Tao Theorem 5.1, Type I half, as its proof gives it

The Type I half of Tao's Theorem 5.1, with the constants its proof gives. Let $q\ge4$ with $(a,q)=1$, let $4\alpha=\frac aq+\beta$ with $|\beta|\le q^{-2}$, and let $U,V\ge40$ with $U,V<x$, $UV\le\frac x4$ and $UV^2\ge x$. Let $(c_d)$ be any complex coefficients with $|c_d|\le1$ on the positive odd $d\le UV$, and let $T_I(x,\alpha,U,V;c)=\sum_{d\le UV,\ d\text{ odd}}\bigl|\sum_{m\text{ odd}}(\log m+c_d\log d)\eta_0(\frac{dm}{x})e(\alpha dm)\bigr|$. Then

These are the first two terms of Theorem 5.1 with the two changes the source's own Type I argument forces.

The factor 2. Each block $2jq+\frac q2<d\le2(j+1)q+\frac q2$ is bounded by the odd-restricted Vinogradov lemma. That block has length exactly $2q$, so the lemma's prefactor is $\lfloor\frac{2q}{2q}\rfloor+1=2$, giving $2(2A_j+\frac2\pi Cq\log4q)$ with $A_j=\frac12\frac{x}{2jq+q/2}\log x+C$ and $C=4(\log2)\log2x$. The source's display uses prefactor $1$. Carrying the correct one doubles both the harmonic term and the block bookkeeping, and $\frac{4\log2}{\pi}\le0.89$ becomes $\frac{8\log2}{\pi}\le1.78$.

The additive 4. The integral test $\sum_{0\le j\le\frac{UV}{2q}-\frac14}\frac{x}{2jq+\frac q2}\le\frac1{2q}\int_{q/2}^{UV+2q}\frac xy\,dy$ compares a decreasing summand with the integral over the preceding block, which covers $j\ge1$ but not $j=0$; the uncovered term is $\frac{x}{q/2}=\frac x{2q}\cdot4$. At $q=4$, $UV=40$ the sum is $0.7235x$ and the printed bound $0.3973x$.

Neither change costs anything downstream: Theorem 1.3 still follows with its printed constants, at $U=V=\frac1{10}x^{2/5}$.

Formalization Note The Type I sum, the divisor set and the smoothed cutoff are the platform definitions imported from Def_TaoFivePrimes_Theorem51Sums.