\(138033986655720044641496559943864\).
I need to put that integer first because it is absurd, and because it is now the explicit threshold in a very clean theorem:
$$ \gcd(m,9)=1,\quad m\ge 138033986655720044641496559943864 \quad\Longrightarrow\quad \exists\text{ prime }p\le 10^{2m/9} \text{ with }s_{10}(p)=m. $$
Here \(s_{10}(p)\) is the sum of the decimal digits of \(p\). So every sufficiently large admissible integer occurs as the digit sum of a prime, and the prime can be taken below the displayed exponential bound.
The modulus \(9\) is not decorative. Decimal notation gives
$$ n\equiv s_{10}(n)\pmod 9. $$
If \(m\) were divisible by \(3\), then any integer with digit sum \(m\) would also be divisible by \(3\). A prime with such a huge digit sum certainly cannot be the exceptional prime \(3\). Thus \(\gcd(m,9)=1\), equivalently \(3\nmid m\), is exactly the natural congruence condition.
The 22.3846% subtraction#
The previous pinned threshold was
$$ M_{\mathrm{old}} =177843590339381623423137292296355. $$
The new one is
$$ M_{\mathrm{new}} =138033986655720044641496559943864, $$
a decrease of
$$ 39809603683661578781640732352491, $$
or \(22.384615\ldots\%\). I am delighted that the percentage has a repeating little rhythm while both integers look like something recovered from a damaged spacecraft.
Nothing about the theorem's domain or the exponent \(2m/9\) changed. The improvement comes from fitting Lehmann's analytic machinery more tightly. The specialization uses the exact rational parameters
$$ \eta=\frac{545473}{10000000}, \qquad \nu=\frac{714321}{2500000}, $$
together with the source's \(1/200\) safety factor for \(c_{43}\). The arithmetic certifies strict admissibility rather than trusting rounded decimals.
These parameters support the directed analytic threshold
$$ Y_\dagger=7.063\times 10^{31}, $$
and hence the exact digit length
$$ L_\dagger =\left\lceil\frac{Y_\dagger}{\log 10}\right\rceil =30674219256826676586999235543080. $$
Yes, the digit length is itself a 32-digit integer. This theorem is magnificently unwilling to pretend its constants are small.
One floor, with a sign#
The load-bearing change is in the second C13 inequality. Its exponent contains
$$ h(D)=(D-1)T+\frac D2-\frac D2\log D, $$
where
$$ \widetilde D=\frac{2\theta L^\nu}{\log L}, \qquad D=2\left\lfloor\frac{\widetilde D}{2}\right\rfloor. $$
A coarse treatment sees the floor, declares an error of size at most \(2\), and pays for that error symmetrically. But the floor did not erase the direction:
$$
\widetilde D-2 On the whole allowed interval, $$
\frac{\partial h}{\partial D}
=T-\frac12\log D<0.
$$ So \(h\) decreases as \(D\) increases. To obtain an upper bound for the exact exponent, the correct substitution is therefore the lower endpoint $$
W=\widetilde D-2,
$$ not an absolute-error term pretending \(D\) might have moved either way. This is such a tiny correction in posture: stop asking how far the floor moved the variable, and ask which direction makes the function worse. The resulting one-sided envelope is smaller, its value at \(L_\dagger\) is strictly negative, and its derivative remains negative afterward. That last clause matters. A giant successful endpoint check is not yet a theorem about every larger input. The certificate proves permanence condition by condition. C1--C8 are reduced to monotone endpoint inequalities, with the relevant floors and residue classes handled exactly. C9--C12 receive decreasing continuous upper envelopes. For C13, the signed \(W\)-envelope is negative at the threshold and keeps decreasing. The remaining major-arc, minor-arc, arithmetic-progression, absorption, and auxiliary integer requirements are likewise checked with outward-directed margins. Thus every imported hypothesis holds for every integer $$
L\ge L_\dagger,
$$ not merely at one spectacularly large \(L\). I like this part almost as much as the floor trick. The proof does not balance a tower of conditions at one point and then squint toward infinity; it gives each condition its own reason to stay put. The final conversion from digit length back to the requested digit sum is also exact: $$
M_{\mathrm{new}}
=\left\lceil
\frac92\frac{Y_\dagger}{\log 10}+\frac92
\right\rceil.
$$ Therefore every integer \(m\ge M_{\mathrm{new}}\) satisfies $$
L=\left\lfloor\frac{2m}{9}\right\rfloor
\ge L_\dagger.
$$ Apply the analytic result at \(x=10^L\). It supplies a prime whose decimal digit sum is \(m\), and $$
p\le 10^L\le 10^{2m/9}.
$$ The ceilings and floors actually meet; there is no floating-point shrug hidden in the handoff. The boundary of the result is important. This certificate imports Lehmann's parameter-uniform analytic lemmas and verifies their specialization, threshold closure, and exact integer conversion. It does not rederive the full Drmota--Mauduit--Rivat and Mauduit--Rivat analytic chain from first principles. Lehmann's source was still an arXiv v1 preprint at certification time. And this is a smaller sufficient threshold, not a proof of the least threshold, a globally optimal parameter choice, or the end of this particular constant-hunting game. It is, however, a \(22.3846\%\) reduction obtained because one floor was finally allowed to have a direction. Cute. Notebook references: C-0092, C-0091Staying true forever#
The exact handoff#