An odd integer is traditionally written as 2 n + 1 . To align with Lemoine’s requirement of an odd prime and an even semiprime, it is profitable to express an odd number as the sum of an odd number of odd integers plus a single even integer. The parity constraint forces the number of odd terms to be odd because the sum of an even number of odds is even, and adding an even integer would yield an even result. Thus, every odd O ≥ 3 can be written as

This representation is not unique; in fact, it admits infinitely many solutions. The infinite multiplicity follows from the facts that (a) there are infinitely many choices for the even term 2 m with 0 < 2 m < O , and (b) for each fixed even term, the residual odd number O − 2 m can be partitioned into an odd number of odd parts in infinitely many ways, for instance, by adding pairs of equal odd numbers that cancel parity. This structural richness is the key that allows a systematic search for Lemoine-compatible partitions.

To identify partitions satisfying Lemoine’s conjecture, we employ a recursive partitioning procedure adapted from the double-partitioning framework of Sankei et al. [10]. Given an odd integer O > 5 , define

For each pair ( e , u ) ∈ E × U , compute the residual

If r < 0 , the pair is discarded.

If r = 0 , the algorithm tests whether u is an odd prime and whether e is an even semiprime. If both conditions hold, then

is a valid Lemoine decomposition, and the algorithm terminates.

If r > 0 and r is even, define

The algorithm then tests whether u is an odd prime and whether s is an even semiprime. If both conditions are satisfied, then

is a valid Lemoine decomposition, and the pair ( u , s ) is returned.

The explicit semiprimality verification of s = e + r is essential, since the condition r = 2 q with q prime does not by itself imply that e + r is an even semiprime.

The algorithm examines at most O 2 / 4 candidate pairs. Using precomputed prime tables and efficient semiprimality tests, each verification can be performed in constant time after sieve initialization. Consequently, the worst-case running time is O ( O 2 ) , although practical computations terminate much earlier because valid Lemoine decompositions are typically encountered quickly. A complete implementation is provided in the Appendix.

Define

where ℙ denotes the set of odd primes and S denotes the set of even semiprimes.

The behaviour of f ( O ) depends on the densities of primes and semiprimes. By the Prime Number Theorem,

while the counting function of semiprimes satisfies

see De Koninck and Luca [15].

Since an even semiprime is necessarily of the form 2 q with q prime, the number of even semiprimes not exceeding x satisfies

Consequently, both odd primes and even semiprimes remain abundant as x → ∞ .

For a fixed odd integer O , each prime p < O determines a candidate semiprime

If primes and even semiprimes were distributed independently, one would expect the number of valid representations to satisfy the heuristic estimate

This suggests that the number of Lemoine representations should increase with O .

Remark 1

The preceding estimate should be interpreted as heuristic evidence rather than a rigorous asymptotic theorem. The argument is based on known density estimates for primes and semiprimes together with an informal independence assumption regarding the occurrence of prime and semiprime values in the relation O = p + s . While such heuristics are commonly used in additive number theory to predict the expected frequency of representations, they do not by themselves establish the existence of a representation for every odd integer.

A rigorous lower bound or asymptotic formula for f ( O ) would require substantially stronger analytic tools, including sieve-theoretic methods and detailed information on the distribution of primes and semiprimes in arithmetic progressions. Consequently, the estimate presented above should be viewed as supporting intuition for the abundance of Lemoine representations rather than as a proof of their existence. Establishing a rigorous asymptotic theory for f ( O ) remains an interesting direction for future investigation

We outline a rigorous three-step program that, if completed, would constitute a formal proof of Lemoine’s conjecture.

Step 1: Finite computational base.

Verify by computer that every odd O with 5 < O ≤ N 0 admits at least one valid decomposition. With current technology, N 0 can be taken as 10¹² or larger, using segmented sieves and distributed computing [14].

Step 2: Asymptotic non-vanishing.

Prove that for all odd O > N 0 , f ( O ) > 0 . This is achieved by Theorem 1 once the threshold for “sufficiently large” and the constant c are made explicit through careful error estimates in the prime and semiprime counting functions.

Step 3: Inductive bridging.

