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.

The Farey sequence page 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/rs/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.

Those same points draw two further pictures of the one node set. The sunburst is the visible lattice points themselves, the (a, b) with gcd(a, b) = 1 and abs a, abs b <= n, joined in angular order into a closed polygon, which reads the lit set as directions rather than as points of the line. The resonance diagram is the spike profile, a spike at the node a/b of height 1/b. The stack is that diagram up to the floor. Proved, from the brightness law: normalising gives the node a/b the height floor(Q/b)/Q, which lies in (1/b - 1/Q, 1/b] at every depth and equals 1/b exactly when b divides Q (lab/rs/farey-discrepancy).

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.

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/rs/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/rs/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.

The meter on a digit design

A digit design S_F is the set of whole numbers whose every digit lies in a digit set F, of dimension alpha = log |F| / log base. Restricting the meter to one needs a convention, and the strict one is taken here: F_Q(S_F) is the set of reduced fractions a/b with 0 < a <= b <= Q and both a and b in S_F. The weaker denominator convention - b in S_F, a free - is measured beside it, and the unrestricted F_Q is the control. Neither is a set the stack produces: Farey order is the stack still holds and every design gives the same stack at fixed Q, so F_Q(S_F) is a filter laid over the stack's nodes by hand, and all that is at stake is what the Franel-Landau functional reads on it. No Mobius sum appears anywhere in this section, so nothing in it bears on the square-root conjecture of mobius, whose theta(F) is a Mertens exponent and belongs to that page alone; the exponents here are called e_2 and e_1 and are this section's own.

Each convention carries a count that never enumerates a fraction. The denominator convention has card = sum_{b in S_F, b <= Q} phi(b). The strict one has card = sum_{b in S_F, b <= Q} phi_F(b) with phi_F(b) = #{a in S_F : a <= b, gcd(a,b) = 1} = sum_{d | b} mu(d) * #{multiples of d in S_F up to b}, inclusion-exclusion over the divisors of b. Proved. Both are sieved independently of the enumeration and match it at every rung of every table below, which is the control that says the object measured is the object defined; the largest check is 9538759028 nodes on the base 3 control at Q = 3^11. Verified by lab/rs/farey-discrepancy design.

Rungs are powers of the base, so the design's set is self-similar at every rung. Every exponent below is one ratio between consecutive rungs, e_2 = ln(S2(Q')/S2(Q)) / ln(Q'/Q) and e_1 the same for S1, and nothing is fitted.

Write D_Q = #{b in S_F : b <= Q} ~ Q^alpha for the denominators the design supplies and card F_Q(S_F) ~ Q^e for the nodes. Square-root cancellation in the denominators is a node-count error of order sqrt(D_Q), which puts e_2 at alpha - e, and Cauchy-Schwarz on S1 <= sqrt(card * S2) then caps e_1 at alpha/2. Franel and Landau are this pair at alpha = 1, e = 2: e_2 <= -1 and e_1 <= 1/2, both under RH, both caps on a limsup and never values. The control misses them freely at a single rung - its e_2 wanders from -0.463 to -1.017 and its e_1 from +0.236 to +0.673 across the base 3 ladder - so a lane sitting off a cap at one rung shows nothing by itself. The ladders run to Q = 3^11 = 177147 and Q = 10^5, each with its own control, and the top rung reads:

The denominator lanes sit in the control's own band and the strict lanes sit on the mass. Verified. Both denominator rows put e_2 at -0.959 and -0.899 against alpha - e = -1.000 and e_1 at +0.336 and +0.585 against caps of 0.315 and 0.477, every one of the four inside the range the control walks. Both strict rows instead put e_2 and e_1 on exp card itself, agreeing with it to two decimals: +1.259 and +1.262 against +1.263, +1.904 and +1.906 against +1.908. Sums that ride the mass are sums with no cancellation at all, and the normalised readings settle:

The strict lane holds those constants from Q = 2187 at base 3 {0,1}, card running 4286 to 1080458, and from Q = 10000 at base 10 without 9, card running 11890654 to 963170938: two figures of S1/card and of S2/card fixed across each span, and no more of either claimed, the base 10 rung below moving the first figure of S2/card from 4.1e-5. Verified.

At base 3 {0,1} the last column names an exact interval, a sixth of the line. Proved. If b has its leading base 3 digit at position level then 3^level <= b <= (3^(level+1) - 1)/2. A numerator whose own leading digit sits at level gives a/b >= 2 * 3^level/(3^(level+1) - 1) > 2/3; one whose leading digit sits at level - 1 or below gives a <= (3^level - 1)/2 < b/2, so a/b < 1/2. The closed interval [1/2, 2/3] is therefore empty in F_Q(S_F) at every Q, and a set that misses a fixed interval of positive length does not equidistribute. The widest gap the meter finds contains that interval at every finite Q and shrinks onto it from outside, reading 0.16827, 0.16720, 0.16684, 0.16673, 0.16669, 0.16667 at Q = 3^6 .. 3^11 and starting at 0.49931, 0.49977, 0.49992, 0.49997, 0.49999, 0.50000.

Base 10 without 9 has no gap to argue from. Conjecture. Its widest gap falls like 1/Q, reading 0.01136, 0.00113, 0.00011, 0.00001 at Q = 10^2 .. 10^5 against the control's 0.01000, 0.00100, 0.00010, 0.00001 on the same rungs, so the strict set there is as fine-grained as the full Farey sequence and the base 3 argument does not transfer. That it fails to equidistribute rests on the settled constants alone, which is a reading and not a proof.

The denominator convention keeps the transplanted shape. Conjecture. Restricting the denominator moves the mass from Q^2 to Q^(1+alpha) and the count error from sqrt(Q) to sqrt(D_Q) = Q^(alpha/2), and the two moves cancel in S2: S2*Q reads 0.8926 at base 3 {0,1} and 0.8536 at base 10 without 9 against the control's 0.6782 and 0.6684 at the same top rungs, a constant factor and not a rate. The transplanted Landau reading S1/Q^(alpha/2) is flat where the control's S1/sqrt(Q) is falling, at 0.243, 0.281, 0.267, 0.268, 0.274 for Q = 3^7 .. 3^11 and 0.213, 0.222, 0.207, 0.265 for Q = 10^2 .. 10^5. So the conjecture is S2 = O(Q^(-1+eps)) and S1 = O(Q^(alpha/2+eps)) on the thinner set. Normalising against the node count instead would demand e_2 = -(1+alpha)/2, which is the wrong yardstick: it agrees with alpha - e on the control only through the accident sqrt(Q) = card^(1/4) at e = 2, and the denominator lane is not beating a shape by missing it.

The honest cap below covers this section too. Nothing here reaches further than the meter above it. What it adds is negative and clean: digit restriction of the denominator is invisible to the shape, and digit restriction of both coordinates destroys equidistribution outright at base 3 {0,1} and appears to at base 10 without 9. The denominator lane's invisibility is measured here and priced in the restricted Franel identity, where its exponential sum turns out to have a closed form in Mobius sums.

The restricted Franel identity

The identity behind the denominator lane has a note of its own, the restricted Franel identity: the exponential sums as dilated Mertens sums, Franel's identity in its Fourier and rank forms, the ceiling it gives, the dilate automaton, the converse under (U') and under (SR), the mean-square equivalence, the surrogate B(Q), the strict set's divisor form, and the repunits.

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/py/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/py/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 Franel and Landau, 1924 and the meter establish, and what the mean-square equivalence carries one digit design over, 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 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/rs/farey-discrepancy). Verified.
• Under the transparent convention Q = 3^level, 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^level 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/py/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/py/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/py/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/py/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.