The stack is an RH-observable

Lay the same fractal grid on the unit square at many scales at once - scale n puts its cell boundaries at x = k/n - drop the opacity and add the layers up. The result is a moire, and a bright point is one that many scales agree on. The question this page answers is what the bright points are, and the answer is not decorative: the lit nodes are the Farey fractions, the amount of new structure each scale contributes is Euler's totient phi(n), and how evenly those nodes spread is - by a pair of theorems from 1924 - literally equivalent to the Riemann hypothesis.

Every claim below carries a tag. Proved means derived here from definitions. Verified means recomputed from scratch, or checked against the published literature. Nothing on this page is a conjecture. The Farey demo builds the stack scale by scale, lights the Farey fractions, and shows phi(n) novelty peaking at the primes.

Where the lines land

Stack the scales n = 1..N. A point a/b in lowest terms receives a grid line from exactly the scales that are multiples of b, so over 1..N its brightness is floor(N/b). Proved, and Verified by direct simulation at N = 30: building the stack node by node and comparing every node's hit count against floor(30/b) gives no mismatch anywhere (lab/farey-discrepancy).

Brightness therefore falls as one over the denominator, which is the Stern-Brocot ordering of the rationals. The top of the table at N = 30:

The lit nodes are also exactly the lattice points visible from the origin, since a/b is in lowest terms precisely when gcd(a,b) = 1. That is the "lighthouse" reading of the picture. Proved. The density of visible points is 6/pi^2 - the same constant, and the same base-blindness, discussed in what base 3 hides, where it is measured as 0.608042 on a 3000 x 3000 grid. Verified, by recounting that grid.

Primes are the maximally novel scales

The nodes scale n introduces for the first time are the fractions a/n with gcd(a,n) = 1, since any a/n that reduces was already lit by the smaller scale it reduces to. There are exactly phi(n) of them. Proved, and Verified by set difference over the stack for n = 2..30:

n    2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
new   1 2 2 4 2 6 4 6 4 10 4 12 6 8 8

n   17 18 19 20 21 22 23 24 25 26 27 28 29 30
new  16 6 18 8 12 10 22 8 20 12 18 12 28 8

Every count equals phi(n), and the running maxima 1, 2, 4, 6, 10, 12, 16, 18, 22, 28 occur at n = 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. The reason is one line: phi(n) = n - 1 if and only if n is prime, because every one of 1..n-1 is coprime to n exactly when n has no smaller factor. Proved, and Verified by testing the equivalence against trial division for all n up to 200.

So primality is readable off the picture. Stack 1..n-1, then add scale n, and count what appeared: n - 1 new nodes means n is prime, fewer means composite. Proved (it is the previous claim restated). A composite scale mostly re-lights nodes its own divisors already drew - scale 30 adds only 8 new lines, the rest of its grid falling on lines from 1, 2, 3, 5, 6, 10 and 15.

Franel and Landau, 1924

Over scales 1..Q the stack lights exactly the reduced fractions of denominator at most Q: the Farey sequence F_Q. Its size in (0,1] is m = sum_{k<=Q} phi(k). Proved, and Verified by generating F_Q through the next-term recurrence and comparing its length with the totient sum at Q = 10, 30, 60 (lab/farey-discrepancy, which runs the same comparison at Q = 10, 30, 60, 125).

Write rho_1 < ... < rho_m for those nodes and delta_j = rho_j - j/m for how far each one sits from perfect equidistribution. Then:

• Franel (1924) proved that sum_j delta_j^2 = O(Q^(-1+eps)) for every eps > 0 is equivalent to the Riemann hypothesis.
• Landau (1924), in a note published immediately after Franel's, proved the same for sum_j |delta_j| = O(Q^(1/2+eps)).

Verified against the literature: both statements, with the original 1924 citations to the Göttingen Nachrichten, are the standard Franel-Landau formulation, and are reproduced in Edwards, Riemann's Zeta Function, chapter 12.

Put the two halves together. The nodes whose discrepancy Franel and Landau are talking about are the nodes the stack draws - not an analogue of them, the same set. So the question "how evenly are the bright points spread?" is not related to the Riemann hypothesis; at this level of precision it is the Riemann hypothesis. Proved, given the identification above, which is what the first two sections establish.

The meter reads what RH predicts

Both sums are computable. Generating F_Q exactly and measuring, with S2 = sum delta_j^2 and S1 = sum |delta_j|:

Verified by lab/farey-discrepancy. S2*Q flattens near 0.656 and the local exponent walks to -1, which is the Franel condition; S1 stays under its Q^(1/2) envelope and its own local exponent runs between 0.27 and 0.43, under the Landau threshold of 0.5. The node count matches sum phi(k) exactly at every rung, which is the control that says the object being measured really is the stack's node set.

