(10^9\) sieve:

python3 runs/verify_erdos9_wave5d.py

If the three downloaded PDFs are present under the filenames shown in the checker:

python3 runs/verify_erdos9_wave5d.py --source-dir /path/to/source-pdfs

The final run on this VM used about 1.5 GiB peak resident memory and ended with:

CRT_WITNESS_OK ... one_power_candidates_checked=1393 unequal_pairs_checked=968132
FOUR_PRIME_LOCAL_LEMMA_OK ... min_good_mod_3_5_7=28 min_good_mod_3_5_7_13=25
RANGE_CHECK_OK ... limit=1000000000 minimal_uniform_exponent_cap=28 exceptions=[1, 3]
ALL_CHECKS_PASSED ... seconds=73.82

PARTIAL: verified an explicit 420-digit member and infinite CRT family in \(A\), proved computationally that \(A\cap[1,10^9]=\{1,3\}\) with sharp exponent cap 28, and ruled out fixed local covering obstructions using at most four odd primes; positive upper density remains open.