Positivity tests at generating-function singularities

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The attempted mechanism

If every sufficiently large integer is representable, subtracting the generating function of an integer tail from the representation generating function leaves nonnegative coefficients. This gives positive-semidefinite matrices at arbitrary collections of angles, apparently stronger than individual progression counts. The idea was to contradict this positivity using the exact power tracks and their singularities.

The calculation below proves that every fixed finite test involving rational or irrational limiting angles and $O(t)$ angular offsets reduces to the existing progression first moments. For the quintic problem, the strict first-moment floor makes every such matrix eventually strictly positive definite, without assuming cofiniteness. This rules out that analytic test, not a generating-function argument of growing complexity.

1. The exact boundary limits

Fix $r\ge3$. Let

$$A(z)=\sum_{n\in\mathcal P_r}z^n,\qquad U(z)=\sum_{n\ge1}R(n)z^n,$$

where $R(n)$ counts unordered representations with between one and $r$ positive $r$-powerful summands. Write $m$ for probability Haar measure on $\widehat{\mathbb Z}$, $\gamma=\Gamma(1+1/r)$, and define finite measures

$$\kappa=\sum_{d\in\mathcal C_r}d^{-1/r}(x\mapsto dx^r)_*m, \qquad \mu=\frac{\gamma^r}{r!}\kappa^{*r}. \tag{1}$$

The canonical weight makes the series converge. For $h\in\mathbb Q/\mathbb Z$, use $\widehat\mu(h)=\int\chi_h(z)\,d\mu(z)$, where $\chi_h$ is the corresponding profinite additive character.

For fixed $h=a/q$ and fixed real $u$,

$$\lim_{t\downarrow0}t\,U(e^{-t+i(2\pi h+tu)}) =\frac{\widehat\mu(h)}{1-iu}. \tag{2}$$

Indeed, the canonical expansion of $A$ and a Riemann sum on root residues modulo $q$ give

$$t^{1/r}A(e^{-t+i(2\pi h+tu)})\longrightarrow \gamma(1-iu)^{-1/r} \sum_d d^{-1/r}\frac1q\sum_{b\bmod q}e(adb^r/q).$$

The dominating bound for the $d$-term is

$$t^{1/r}\sum_{x\ge1}e^{-tdx^r}\le\gamma d^{-1/r},$$

obtained by integrating a decreasing function. It justifies the infinite core sum. The finite multiset cycle-index formula gives, uniformly in the argument of $z$ with $|z|=e^{-t}$,

$$U(z)=A(z)^r/r!+O_r(t^{-(r-1)/r}). \tag{3}$$

Every other monomial in the power sums $A(z^j)$ has at most $r-1$ factors, each $O_r(t^{-1/r})$. Equations (1)–(3) prove (2).

For every fixed irrational $\alpha$ and real $u$,

$$t\,U(e^{-t+i(2\pi\alpha+tu)})\longrightarrow0. \tag{4}$$

For each fixed core, the polynomial sequence $e(\alpha d x^r)$ has vanishing mean. This follows by induction using van der Corput differencing, starting from the geometric-sum bound for an irrational linear phase. Partial summation with the weight $e^{-(1-iu)tdx^r}$ gives the corresponding zero limit for $t^{1/r}A$. The same core majorant and (3) prove (4).

Fixed-core lattice counting also identifies

$$\lim_{X\to\infty}\frac qX \sum_{\substack{n\le X\\n\equiv a\ (q)}}R(n) =q\mu(a+q\widehat{\mathbb Z}). \tag{5}$$

The archimedean volume is $\gamma^r$, and the core weights are summable. Repeated summands and shorter representations contribute $o(X)$.

2. Positive definiteness adds no fixed-scale condition

If $S_r$ were cofinite, then for some fixed $N$ the function

$$H_N(z)=U(z)-\frac{z^N}{1-z}$$

would have nonnegative coefficients. Therefore, for any finite collection of angles $\theta_j(t)$, the Hermitian matrix

$$B_t(i,j)=tH_N(e^{-t+i(\theta_j(t)-\theta_i(t))}) \tag{6}$$

would be positive semidefinite. Its quadratic form is

$$t\sum_{n\ge0}[z^n]H_N(z)e^{-tn} \left|\sum_j v_j e^{in\theta_j(t)}\right|^2.$$

Take $\theta_j(t)=2\pi h_j+tu_j$, with rational $h_j$. Equations (2) and the elementary limit for $z^N/(1-z)$ show that the limiting kernel is

$$K(h,u)=\frac{\widehat\nu(h)}{1-iu},\qquad \nu=\mu-m. \tag{7}$$

This kernel is positive definite if and only if $\mu\ge m$. For sufficiency, its quadratic form is

$$\int_{\widehat{\mathbb Z}}\int_0^\infty e^{-x} \left|\sum_jv_j\chi_{h_j}(z)e^{iu_jx}\right|^2\,dx\,d\nu(z).$$

For necessity set all $u_j=0$ and take $h_j=j/q$. Discrete Fourier diagonalization gives the eigenvalues $q\mu(a+q\widehat{\mathbb Z})-1$. Nonnegativity for every $q,a$ means $\nu$ is nonnegative on every clopen cylinder, hence is a positive measure. By (5), this is precisely the condition that every progression first moment be at least one.

At the critical number of summands, the archimedean factor is the positive Laplace kernel $(1-iu)^{-1}$. It therefore adds no condition to positivity of the profinite measure.

3. Strict blindness in the quintic case

Suppose for some $\delta>0$ that every progression mean is at least $1+\delta$. Then $\nu\ge\delta m$. Fix $N$ and any finite collection of distinct pairs

$$(\alpha_j,u_j)\in(\mathbb R/\mathbb Z)\times\mathbb R, \qquad \theta_j(t)=2\pi\alpha_j+tu_j.$$

The actual matrices (6) are strictly positive definite for sufficiently small $t$, whether or not $S_r$ is cofinite.

Partition the indices according to rational differences of their $\alpha_j$. Irrational differences give zero limiting entries by (4). Within a block write $\alpha_j=\alpha_0+h_j$, with rational $h_j$. The limiting quadratic form is bounded below by

$$\delta\sum_h\int_0^\infty e^{-x} \left|\sum_{j:h_j=h}v_j e^{iu_jx}\right|^2dx.$$

Within each equal-$h$ group, distinct pairs have distinct $u_j$. Linear independence of these exponentials makes this expression strictly positive for every nonzero vector. Entrywise convergence of a fixed finite matrix then proves the assertion.

The quintic lower bound exceeds $2.468$, so it supplies such a $\delta$ unconditionally. Lower-order terms cannot overcome this positive limiting spectral margin in a fixed test.

The qualifications are essential: $N$, the matrix size, and the pairs $(\alpha_j,u_j)$ are fixed. Growing denominators, increasing matrix size, or other frequencies depending on $t$ are not covered. No statement about nonnegativity of the actual coefficients, or absence of missing integers, has been proved.