Sparse sequences with the same congruence and major-arc data

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Status and scope

The results below are proved. They concern synthetic integer sequences, not representation functions of a diagonal form. In particular, they do not prove either density conclusion for Erdős 940.

The construction preserves the exact cubic box congruence distribution modulo a modulus exceeding the square of the root scale, all fixed congruences, the joint profinite limit, and all the estimates in the major-arc note. Nevertheless its support has density zero at the output scale and its logarithmic multiplicity diverges. Thus the missing integer-fiber estimate cannot follow from these data alone. The construction does not preserve the factorization into three cubic Weyl sums; exploiting that factorization on the minor arcs remains possible.

Congruence-preserving sampling

Lemma. Let $R$ be a nonnegative integer sequence supported in $\{0,\ldots,B\}$, where $B\ge1$, and put $T=\sum_nR(n)>0$. For arbitrary positive integers $D,K$, there is a nonnegative integer sequence $\widetilde R$, supported inside $\operatorname{supp}R$, such that

$$\sum_{n\equiv a\pmod D}\widetilde R(n) =\sum_{n\equiv a\pmod D}R(n)\qquad(a\bmod D), \tag{1}$$ $$|\operatorname{supp}\widetilde R|\le T/K+D, \qquad \sum_{\widetilde R(n)\ge K}\widetilde R(n)\ge T-KD, \tag{2}$$

and, writing $e(t)=\exp(2\pi it)$,

$$\sup_{\alpha\in\mathbb R/\mathbb Z} \left|\sum_n(\widetilde R(n)-R(n))e(n\alpha)\right| \ll\sqrt{KT\log(2BT)}. \tag{3}$$

Every such sequence also satisfies

$$\frac1T\sum_n\widetilde R(n)\log\widetilde R(n) \ge\log K-\log(1+KD/T), \tag{4}$$ $$\sum_n\widetilde R(n)^2\ge\frac{KT}{1+KD/T}. \tag{5}$$

As usual $0\log0=0$.

Proof. Write the mass of class $a$ as $M_a=m_aK+r_a$, with $0\le r_a<K$. When $M_a>0$, sample independently $m_a$ outputs from the conditional law $R(n)/M_a$ in that class, assigning weight $K$ to each draw. If $r_a>0$, make one further independent draw with weight $r_a$. Add weights when several draws choose the same output.

Each class has exactly its prescribed mass, and the number of draws is at most $T/K+D$. The total weight of the remainder draws is less than $KD$; every weight-$K$ draw contributes to a fiber of size at least $K$. This proves (1) and (2) for every realization. Also $\mathbb E\widetilde R(n)=R(n)$.

Let $Z(\alpha)$ denote the sum in (3). The independent draws have weights $w_\ell\le K$, and $\sum_\ell w_\ell^2\le KT$. The elementary Hoeffding bound for the real and imaginary parts gives

$$\Pr(|Z(\alpha)|\ge u)\le4\exp(-u^2/(4KT)). \tag{6}$$

For completeness, the real part contributed by a weight-$w$ draw lies in an interval of length $2w$; applying the exponential-moment bound for a centered variable in an interval, followed by Markov's inequality, gives the real tail. Use threshold $u/\sqrt2$ for each part and a union bound to obtain (6).

Deterministically $\|Z'\|_\infty\le4\pi BT$, since both sequences have mass $T$ and support in $[0,B]$. On a uniform grid of $L=\lceil8\pi BT\rceil$ frequencies, take $u=2\sqrt{KT\log(8L)}$. Equation (6) and a union bound give probability at most $1/2$ of a grid violation. The nearest grid point changes $Z$ by at most $1/4$. Some realization therefore satisfies (3).

Finally, the probability law $\nu(n)=\widetilde R(n)/T$ has entropy at most $\log(T/K+D)$. The identity $T^{-1}\sum\widetilde R\log\widetilde R=\log T-H(\nu)$ proves (4), and Cauchy--Schwarz proves (5). $\square$

Application to the cubic box distribution

Put $F=x^3+2y^3+4z^3$, take $N=2^k\to\infty$, and let $R_N$ count representations with $1\le x,y,z\le N$. Thus $T=N^3$ and $B=7N^3$. Choose integers $m_N\to\infty$ so slowly that

$$L_N=\operatorname{lcm}(1,\ldots,m_N)\le\sqrt{\log(2N)}.$$

For example, take the largest $m\le\lfloor\log N\rfloor$ satisfying this inequality. Set

$$D_N=N^2L_N, \qquad K_N=\left\lfloor N/\log(2N)\right\rfloor.$$

For sufficiently large $N$, $K_N\ge1$ and $K_ND_N/N^3\le1/\sqrt{\log(2N)}\to0$. The lemma gives

$$|\operatorname{supp}\widetilde R_N|\ll N^2\log N=o(N^3), \tag{7}$$ $$N^{-3}\sum_n\widetilde R_N(n)\log\widetilde R_N(n) \ge\log N-\log\log N-O(1), \tag{8}$$ $$\sum_n\widetilde R_N(n)^2\gg N^4/\log N. \tag{9}$$

