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Math-583 · mathematics · Number theory

Lemoine's Conjecture

Status Open · unsolvedmath.comLLM status Open unverified

Statement

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

An even semiprime is a product of exactly two primes counting multiplicity, one of which is 2; it therefore has the form $2q$ with $q$ prime. The equation is $n=p+2q$ for primes $p,q$, where $n>5$ is odd. Consequently $p$ is odd; the prime $q$ is allowed to equal 2. [S1] [S2]

Statement recorded in the supplied source.
Can every odd integer greater than 5 be expressed as the sum of an odd prime and an even semiprime? [S1]
Residual task reported by the archive.

The embedded review (2026-08-17) records: Prove that every odd n>5 has n=p+2q with p,q prime, or find a counterexample. [S1]

Short English statement

Can every odd integer greater than five be obtained by adding an odd prime to twice a prime?

Sources

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Statement licence: CC-BY-4.0 · Source file

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Math-583-A1 · Exploration · LLM-generated · gpt 6 astra ultra ·

gpt 6 astra ultra attempt on Math-583

Lemoine's Conjecture

Definitions and mathematical statement

Shared notation.
Unless a section says otherwise, ${\mathbb{N}}=\{1,2,\ldots\}$, ${\mathbb{Z}}$ is the ring of integers, ${\mathbb{Q}},{\mathbb{R}},{\mathbb{C}}$ are the rational, real, and complex fields, $[N]=\{1,\ldots,\lfloor N\rfloor\}$, and $\log$ is the natural logarithm. For real functions of a parameter tending to infinity, $f\ll g$ and $f=O(g)$ mean $|f|\leq Cg$ eventually for some fixed $C$; $f\gg g$ reverses this comparison; $f=o(g)$ means $f/g\to0$; $f\sim g$ means $f/g\to1$; and $f\asymp g$ means both order bounds. A subscript on $O$ or $\ll$ specifies allowed constant dependence. The expression $n^{o(1)}$ in an upper bound means growth smaller than $n^\epsilon$ for each fixed $\epsilon>0$. Local definitions govern any exception, finite-field convention, or use of the same symbol for another object. An asymptotic claim is not automatically an effective algorithm.

An even semiprime is a product of exactly two primes counting multiplicity, one of which is 2; it therefore has the form $2q$ with $q$ prime. The equation is $n=p+2q$ for primes $p,q$, where $n>5$ is odd. Consequently $p$ is odd; the prime $q$ is allowed to equal 2. [S1] [S2]

Statement recorded in the supplied source.
Can every odd integer greater than 5 be expressed as the sum of an odd prime and an even semiprime? [S1]
Residual task reported by the archive.

The embedded review $2026-08-17$ records: Prove that every odd n>5 has n=p+2q with p,q prime, or find a counterexample. [S1]

Short English statement

Can every odd integer greater than five be obtained by adding an odd prime to twice a prime?

Research attempt

Scope and route.
The statement includes $n=7$ and allows $q=2$, so even semiprime four must not be omitted. Juhász's primary computational paper [S2] was inspected on 1 October 2026. Its finite-verification setting motivates an exact counting function, but does not replace control of all odd integers. The attempt separates local admissibility from the analytic estimate required for a positive representation count.
An exact finite representation count.
For an odd integer $N>5$, define \[ r(N)=\sum_{\substack{q\le(N-3)/2\\q\ { m prime}}} 1_{N-2q\ { m prime}}. \] The term $p=N-2q$ is then an odd prime at least three. This counts ordered roles $(p,q)$ rather than unordered pairs, since the coefficient two distinguishes their roles. The target is exactly $r(N)>0$. For $N=7$, the pair $(p,q)=(3,2)$ gives the unique representation; omitting $q=2$ would incorrectly manufacture a counterexample.

For a fixed cutoff $X\ge N$, put $S_X(\alpha)=\sum_{p\le X}(\log p)e^{2\pi i p\alpha}$. Orthogonality of the exponentials gives the exact weighted identity \[ R(N)=\int_0^1S_X(\alpha)S_X(2\alpha)e^{-2\pi iN\alpha}\,d\alpha =\sum_{p+2q=N}\log p\log q. \] All weights are positive, so $R(N)>0$ is equivalent to $r(N)>0$. Replacing either factor by its absolute square changes the equation being counted and cannot prove positivity of this coefficient.

Local congruences do not forbid a representation.
For an odd prime $\ell$, choose a nonzero residue $q$ modulo $\ell$ and require $p=N-2q$ also nonzero. If $\ell\mid N$, all $\ell-1$ nonzero choices work. Otherwise exactly one of them, $q=N/2$, is excluded, leaving $\ell-2$ choices. Both counts are positive. At two, for odd $N$ the odd-prime model allows $p=q=1$ modulo two, so it too is admissible. The exceptional small prime choice $q=2$ is handled by the exact count rather than the all-units local model.

Relative to the independent unit densities these counts produce the formal local product \[ \mathfrak S(N)=2\prod_{\ell>2}\left(1-\frac1{(\ell-1)^2}\right) \prod_{\substack{\ell\mid N\\\ell>2}} \frac{\ell-1}{\ell-2}>0. \] The infinite product is positive because the sum of its deficits converges. The archimedean length for $p+2q=N$ is $N/2$, suggesting a weighted main term $\mathfrak S(N)N/2$. This is a local prediction, not a derived asymptotic for $R(N)$.

A precise sufficient estimate and the missing cancellation.
Partition $[0,1]$ into a chosen major-arc set $\mathfrak M$ and its complement $\mathfrak m$. If one could prove, uniformly for all large odd $N$, that the major-arc integral equals $\mathfrak S(N)N/2+o(N)$ and the absolute value of the minor-arc integral is $o(N)$, positivity would follow from the uniform positive lower bound on $\mathfrak S(N)$. An explicit threshold plus a finite check below it would settle the exact target. Neither estimate is proved in this notebook.

The elementary absolute-value bound does not suffice. Cauchy--Schwarz and orthogonality give \[ \int_0^1|S_X(\alpha)S_X(2\alpha)|\,d\alpha \le\sum_{p\le X}(\log p)^2\le X\log^2X. \] The equality of the two quadratic norms follows by the change of variable $\beta=2\alpha$ and periodicity. The resulting bound is larger than the predicted order-$N$ main term when $X\asymp N$. It loses the oscillatory cancellation depending on the specified $N$. An average estimate allowing a sparse exceptional set would also not prove positivity for every odd integer.

Executed finite verification.
An integer sieve is used to evaluate $r(N)$ for every odd $7\le N\le9999$; all counts are positive. Selected exact values are \[ \begin{array}{c|rrrrrr} N&7&9&11&99&999&9999\\\hline r(N)&1&2&2&9&39&235 \end{array} \] The script checks primality of the actual summands, not merely congruence admissibility. Files: checks_second.py and results_second.json under _Local/Erdos_problems/UnsolvedMath/attempt_audit_2026_10_01/number_theory/.
Remaining gap and outcome.
The Fourier identity and local counts are exact; their combination does not by itself yield a positive Fourier coefficient. A pointwise analytic estimate or another uniform prime-producing argument is missing. Outcome: UNRESOLVED. The finite range has been verified, with no assertion about an unbounded tail.

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