Signed Dilation Designs Need Many Primes

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This page's result is Plasma AI's, author-recorded and without independent review. The prime dilation page adapts the alternating cube of Tao and Teräväinen to Problem 249: $2^K$ prime dilations with signs cancel the first $K$ levels of the shift tail identically, and the Chinese remainder condition on $2^K$ primes limits $K$ to about $\log_2\ln x$ at scale $x$ (its bound (7)). Proposition 1 below applies to every signed design satisfying its hypotheses. Cancelling $K$ levels for every function needs $K+1$ distinct primes larger than $K$. Thus the modulus exceeds $K^{K+1}$. If it is at most the averaging scale $x$, this cancels only $o(\log x)$ levels. In a method whose remaining tail bound requires a cutoff of order $\log x$, improving the design alone therefore leaves levels to estimate.

Designs

A design is a nonempty finite set $V$ with, for each $v\in V$, a prime $p_v$, an integer shift $s_v$ and a nonzero integer weight $c_v$, the pairs $(p_v,s_v)$ being distinct. An integer $n$ is admissible when $n\equiv s_v\pmod{p_v}$ for every $v$; then $N_v=(n-s_v)/p_v$ is an integer and

$$n+p_vh-s_v=p_v(N_v+h)\qquad(h\ge1).$$

The design cancels level $h$ when the weighted multiset $\{p_vh-s_v\}$ vanishes, that is

$$\sum_{v:\ p_vh-s_v=m}c_v=0\qquad\hbox{for every integer }m.$$

Then $\sum_vc_vf(n+p_vh-s_v)=0$ for every function $f$ and every $n$, which is the cancellation the cube uses in its Proposition 2 with $p_\epsilon=p_0+\sum_k\epsilon_kv_k$ and $s_\epsilon=\sum_kk\epsilon_kv_k$. Merging two vertices with the same pair $(p,s)$ adds their weights, so distinct pairs lose no generality.

Proposition 1 (L219). Let a design cancel levels $1,\dots,K$, let every $p_v$ exceed $K$, and suppose an admissible integer exists. Then the primes $p_v$ take at least $K+1$ distinct values. Consequently their product $Q$, the modulus of the admissible class, satisfies $Q>K^{K+1}$.

This is the claim L219. Its [[theory/irrationality/L219_dilation_design_prime_count/_proof|canonical proof]] observes that a vertex must share a fiber with some other vertex at every cancelled level, that these partners are distinct because two vertices with different primes share a fiber at most at one level, and that two partners with the same prime $p'$ would make $p'$ divide the product of $p_v-p'$ and a difference of levels, which is impossible for primes above $K$.

The bound is about counting primes, not vertices: vertices may share a prime when their shifts agree modulo it. The cube attains cancellation of $K$ levels with $2^K$ distinct primes, and it is not optimal. Ignoring primality, give weight $+1$ to the integer points $(0,0),(4,10),(5,9)$ and weight $-1$ to $(1,1),(6,12),(2,6)$, and let each point $(a,b)$ play the role of $(p_v,s_v)$. These six points cancel levels $1,2,3$ in the sense above and have six distinct first coordinates; at each level their fibers pair the plus and minus points along one of three perfect matchings. The affine map $(a,b)\mapsto(43+30a,30b)$ gives the actual prime design

$$\begin{array}{c|ccc} c_v=+1&(43,0)&(163,300)&(193,270)\\ c_v=-1&(73,30)&(223,360)&(103,180). \end{array}$$

All six first coordinates are distinct primes greater than three. The Chinese remainder theorem supplies admissible $n$, even with the stronger conditions $n\equiv s_v\pmod{p_v^2}$. Thus three-level prime designs can use six vertices instead of the cube's eight. Check 8 of the evidence verifies the primality and cancellation, and verifies a twelve-point integer configuration cancelling four levels. Primality is not established for that four-level configuration. These examples do not determine the minimum number of primes for general $K$.

Consequences for the window

The phase of the binary tail at $n$ depends on the totients at about $\log_2n$ consecutive levels. The dilation page's averaged estimate takes $n$ in one class modulo $Q$ inside $[x/2,x]$ with many representatives, so $Q\le x$, and its prime-error analysis needs every dilation prime to exceed the number of levels. Under those conditions Proposition 1 gives $(K+1)\ln K<\ln x$, so

$$K<(1+o(1))\frac{\ln x}{\ln\ln x},\qquad \frac K{\log_2x}\longrightarrow0 .$$

Consequently a cutoff $H\asymp\log x$ leaves $H-o(\log x)$ levels uncancelled by this mechanism. The complete absolute tail estimate on the dilation page requires such a cutoff in its specified regime, including its common-divisor normalization. Its bound (8), with its extra hypotheses on the averaging modulus, gives integrality below about $(2-\varepsilon)\ln\ln x$ for each vertex and level separately outside a proportion $O(1/\ln\ln x)$ of the class. It does not give simultaneous integrality for all vertices and levels as their number grows. This is a bound on one mechanism, cancellation of levels for every function by dilations, under the stated conditions; it does not exclude cancellation that uses the values of $\varphi$, a single lucky representative of a class with $Q>x$, or a phase estimate for the remaining levels. It is consistent with the [[research/totient_binary_new_attack/perturbation_barrier|perturbation comparison]], whose separate scope is empirical distributions of shorter windows up to $o(1)$ error.