The congruence marginals agree exactly modulo $N^2$, and eventually modulo $N^2q$ for every fixed $q$. In particular, every residue modulo $N$ has mass exactly $N^2$, by the special cubic's dyadic uniformity. No uniformity or bounded second moment modulo $N^2$ is asserted.

With $W_N=\sum_nR_N(n)e(n\alpha)$ and $\widetilde W_N=\sum_n\widetilde R_N(n)e(n\alpha)$, (3) gives

$$\|\widetilde W_N-W_N\|_\infty\ll N^2. \tag{10}$$

Thus the normalized Fourier transforms differ by only $O(N^{-1})$ uniformly, and agree exactly at every rational frequency whose denominator divides $D_N$.

The joint profinite limit also agrees

Let

$$\rho_{N;q}(a)=\frac q{N^3}\sum_{n\equiv a\pmod q}R_N(n).$$

If $q\le N/2$, each root residue occurs between $N/q-1$ and $N/q+1$ times. Comparing each residue triple with the complete residue box gives

$$(1-q/N)^3\rho_q(a)\le\rho_{N;q}(a) \le(1+q/N)^3\rho_q(a), \tag{11}$$

where $\rho_q$ is the complete density from the profinite note. Hence

$$\|\rho_{N;q}-\rho_q\|_2\le5(q/N)\|\rho_q\|_2 \le5\sqrt{\mathfrak S}\,q/N. \tag{12}$$

The norm here uses normalized counting measure modulo $q$.

Exact marginal preservation at $q=L_N$ and the proved profinite convergence show

$$\|\widetilde\rho_{N;L_N}-\rho\|_{L^2(\widehat{\mathbb Z})} \ll L_N/N+m_N^{-1/3}\longrightarrow0. \tag{13}$$

In particular, the joint congruence second moments tend to $\mathfrak S$, and their entropy deficits are at most $\log\mathfrak S+o(1)$. This bounded coarse entropy coexists with (8).

Every established major-arc estimate survives

For the arcs $\mathfrak M(N,Q)$ in the major-arc theorem, $1\le Q\le N^{1/2}$,

$$|\mathfrak M(N,Q)|\le2Q^2/N^3, \qquad \int_{\mathfrak M}|W_N|^2\ll N^3.$$

Equation (10) implies $\int_{\mathfrak M}|\widetilde W_N-W_N|^2\ll NQ^2$. Cauchy--Schwarz therefore gives

$$\left|\int_{\mathfrak M}|\widetilde W_N|^2 -\int_{\mathfrak M}|W_N|^2\right| \ll N^2Q+NQ^2\ll N^2Q^2. \tag{14}$$

Consequently exactly the same asymptotic holds:

$$\int_{\mathfrak M(N,Q)}|\widetilde W_N|^2 =N^3\mathfrak J\mathfrak S+ O(N^3Q^{-2/3}+N^2Q^2). \tag{15}$$

The pointwise approximation survives too: (10) and $W_N=M_N+O(QN^2)$ imply $\widetilde W_N=M_N+O(QN^2)$, with the same main function $M_N$. Thus its integrated squared approximation error remains $O(NQ^4)$.

In contrast, put $E_N=\{n:\widetilde R_N(n)\ge K_N\}$ and $P_{E_N}(\alpha)=\sum_{n\in E_N}e(-n\alpha)$. Then

$$|E_N|\ll N^2\log N, \qquad \sum_{n\in E_N}\widetilde R_N(n)=N^3+O(K_ND_N).$$

The major-arc correlation is $O(N^{5/2}\sqrt{\log N})$ by Cauchy--Schwarz and (15). Subtracting it from the full Fourier integral gives

$$\int_{\mathfrak M(N,Q)^c}\widetilde W_N P_{E_N} =N^3+O(K_ND_N+N^{5/2}\sqrt{\log N})=(1+o(1))N^3. \tag{16}$$

So essentially all the mass of the heavy fibers can lie in the missing minor-arc correlation, while the preceding congruence and major-arc estimates remain valid. An argument for the actual polynomial must use additional information that excludes this construction; none is supplied by this lemma.