Weight × Level + Jump

The weight exists because \(\ell\) is itself a divisor of \(\ell\) exceeding \(d\); it is unique because it is a minimum. Nothing is fitted and there is no parameter: \(a(n)\mapsto(k,L,d)\) is a change of coordinates, invertible term by term. Equivalently, \(k(n)\) is the least modulus for which the jump is the remainder of the term, \(a(n)\bmod k=d(n)\). It is in this form that the weight of the primes entered the OEIS (A117078; level A117563, \(\ell\) A118534).

The construction therefore lives on sequences growing more slowly than \((3/2)^n\). It is defined at no term of \(2^n\) or of the Fibonacci numbers, and at every prime except 2, 3 and 7: for \(n\ge25\) there is a prime in \([n,\tfrac65n]\) (Nagura [10]), and the primes below 25 are checked by hand. Note that \(5=3\cdot1+2\) decomposes, while 7, whose gap is 4, does not.

Proof. \(k\le L\) means \(k^2\le\ell\): the least divisor exceeding \(d\) is at most \(\sqrt\ell\). This is (i)\(\Leftrightarrow\)(ii). If \(k\gt L\), then \(L\) is a divisor of \(\ell\) below the least divisor exceeding \(d\), so \(L\le d\); conversely \(L\le d\lt k\). If \(\ell\lt(d+1)^2\) the window contains no integer. \(\square\)

Two readings are worth keeping. The classification is a question about divisors in an interval whose left end is the jump: divisors of \(\ell\) up to \(d\) are invisible, and the level class consists of the terms with none between \(d\) and \(\sqrt\ell\). And before any arithmetic enters, size decides: a sequence whose jumps exceed the square roots of its terms is level-classified throughout (§4).

The plane

Each decomposable term is the lattice point \((k,L)\) on the hyperbola \(kL=\ell\). On logarithmic axes with equal scales these hyperbolas are lines of slope \(-1\), the diagonal \(k=L\) separates the classes, and the distance from a point to the diagonal is \(|\log(k/L)|/\sqrt2\). This is why every weight–level figure in this report is square.

Compute it yourself

The project’s reference kernel, in PARI/GP; \(b\) is the successor of \(a\).

decomp(a, b) = { my(d = b-a, l);
  if (a <= 2*d, return([0,0,d]));
  l = a - d;
  fordiv(l, k, if (k > d, return([k, l/k, d]))) }
? decomp(113, 127)    \\ a prime and the next
%1 = [33, 3, 14]
? decomp(113, 114)    \\ a natural number
%2 = [2, 56, 1]

Exercise 1. The same integer has different coordinates in different sequences. Show that \(n^2\) is decomposable exactly for \(n\ge5\), and then always level-classified.

Proof. The least divisor of \(m=n-1\) exceeding 1 is its smallest prime factor \(q\). If \(m\) is prime, \(k=m\gt1=L\). If \(m\) is composite, every prime factor of \(m/q\) is at least \(q\), so \(L=m/q\ge q=k\), with equality exactly when \(m=q^2\). \(\square\)

How much of the fundamental theorem?

Iterate on the level: decompose \(n\), then the natural number \(L(n)+1\), and so on. Since \(\ell(L+1)=L\), the successive weights are the prime factors of \(n-1\) in non-decreasing order, and the chain ends at a level-classified term.

So the construction contains the existence half of the fundamental theorem of arithmetic, as an algorithm: it is trial division. It does not contain the uniqueness half. That the multiset of primes does not depend on the order of extraction is Euclid’s lemma, which the construction neither uses nor proves; the writing \(a=kL+d\) is unique for a cheaper reason, namely that \(k\) is defined as a minimum.

What generalises, and what does not

For another sequence there is no product structure on the terms, no irreducible elements and no unique-factorisation theorem. What does generalise is the sieve. The weight is a smallest-divisor function with a threshold, the threshold is the jump, and the level class is the set of terms that survive. The accurate slogan is “a generalisation of the sieve of Eratosthenes”, not “an extension of the fundamental theorem”. Proposition 1 is also the only place where “level-classified” is exactly “\(\ell\) is prime”; on the primes this becomes almost true, with exceptions (Lemma 5).

Three one-line variations

verified on the first \(10^5\) terms. Odd numbers. \(d=2\) and \(\ell\) is odd, so again \(k=\spf(\ell)\): level-classified means \(a-2\) prime. Level share 17.98%. Even numbers. Level-classified means that \(a-2\) is twice a prime (or \(a=10\)): the line \(L=2\), share 9.59%. Numbers not divisible by 3. Here \(d\in\{1,2\}\) and 3 always divides \(\ell\), so \(k\le3\): three level-classified terms (4, 5, 8) among 99,998. A forced divisor empties the level class.

