Axler's global fourth- and fifth-order prime intervals improve sharply

What is proved

The theorem improves Axler's fourth- and fifth-order prime intervals on the identical range $x\ge2$. The interval is open at $x$ and closed at its right endpoint.

The sharp order-four gap

For $G_n(x)=x(1+B_n/\log^n x)$, the order-four derivative has one low critical point inside the gap $11<x<13$. The certificate proves $G_4(x)>13$ there and strict increase afterward.

An exhaustive sieve through $17{,}051{,}887$ compares every consecutive prime gap. The unique largest requirement occurs at $1327<1361$; the runner-up $113<127$ is lower by more than $6.6216$.

Why B4 is globally least

At $x=1327$, the right endpoint is exactly $1361$, so the proposed interval reaches the next prime with equality. Any smaller coefficient leaves the entire interval below $1361$.

This single prime-free interval proves actual full-range optimality for $B_4$, independently of the later analytic splice.

Bridges and numerical cells

Axler's order-three theorem carries the order-four interval from the finite sieve into the corrected Fiori--Kadiri--Swidinsky range. Once order four is established, it similarly carries order five into that range.

For a two-sided theta-error row $(L_i,\epsilon_i)$ and $y=x+B_nx/L^n$, positivity of $\theta(y)-\theta(x)$ follows from an exact cell inequality. The first Table 2 theta exponent is repaired from Proposition 17 before all 236 cells are compared.

The active cell and analytic tail

For both orders the unique active numerical cell is $2075\le L<2100$. Its order-four requirement is below $B_4$, while its order-five right-end supremum is exactly $B_5$. Because the cell is open at $L=2100$, that supremum is not attained; the next row gives a strict positive knot margin.

Beyond $L=10^8$, C-0053's absolute $\psi$ estimate and the standard prime-power decomposition give an absolute theta error. The certificate proves the error and its $L^n$ multiples decrease, so the prime-producing theta increment remains positive forever.