Weighting the stack by Mobius

Give scale n the weight mu(n) instead of weight one and the same stack renders a different arithmetic function: the node a/b collects mu over the scales that are multiples of b, so its brightness is Sum_{k <= N/b} mu(kb) = mu(b) * Sum_{k <= N/b, gcd(k,b) = 1} mu(k), a Mertens-type sum over the integers coprime to b. Proved, by the same divisor count that gives floor(N/b) in the unweighted stack. It is not M(floor(N/b)): the two agree at only 64 of 200 denominators at N = 200 (lab/mertens-meter), and coincide at b = 1, where the node reads M(N) exactly.

That makes the picture a Mertens meter rather than a Farey one, and the oscillations of M(x)/sqrt(x) are where the nontrivial zeta zeros live, by the explicit formula. Sampling M(x)/sqrt(x) in log-space and taking the power spectrum puts peaks at the first eight zeros:

Verified by lab/mertens-meter, and weaker than everything above it on this page: M(x)/sqrt(x) for x = 1..50000 from a linear Mobius sieve, resampled uniformly in log x on 8192 points, Hann-windowed, the real FFT power spectrum read as gamma = 2 pi f, local maxima above three times the band median over 8 < gamma < 55. The bin width is 0.5806, so every error in the table sits inside one bin. The honest cap below covers this section too, and covers it harder: the zeros are known to far greater precision than a moire can reach, so what the picture buys is a rendering, not a measurement.

The honest cap

An observable is not a handle. What the last two sections establish is that this picture renders a genuinely RH-equivalent object, which is a real upgrade over the vaguer "fractals and zeta both have self-similar structure" gestures. What it does not do is supply any route to a proof. The Riemann hypothesis is already checked numerically far beyond any range this or any other meter can reach, so the table above can only ever illustrate the expected behaviour - it is consistent with RH, it is not evidence for it, and no amount of extra Q changes that. Scored here, the link quality is 6 out of 10 and the meter's tractability 0, and both numbers deserve to be stated together: the connection is exact, and no renderer reaches it - an attack must come through the equidistribution toolkit, never through a picture.

Two instructions this page hands the rest of the tree. First: the toolkit flows both ways. The window at dimension one in coprime is a discrepancy statement about a discrete arithmetic set and so is Franel-Landau, so the equidistribution methods that attack one are the methods the other needs - that kinship in technique is why an RH equivalence sits on a page of this tree, and it marks the one honest route: theorems, not renders. The verdict above is final for the meter alone; it caps what a picture can claim, never what a proof may attempt. Second: this page renders the Farey set without owning it. The rule whose ORBIT is the Farey set is the mediant, (a/b, c/d) -> (a+c)/(b+d), with the Gauss map x -> {1/x} as its continued-fraction twin - simple local rules with emergent complexity, exactly this project's own principle, and they carry the Stern-Brocot and GL_2(Z) symmetry that base-q digit restriction does not. Mayer's theorem lives there: the Selberg zeta function of the modular surface is the Fredholm determinant of the Gauss-Kuzmin-Wirsing transfer operator, a genuine fractal-dynamics-to-zeta bridge. Two cautions travel with it - that is Selberg zeta and not Riemann zeta, and its RH-analogue is known for unrelated reasons; and the alphabet is infinite, so every finite-state tool on this tree needs rebuilding there.

Farey order is the stack, not the design

• There is no design-specific Farey sequence, and there never was one to find. Refuted.
• The stack's lit set at maximum scale Q is exactly {a/b : 1 <= a <= b <= Q, gcd(a,b) = 1}, because a boundary coordinate k/n reduces to a/b and reappears at every scale divisible by b.
• Farey order is therefore Q, the maximum stacked grid scale. Fill count plays no part, and every design gives the same Farey sequence at fixed Q. Proved from the construction.
• Brightness hits(a/b) = floor(Q/b) is checked by literal stacking at Q = 30 on all 278 lit fractions and up to Q = 125 (lab/farey-discrepancy). Verified.
• Under the transparent convention Q = 3^L, the geometric side length, the Landau discrepancy D_Q = sum_i |f_i - i/m| reads 0.166667, 0.549206, 1.150760, 2.118500, 3.187070 at Q = 3, 9, 27, 81, 243, with m = 4, 28, 230, 2020, 18056. Both generation routes agree exactly, a Farey next-term recurrence being the independent cross-check; the rows have no generator in lab/. Conjecture.
• D_Q/sqrt(Q) stays inside [0.0962, 0.2354] and reads 0.2045 at Q = 243. The adjacent log slope falls 1.085, 0.673, 0.556, 0.372; the all-five log-log fit is 0.660 and the last-three fit 0.464. Consistent with O(Q^{1/2+eps}), discriminating nothing: five nested deterministic points cannot test a statement quantified over every positive epsilon.
• The Q = 3^L map is a comparison convention chosen here, not a mapping the tree defines. Mapping Q to fill count would be arbitrary and was explicitly rejected.