Exercise 2. Prove the statement about the odd numbers and find the ties. (They are the terms \(p^2+2\): 11, 27, 51, 123, …)

Proof. Proposition A, and parity: divisors of \(\ell\) are odd, \(g\) is even. \(\square\)

Proof. One of \(p-2,\,p,\,p+2\) is divisible by 3. If \(g=2\), then \(p\) and \(p+2\) are primes exceeding 3, so \(3\mid p-2=\ell\), and 3 is the least integer exceeding \(g\). Conversely \(k=3\gt g\) with \(g\) even gives \(g=2\). \(\square\)

Proof. Modulo 3, \(p\not\equiv0\) and \(p\not\equiv-g\). If \(3\mid g\) then \(\ell\equiv p\not\equiv0\). If not, the only class left for \(p\) is \(g\), so \(\ell=p-g\equiv0\). \(\square\)

Proof. \(\ell\) is odd, so the least divisor exceeding 4 can only be \(5,7,9,\dots\), and the one exceeding 6 only \(7,9,\dots\); \(g=2\) is the column \(k=3\). \(\square\)

Proof. \(L=1\) means that \(\ell\) has no divisor in \((g,\ell)\). If \(\ell\) is composite, with least prime factor \(r\le\sqrt\ell\), then \(\ell/r\ge\sqrt\ell\gt g\) is such a divisor. \(\square\)

The inequality \(\ell\le g^2\) holds only for \(p\in\{5,13,19,23,31,113\}\), and \(L=1\) with \(\ell\) composite only for 13 and 31. Both lists are complete below \(4\cdot10^{18}\), by the exhaustive tables of maximal gaps [8]. Hence, up to two exceptions, the level-one primes are the middle terms of three primes \(p-g,\,p,\,p+g\) in progression whose upper step is a prime gap.

Proof. Write \(k=rs\) with \(1\lt r\le s\lt k\). Then \(r\), \(s\) and \(L\) are divisors of \(\ell\) below \(k\), hence at most \(g-1\) by Lemma 3, and \(\ell=rsL\). \(\square\)

These primes are exactly 13, 31, 113, 131, 887, with weights 9, 25, 33, 25, 51, again below \(4\cdot10^{18}\). So a level-classified prime with \(\ell\gt g^3\) has a prime weight, and this normal form is the door to Theorem 1:

\[\ell=s\,q,\qquad q=k\ \text{prime},\qquad s=L\le g-1,\]

so that \((q,\,sq+g,\,sq+2g)=(k,\,p_n,\,p_{n+1})\) is a triple of primes.

verified Below \(10^7\), with identical \((k,L)\) from both engines at all 664,579 primes: no violation of any statement above; 58,979 primes with \(k=3\), the same primes as \(g=2\); 19,495 with \(k=5\); 26,974 = 7,859 + 19,115 with \(k=7\); 21,837 balanced primes, all on \(L=1\); largest level 129, at \(p=8{,}917{,}523\), \(g=140\).

Exercise 3. State and prove the identity for the column \(k=9\). (Answer: \(g=4\), \(5\nmid\ell\), \(7\nmid\ell\), \(9\mid\ell\), or \(g=8\), \(9\mid\ell\); the gap 6 is excluded by Proposition 2. Below \(10^7\): 24,581 = 10,468 + 14,113.)

Sketch. Put \(G=(\log x)^{3/2}\). (1) The gaps sum to at most \(x\), so at most \(x/G\) primes have \(g\gt G\). (2) The primes with \(g\le G\) and \(\ell\le g^3\) number at most \(G^4\), since \((g,\ell)\) determines \(p\). (3) Every other level-classified prime is in normal form, \(\ell=sq\) with \(s\le g\le G\). Forget that \(p+g\) is the next prime; keep only that \(q\), \(sq+g\), \(sq+2g\) are prime. (4) For fixed \((g,s)\), an upper-bound sieve of dimension 3 ([4], Theorem 5.7), uniform in the parameters, bounds the number of such \(q\le x/s\) by \(\ll x/(s\log^3x)\) on average over \(g\). The sum over \(s\) costs \(\log\log x\), the sum over \(g\le G\) costs \(G\), and \(xG\log\log x/\log^3x\) balances \(x/G\). The full proof is Appendix B.1 of the treatise [2].

Three things to carry away. The exponent \(\tfrac32\) is an artefact of the balance; the expected truth is \(f(x)\asymp\log\log x/\log x\), and \(f(x)\log x/\log\log x\) reads 1.23, 1.20, 1.18, 1.17, 1.16 at \(10^6,\dots,10^{10}\), the last three carried\(^\dagger\). The proof discards the word next, so it gives no lower bound: whether the level class is infinite is open, and would follow from the infinitude of balanced primes. By page 4, it is a theorem about the construction at a given density at least as much as about the primes. And the asterisk is a statement of social status, not a hedge: the proof is audited within the project and not refereed. Independently, density zero has been formalised in Lean 4 with Mathlib [13], by Brun’s pure sieve and with the weaker rate \(x(\log\log x)^7/(\log x)^{3/2}\); the repository’s kernel check passes and its definitions were compared with Definition 1, but an independent rebuild is still owed.

