Gap Diagonal Identity
$D(n)$ is the triangle-weighted mean of $\prod_{p \mid g(g+1)\cdots(g+n)}\!\left(1-\frac{1}{p}\right)$ over the gap $g$, for every $n$ — proved below.
For the lattice $\{(r,M) : 1 \le r \le M \le G\}$, the fraction of pairs on the diagonal $|M-r|=g$ that are coprime equals $\varphi(g)/g$ — a direct consequence of Euler's totient function. The identity is classical; reading the lattice as a family of diagonal strips, each carrying its own density, gives a geometric way to decompose and visualize C(n).
At $n=0$, diagonal strip $g$ has coprime density exactly $\varphi(g)/g$ — a classical fact.
As $n$ grows, the block-coprime condition imposes additional constraints: a pair $(r, r+g)$ must
now satisfy $\gcd(r, r+g+j) = 1$ for $j = 0, \ldots, n$. This modifies the local density
on each strip by eliminating extra residue classes at each prime dividing $g, g+1, \ldots, g+n$.
The formula below gives the block-coprime density restricted to diagonal strip $g$ at depth $n$:
This per-strip formula describes each diagonal individually, and averaging it over g recovers D(n) = C(n)/ζ(2) in the limit, for every n — an identity, since an average is a weighted sum rather than a joint event, so no independence between strips is needed. The proof works prime by prime: among the residues of g mod p, a fraction 1−min(n+1,p)/p avoid the block entirely (local factor 1), and the rest carry the factor 1−1/p, so the mean local factor is (1−νp/p) + (νp/p)(1−1/p) = 1−νp/p² with νp = min(n+1,p) — D(n)'s Euler factor. At n=1 this reduces to the classical mean value of φ(g)φ(g+1)/(g(g+1)) = ∏p(1−2/p²). Shared prime factors between different strips do matter for joint events, though — that dependence is computable in closed form, and Page 6 works it out.
A gap $g$ selects four lines: the difference pair $a: M-r=g$ and $b: r-M=g$, and the sum pair $c: r+M=g$ and $d: r+M=2(G+1)-g$. They are not on equal footing.
$a$ and $b$ are identical, always. The transpose $(r,M)\mapsto(M,r)$ carries one onto the other and $\gcd$ is symmetric, so the two branches agree cell for cell — same point count, same coprime count, for every $G$ and $g$.
$c$ and $d$ are not. The reflection $(r,M)\mapsto(G+1-r,\,G+1-M)$ is a bijection between them, so they always carry the same number of lattice points, $g-1$ each — but it does not preserve $\gcd$, and the coprime counts differ for about 70% of pairs $(G,g)$. Strip $c$ is the exact one: $r$ runs over $1,\dots,g-1$, a full period of $\gcd(\cdot,g)$ minus the single point $r=g$, so it holds exactly $\varphi(g)$ coprime points. Strip $d$ samples a window of only $g-1$ residues out of the far longer period $2(G+1)-g$, so it is a partial sample and lands wherever that window happens to fall.
This is why the pooled figure can mislead, and why the export now prints each branch separately. At $G=23,\ g=11$: $a=11/12$, $b=11/12$, $c=10/10$, $d=10/10$ — here $c$ and $d$ happen to agree because both sums, $11$ and $37$, are prime. At $G=121,\ g=61$ they do not: $c$ has sum $61$ and $d$ has sum $183=3\cdot 61$, giving $\varphi(183)/183 = \tfrac{2}{3}\varphi(61)/61$.
The observation that a strip ratio sits close to $1$ exactly when the gap is prime is not a coincidence, and it is sharper than it first looks. Since $\varphi(g)/g = \prod_{p \mid g}(1 - 1/p)$, the density can only reach $1 - 1/g$ when $g$ has a single prime factor and that factor is $g$ itself:
And the separation is wide, not marginal. Every composite $g$ has a prime factor $q \le \sqrt{g}$, so $\varphi(g)/g \le 1 - 1/\sqrt{g}$, while a prime sits at $1 - 1/g$. Nothing composite ever reaches the band above $1 - 1/\sqrt{g}$ — so a strip whose density lands in that band is prime, with no further test. At $g = 97,\,98,\,\dots,\,103$ the primes read $0.98969$, $0.99010$, $0.99029$ and the composites read $0.4286$, $0.6061$, $0.4000$, $0.3137$. The gap is not subtle.
On the sum strip $c$ this becomes an exact, finite test rather than an asymptotic one. Strip $c$ holds precisely $\varphi(g)$ coprime points among its $g-1$ lattice points, so $c$ is completely lit if and only if $g$ is prime — no error term, no limit, readable straight off the picture at any $G$.
Strip $d$, with sum $s = 2(G+1)-g$, is the weaker witness. If $s$ is prime then $d$ is fully lit; and if $s$ has any prime factor $q \le g-1$ then $d$ cannot be fully lit, since a window of $g-1$ consecutive integers must contain a multiple of $q$. But the converse fails: $d$ can be fully lit with $s$ composite whenever every prime factor of $s$ exceeds $g-1$ and the window happens to miss their multiples — 253 such pairs occur with $G < 200$, the smallest being $G=5,\ g=3,\ s=9$. So $c$ decides primality and $d$ only points at it.
- Geometric decomposition: reading the lattice as diagonal strips, each with its own density φ(g)/g, gives a visual explanation for why C(n)'s Euler product looks the way it does — each prime p contributes most strongly through the diagonals it divides.
- Classical foundation: the density φ(g)/g on diagonal g is a direct consequence of Euler's totient formula and periodicity.
- Hardy–Littlewood parallel: the per-strip block-coprime density is structurally similar to the local factors in Hardy–Littlewood's singular series for prime constellations. The Hardy–Littlewood conjecture itself remains unproven, and the parallel here is suggestive rather than a formal connection.
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