The stack is an address, not a construction

Can a stack be created immediately, without stacking? The answer is yes, exactly, and the boundaries of the yes are theorems of their own; every number in this section is regenerated by lab/carpet-stack-address.

Everything layer n does at a rational point x = (a_1/q, a_2/q) depends only on r = n mod 2q: n is odd iff r is odd, and floor(n*a/q) is odd iff (r*a) mod 2q >= q, since n*a mod 2q = q*(floor(n*a/q) mod 2) + (n*a mod q). So the odd-carpet stack's brightness is a residue count with the N-dependence in closed form,

B_N(x) = ceil(N/2) - Sum_{r in S(x), r <= N} (floor((N-r)/2q) + 1)

with S(x) the bad residues, and the per-point cost depends on q alone, never on N. Proved. The line-stack's own form is the floor(N/b) at the top of this page, O(1) per node. Verified by two generators sharing no code in lab/carpet-stack-address, one stacking literally and one forbidden to loop over layers: identical Farey digests at N = 55 (940 nodes, brightness sum 1540 = N(N+1)/2 landed by count), sha256-identical 512 x 512 renders by three routes, all 48 probes equal at N = 55 and 5555, the closed form against literal stacking at N in {1, 2, 55, 5555, 19945, 19946, 19947, 40001} with zero mismatches, and a stack of 5*10^17 layers - N = 10^18 - evaluated in a tenth of a second by both implementations, exactly, values agreeing digit for digit.

The scope is part of the result, each boundary proved. Per-point only: an R x R raster costs R^2 writes no matter what. Exact representations only: on a point supplied as a real oracle the value is undecidable at the discontinuity set {n*x integer}, while an irrational with a known continued fraction stays computable by the Ostrowski recursion - the obstruction is representation, not irrationality. Finite N only: membership in the infinite-depth limsup set is not decidable. And unweighted only: the Mobius-weighted node of this page carries the Mertens-type sum Sum_{k <= N/b} mu(kb), M(N) at b = 1, and no polynomial-time algorithm for the Mertens function at binary input is known, the best standing near x^(2/3) (Deleglise-Rivat 1996) - the one value on this page without an immediate form, an open computational status and explicitly not a hardness result.

What immediacy does not buy is the RH question, and the reason is sharp. The Franel-Landau functional needs each node's rank, and the rank's own closed form is A(x, Q) = Sum_{d <= Q} mu(d) Sum_{e <= Q/d} floor(x*e) - classical, Verified here at Q = 12, 25, 40 against brute-force enumeration - so the moment the picture is asked where its nodes sit, Mobius enters the formula. Brightness has a mu-free closed form; rank does not; only rank carries the difficulty. Franel's 1924 theorem is the symbolic all-Q reduction of the discrepancy to Mertens-type sums, so the route "generate every frame at once and read off structure" is not unexplored - it is the proof of the equivalence, and it terminates at Mertens. The Mertens meter's natural global readout collapses outright: Sum_{n <= N} M(floor(N/n)) = 1 identically (Proved, classical Mobius inversion; Verified at every N through 20000 with zero breaches, lab/mertens-meter), so the weighted picture aggregates to a constant and informs only where it presupposes M. One steelman deserves its named kill so it is closed: the stack's divisibility incidence array is the Redheffer matrix up to its first column, whose entries were always trivial and whose determinant is M(n), RH iff M(n) = O(n^(1/2+eps)). Immediate entries, untouchable determinant - the same wall this page's honest cap already describes, stated in the highest shape-adjacency object this tree carries.

The complexity frontier runs beside this page, not through it. Deciding a pixel's brightness with every input in binary is in P: the constraint set is a rational polytope in fixed dimension three (the both-even parity branch summed alongside the both-odd), and lattice-point counting in fixed dimension is polynomial (Barvinok 1994). Verified against the literature, with the caution that the tree's O(q) residue pass is polynomial in q and so exponential in bit-length - a unary-input algorithm, the honest name for what runs in lab/carpet-stack-address. The shared scales are the whole engine: moduli 1..N give the picture polynomially many faces and closed-form extrema, maximum brightness on the diagonal and floor(N/b) at b = 1. Destroy the sharing - arbitrary binary moduli, one darkened residue class per layer - and "does any point reach maximum brightness" is Simultaneous Incongruences, NP-complete (Garey and Johnson, SP3); make the ambient dimension part of the input and "is any layer lit at this fixed point" is NP-complete (Lagarias 1985), polynomial at every fixed dimension. Verified against the literature, both at source. Evaluation stays easy here exactly because the stack shares its scales; hardness begins where the sharing ends, one structural parameter away. A proved no-shortcut theorem for this stack could therefore never have separated P from NP: the problem it would bound is already in P, and what remains bindable there is fine-grained or expressibility only. Nor does the yes touch RH, for the reasons above - both halves of the question were category errors, and each points at the true theorem beside it.