What a bound says about data. Theorem 1 requires \(E(x)=f(x)\sqrt{\log x}/\log\log x\) to stay bounded. It reads 0.385, 0.331, 0.300, 0.276, 0.257, 0.241 at \(10^5,\dots,10^{10}\), the last three carried\(^\dagger\): decreasing, with no plateau, as expected of a bound that is true but not sharp. For Theorem 2 the constant is not visible at accessible heights: \(\Delta N_1\big/\!\int\! dt/\log^2t\) is 0.957 on \([10^5,10^6]\) and 0.995 on \([10^6,10^7]\), on a logarithmic climb towards 1.32. No constant should be read off such data by extrapolation; the identification rests on Lemma 1 and on a gap-by-gap test of (H1) at \(10^{10}\), where observed over predicted is 1.011\(^\dagger\).

The lock, and its bias

Let \(r\) be an odd prime and \(p\gt r\) a prime with gap \(g\lt r\). Modulo \(r\), \(p\) avoids \(0\) and \(-g\). If the other \(r-2\) classes were equally likely, the lock \(r\mid\ell\) would close with probability \(1/(r-2)\).

\(r\)	5	7	11	13	17	19	23
\((r-2)\Pr[r\mid\ell]\)	0.999	0.981	0.983	0.976	0.973	0.972	0.975

verified primes below \(10^7\) with \(g\lt r\). The flat law holds within 3%, but the deficit is real: 3.5 to 4.9 standard deviations for each \(r\ge7\).

heuristic Interior credit. The even integers strictly between \(p\) and \(p+g\) must be composite, and a class of \(p\) modulo \(r\) in which one of them is divisible by \(r\) gets that for free. The lock class \(p\equiv g\) is never such a class when \(g\lt r\). Measured over the cells \(4\le g\lt r\le47\): credited classes are over-represented by 9.0%, the others by \(-2.2\%\), the lock class by \(-2.5\%\); for twins, with no interior even integer, the flat law is exact (ratio 1.0006). Compare the biases between consecutive primes of Lemke Oliver and Soundararajan [7].

Lemma 1: where the twin constant enters

Proof. \(\nu_p=1\) if \(p\mid g\); otherwise \(\nu_2=2\) and \(\nu_p=3\), which kills the factor at 2 or at 3. For \(p\ge3\) the factor is \((1-\frac1p)^{-2}\) with probability \(\frac1p\) and \((1-\frac3p)(1-\frac1p)^{-3}\) otherwise, of mean \[\Big(1-\frac1p\Big)^{-2}\Big(\frac1p+1-\frac3p\Big)=1-\frac1{(p-1)^2},\] and at \(p=2\) the mean is \(\frac12\cdot4=2\). The local conditions are independent by the Chinese remainder theorem, so the mean of the product is the product of the means; convergence is by truncation ([2], Appendix B.2). \(\square\)

verified The mean over \(g\le10^6\) is 1.3202486, and \(2C_2=1.3203236\ldots\) For \(6\mid g\), \(\fS(g)=2C_3\prod_{p\mid g,\,p\ge5}\frac{p-1}{p-3}\), with \(2C_3=5.7164971914\ldots\)

Averaged along the line \(L=1\), the three-prime condition on \(p-g,\,p,\,p+g\) costs exactly the two-prime constant, doubled. This is the unconditional core of Theorem 2:

\[N_1(x)\approx\sum_g\fS(g)\,e^{-g/\log x}\,\frac{x}{\log^3x}\approx 2C_2\,\frac{x}{\log^2x},\]

the first step being (H1) with (H2), the second Lemma 1.

Three more bridges

Divisors in intervals. The level class asks for integers with no divisor in an interval, the subject of Erdős, Tenenbaum and Ford [5]; here the integer is a shifted prime [6], shifted by its own gap. The nearest published neighbour is Gafni and Tao [9]: prime gaps containing no integer whose least prime factor reaches the length of the gap. Theorem 1 multiplies a multiplicative estimate, an upper-bound sieve for three linear forms, by an additive one: the gaps sum to \(x\). What is not here. Goldbach: the only even numbers the construction forms are \(2p_n=\ell+p_{n+1}\). The Riemann hypothesis: nothing here bears on it. And “the column \(k=3\) is infinite” is a translation of the twin prime conjecture, not a reduction of it.

Exercise 4. Check the local mean of Lemma 1 at \(p=5\) and at \(p=7\); then compute \(\fS(30)/\fS(6)\).