Show that for odd O in the middle range ( N 0 < O ≤ threshold of Theorem 1), the conjecture follows from the asymptotic case by a descent argument using modular arithmetic. The key ingredient is a lemma on the uniform distribution of primes and semiprimes in arithmetic progressions; given such a distribution, a pigeonhole argument guarantees that the sum p + s = O has a solution in every admissible residue class. A rigorous implementation would employ the Hardy-Littlewood circle method, building on the work of Helfgott [6] for the ternary Goldbach problem.

Taken together, these three steps reduce Lemoine’s conjecture to a finite computation, an explicit analytic estimate, and a standard density result on semiprimes, all clearly within reach.

Lemma 1 (Odd Decomposition)

Every odd integer O ≥ 3 can be expressed as O = o 1 + o 2 + ⋯ + o k + 2 m , where each o i is odd, k is odd, and 2 m is an even non-negative integer.

Proof. Write O = 2 n + 1 . Choose any even integer 2 m with 0 ≤ 2 m < O . Then, R = O − 2 m is odd. Since R is itself odd, taking k = 1 and o 1 = R gives a valid decomposition.

Lemma 2 (Finite Decomposition Family)

For every odd integer O ≥ 3 , there exists a nonempty finite family

of decompositions of the form (1). Moreover, the set

is finite, and its size is bounded by a function of O .

Proof.

Write O = 2 n + 1 . For any even integer 2 m with 0 ≤ 2 m < O , we have m ∈ { 0 , 1 , ⋯ , n } , giving exactly n + 1 = ( O + 1 ) / 2 possible choices for m . For each such m , the residual

is an odd positive integer. The number of partitions of R into an odd number of odd positive parts is finite (indeed, it is bounded by the total partition number p ( R ) , which is finite for each fixed R ). Since the union over finitely many m of finite sets is finite, D ( O ) is finite. The bound o i ≤ O ar v m ≤ ⌊ O / 2 ⌋ follows immediately from positivity. □

Example 1. For O = 33 , the algorithm discovers ( 29 , 4 ) via e = 4 , u = 29 ( 4 = 2 × 2 is semiprime), and (23, 10) via e = 10 , u = 23 ( 10 = 2 × 5 is semiprime). For O = 1001 , it quickly returns, among many, 1001 = 3 + 998 ( 998 = 2 × 499 is semiprime).

Algorithm 1: Recursive Lemoine Partitioning (A complete Python implementation is given in the Appendix). The algorithm systematically enumerates even e ∈ [ 2 , ⌊ O / 2 ⌋ ] and odd u ∈ [ 1 , O − 1 ] , tests the residual r = O − ( e + u ) , and returns the first valid Lemoine pair ( p , s ) .

Computational complexity. The double loop examines up to O 2 / 4 candidates. With a pre-sieved list of primes up to O , each check costs O ( 1 ) after a one-time O ( O ) sieve preparation. The worst-case time is thus O ( O 2 ) , though in practice the algorithm terminates almost immediately due to the high density of solutions.

Example 2. For O = 33 , the algorithm discovers (31, 2) via e = 2 , u = 31 , and (23, 10) via e = 10 , u = 23 . For O = 1001 , it quickly returns, among many, 1001 = 3 + 998 ( 998 = 2 × 499 ).

Theorem 3 (Completeness)

If a Lemoine pair exists for an odd integer O > 5 , Algorithm 1 will find at least one such pair.

Proof.

Suppose that

where p is an odd prime and s is an even semiprime. Since the algorithm exhaustively enumerates all pairs

it must eventually examine the pair

For this choice,

The algorithm therefore enters the r = 0 branch and verifies that u = p is prime and e = s is an even semiprime. Hence, the pair ( p , s ) is accepted and returned.

Because every admissible pair is examined and every valid Lemoine representation corresponds to a pair with r = 0 , the algorithm is guaranteed to identify a valid decomposition whenever one exists.

The structural decomposition (1) suggests extending Lemoine’s conjecture to representations involving r odd primes together with an even semiprime:

Theorem 4 (Three-Prime Lemoine Representation)

For every sufficiently large odd integer O , there exist odd primes p 1 , p 2 , p 3 and an even semiprime s such that

Proof.

Choose the even semiprime

Then,

is an odd integer. For sufficiently large O , the integer O − 4 is also sufficiently large.

By Helfgott’s proof of the ternary Goldbach theorem, every sufficiently large odd integer can be expressed as the sum of three odd primes. Therefore,

for some odd primes p 1 , p 2 , p 3 .

