Research Notes

Why exponent-sieve factors cannot be multiplied

Yagel Bar · 2026-10-20 · Proved finite result

Question and answer

Can the fractions of exponents surviving individual prime divisors be multiplied to obtain the fraction surviving all of them? In general, no. Their periods share factors. On row $k=2$, sieving by just 5 and 7 leaves exactly $7/12$ of exponent classes on the minus face and $8/12$ on the plus face. Multiplication of the individual fractions would give $5/8$ on either face.

These are exact divisibility counts over a finite period. They are not limiting prime frequencies.

Definitions and the exclusion rule

Write $h=p_{k-1}\mathrm{\#}$, $p=p_k$, and $N_{-}(n)=h{p}^n-1$, $N_{+}(n)=h{p}^n+1$, with $k,n\ge 1$. For a prime $q>p$, both $h$ and $p$ are invertible modulo $q$. Thus $q$ divides the minus face exactly when $p^n\equiv h^{-1}(\mathrm{mod}q)$, and the plus face exactly when $p^n\equiv -h^{-1}(\mathrm{mod}q)$.

Let $d_q$ be the multiplicative order of $p$ modulo $q$. The powers of $p$ are distinct over one period of length $d_q$. A target in that orbit therefore excludes one exponent class modulo $d_q$; a target outside it excludes none. For finitely many sieving primes the joint count is obtained by taking the union of all excluded classes over the least common multiple of their periods.

Proof by a complete period

For $h=2$, $p=3$, the order is 4 modulo 5 and 6 modulo 7. Directly listing the powers gives the following exclusions. Residue 0 represents the positive exponents divisible by the period.

On the minus face the two excluded classes never intersect: one requires an odd exponent and the other an even exponent. On the plus face they intersect at $n\equiv 7(\mathrm{mod}12)$. Thus the union sizes are 5 and 4 respectively, leaving 7 and 8 survivors. By contrast, multiplying $1-1/4$ by $1-1/6$ gives $5/8$. This proves that the marginal product is not the joint fraction.

Normalizing by the unrestricted integer survival fraction $(1-1/5)(1-1/7)=24/35$ gives finite corrections $245/288$ on minus and $35/36$ on plus. The marginal-product correction would be $175/192$ on both. All three are rational finite quantities; none is a justified infinite prime-count constant.

Reproduction and limitations

The repository's tools/finite_sieve.py computes the exclusion phases and their union. The complete example can also be reproduced independently with this standalone Python program:

for sign in (-1, 1):
    survivors = [n for n in range(1, 13)
                 if all((2 * 3**n + sign) % q for q in (5, 7))]
    print(sign, survivors, len(survivors), "/ 12")

The table can also be checked without the sieve implementation: evaluate $2\cdot 3^n-1$ and $2\cdot 3^n+1$ for $1\le n\le 12$ and test divisibility by 5 and 7. This checks the literal residues against the integers they describe.

A divisibility exclusion is a compositeness witness only if the divisor is proper. At $n=1$ the minus value is 5 and the plus value is 7: the table counts both as divisible, although both are prime. A full period of residues is also different from a finite exponent window, whose endpoints affect the count.

Extending the finite sieving set does not by itself justify interchanging limits in the sieve bound, exponent-window length and row index. No prime-density law, Poisson uncertainty law or infinitude theorem follows from this calculation.

References and related reading

The paper, §4.1 retains the counterexample where it limits the interpretation of census fits. Its §4.6 discusses the unresolved infinite question. The proof here uses only modular arithmetic and inclusion–exclusion. The program above reproduces the count without external libraries or project files.