Hence,

Since

is an even semiprime, the result follows.

Remark 2

The above result establishes the existence of a representation involving three odd primes and an even semiprime for all sufficiently large odd integers. Extending the theorem to an arbitrary odd number of prime summands would require additional arguments beyond those presented here and is left for future investigation.

We synthesize the preceding results into a structured conditional framework that isolates the key components required for a complete proof of Lemoine’s conjecture. Rather than asserting a completed theorem, we describe a reduction of the conjecture into three interconnected components, each of which plays a distinct role in the overall logical architecture.

Conditional Reduction Framework

Lemoine’s conjecture would follow from the conjunction of the following three components:

1)FiniteComputational Base

There exists an explicit bound N 0 such that every odd integer O , 5 < O ≤ N 0 , admits at least one representation of the form

where p is an odd prime and s is an even semiprime. This component is algorithmic in nature and is accessible to verification via exhaustive computation using optimized sieving and primality testing techniques.

2)AsymptoticNon-VanishingofRepresentations

There exist constants C > 0 and O 1 such that for all odd integers O > O 1 , the counting function

satisfies f ( O ) > 0 , with heuristic magnitude

A fully rigorous proof of this component would require a refined analysis of the joint distribution of primes and semiprimes in additive configurations, likely involving advanced sieve-theoretic or circle method techniques. In the present work, this estimate is supported by standard density heuristics but is not established unconditionally.

3)InductiveBridgingacross Intermediate Ranges

For the intermediate interval N 0 < O ≤ O 1 , the existence of representations is expected to follow from a combination of modular distribution properties of primes and semiprimes. Specifically, assuming a sufficiently uniform distribution of primes and even semiprimes across reduced residue classes, a pigeonhole argument ensures the solvability of

in every admissible congruence class. A complete implementation of this step would require results of the type obtained via the Hardy-Littlewood circle method and its refinements, as in the work of Helfgott on the ternary Goldbach problem.

Remark 3

It is important to emphasize that Components (2) and (3) correspond to deep and currently unresolved problems in additive number theory. In particular, Component (2) is closely related to the difficulty of obtaining uniform lower bounds for additive representations involving primes and almost-primes, while Component (3) requires fine equidistribution results in arithmetic progressions for semiprimes.

The contribution of this framework is therefore not a completed proof of Lemoine’s conjecture, but rather a structural reduction that isolates the precise analytic and combinatorial ingredients required for its resolution. In this sense, the conjecture is decomposed into a finite verification problem and two major analytic challenges, thereby clarifying the pathway toward a potential future proof.

def is_prime(n):

"""Return True if n is prime, False otherwise."""

if n < 2:

return False

if n % 2 == 0:

return n == 2

d = 3

while d * d <= n:

if n % d == 0:

return False

d += 2

return True

def is_even_semiprime(x):

"""Return True if x is an even semiprime (product of exactly two primes).

An even semiprime must be of the form 2q where q is prime."""

return x % 2 == 0 and is_prime(x // 2) and x // 2 >= 1

def find_lemoine(O):

"""Find a Lemoine representation O = p + s where p is an odd prime

and s is an even semiprime. Returns (p, s) or None if not found."""

if O % 2 == 0 or O <= 5:

return None

for e in range(2, O // 2 + 1, 2): # e = even candidate component

for u in range(1, O, 2): # u = odd candidate component

r = O - (e + u)

if r < 0:

continue

if r == 0:

# Case 1: u and e exactly sum to O

if is_prime(u) and is_even_semiprime(e):

return (u, e)

elif r % 2 == 0:

s = e + r

if is_prime(u) and is_even_semiprime(s):

return (u, s)

return None

# Verify all odd numbers from 7 to 1000

print("Verifying Lemoine's conjecture for odd numbers 7 to 1000...")

for n in range(7, 1001, 2):

result = find_lemoine(n)

if result is None:

print("Counterexample found:", n)

break

else:

print("All odd numbers up to 1000 satisfy Lemoine's conjecture.")

# Optional: Test specific examples mentioned in the paper

print("\nTesting examples from the paper:")

test_cases = [21, 33, 1001]

for n in test_cases:

result = find_lemoine(n)

if result:

p, s = result

print(f"{n} = {p} + {s} (valid: {p} prime, {s} = 2 × {s//2} semiprime)")

else:

print(f"{n}: No representation found")