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Source. Sections 3.10–3.13, printed pp. 12–14, physical pp. 18–20 of the selected thesis.

Prime 29

On the $4$-hole, restrict to the first prime-$3$ input and simultaneously to the second and third prime-$5$ inputs. Begin with

$$1,2,4,8^\uparrow, 3\cdot1,3\cdot2,3\cdot4,3\cdot8^\uparrow, 9^\uparrow(1,2),9^\uparrow(4,8^\uparrow). \tag{1}$$

Using the two prime-$5$ targets creates five more packages. The sixteenth is

$$25^\uparrow(1,2,4,8^\uparrow) +25^\uparrow(3\cdot1,3\cdot2,3\cdot4,3\cdot8^\uparrow). \tag{2}$$

The source then displays four cross-packages. With $x$ denoting prior coverage, the first three are

$$\begin{aligned} &7^\uparrow(\_,\_,x,5\cdot1,5\cdot2,5\cdot4) +7^\uparrow(1,2,x,x,x,x),\\ &7^\uparrow(\_,\_,x,5\cdot8^\uparrow,5\cdot3\cdot1,5\cdot3\cdot2) +7^\uparrow(4,8^\uparrow,x,x,x,x),\\ &7^\uparrow(\_,\_,x,5\cdot3\cdot4,5\cdot3\cdot8^\uparrow, 25^\uparrow(1,2,4,8^\uparrow)) +7^\uparrow(3\cdot1,3\cdot2,x,x,x,x). \tag{3} \end{aligned}$$

The fourth is

$$\begin{aligned} &25^\uparrow(9^\uparrow(1,2),9^\uparrow(4,8^\uparrow),\_,\_)\\ &+7^\uparrow(\_,\_,x, 25^\uparrow(x,x,3\cdot1,3\cdot2), 25^\uparrow(x,x,3\cdot4,3\cdot8^\uparrow),\\ &\hspace{35mm}25^\uparrow(x,x,9^\uparrow(1,2), 9^\uparrow(4,8^\uparrow)))\\ &+7^\uparrow(3\cdot4,3\cdot8^\uparrow,x,x,x,x). \tag{4} \end{aligned}$$

After using the first sixteen packages to fill a $17^\uparrow$, (3)–(4) bring the pool to $21$. Two $11^\uparrow$, one $23^\uparrow$, two $13^\uparrow$, and two partially precovered $19^\uparrow$ bring it to $28$. The final source package is

$$\begin{aligned} &7^\uparrow(\_,\_,x,5\cdot9^\uparrow(1,2), 5\cdot(9^\uparrow(4,8^\uparrow)),B)\\ &\qquad+7^\uparrow(9^\uparrow(1,2),9^\uparrow(4,8^\uparrow),x,x,x,x), \tag{5} \end{aligned}$$

where $B$ is the $17^\uparrow$ filled by the first sixteen packages. This gives $29$ packages; deleting atomic $1$ leaves the required $28$.

Prime 31

The target is the deleted modulus-$36$ branch. The section first specifies

$$x\equiv1\pmod4,\qquad x\equiv6\pmod9. \tag{6}$$

Printed p. 13 later says “the branch $2\pmod4$.” That contradicts (6), the opening $1\pmod2$ branch, and the earlier $9\cdot4$ placement. The reconstruction therefore uses (6) and records the later phrase as a local source slip.

The fourteen starting packages are

$$1,2,4,8^\uparrow;\quad 3\cdot1,3\cdot2,3\cdot4,3\cdot8^\uparrow;\quad 9\cdot1,9\cdot2,9\cdot4,9\cdot8^\uparrow;\quad 27^\uparrow(1,2),27^\uparrow(4,8^\uparrow). \tag{7}$$

The first twelve fill three $5^\uparrow$, one chosen as $5^\uparrow(2,4,8^\uparrow,1)$. Put

$$C=5^\uparrow(27^\uparrow(1,2),27^\uparrow(4,8^\uparrow),\_,\_). \tag{8}$$

The first fifteen complete packages fill three $7^\uparrow$. A fourth package is the union of $C$ with

$$\begin{aligned} 7^\uparrow(&5^\uparrow(x,x,3\cdot1,3\cdot2), 5^\uparrow(x,x,3\cdot4,3\cdot8^\uparrow),x,\\ &5^\uparrow(x,x,9\cdot1,9\cdot2), 5^\uparrow(x,x,9\cdot4,9\cdot8^\uparrow), 5^\uparrow(x,x,27^\uparrow(1,2),27^\uparrow(4,8^\uparrow))). \tag{9} \end{aligned}$$

This yields $21$ packages. Three partially precovered $11^\uparrow$, two $13^\uparrow$, two partially precovered $17^\uparrow$, and one each of $19^\uparrow,23^\uparrow,29^\uparrow$ yield $31$; deleting atomic $1$ leaves $30$.

Prime 37

On the first prime-$5$ input of the $8$-hole, restrict first to the first two prime-$3$ children. Owens begins with

$$1,2,4,8,16^\uparrow, 3(1,2),3(4,8),3(3^\uparrow(1,2),3^\uparrow(4,8)), 3(16^\uparrow,\_)+5\cdot3(\_,16^\uparrow). \tag{10}$$

Multiplying the first eight by $5$, filling two $25^\uparrow$, and using six three-input $7^\uparrow$ gives $25$ packages. The stated continuation is two $13^\uparrow$, two thirteen-input $19^\uparrow$, then one each of $29^\uparrow,31^\uparrow$, three $11^\uparrow$, two $17^\uparrow$, and one $23^\uparrow$. It produces $37$ packages; deleting atomic $1$ leaves $36$.

The source supplies no ordered input lists for these completions. In particular, it does not identify which packages can be reused under the second $13^\uparrow$ without duplicating a prime-$5$ exponent family. This is the first sustained continuation built almost entirely from compressed allocations. It adds to, rather than begins, the explicit interfaces already retained for the prime-$19$, prime-$29$, and prime-$31$ selections.

Prime 41 and the count correction

The prime-$41$ target is the second prime-$3$ child in the second prime-$5$ child of the $4$-hole. The source schedule gives

$$4+6+1+1+12+3+2+1+6+1+2+1+2=42 \tag{11}$$

packages before the minimum-modulus deletion: four initial packages, six from $3$ and $9^\uparrow$, the $11$ and $13$ completions, twelve fixed prime-$5$ multiples, three $25^\uparrow$, two partially precovered $19^\uparrow$, $29^\uparrow$, six $7^\uparrow$, $31^\uparrow$, two $17^\uparrow$, $37^\uparrow$, and two partially precovered $23^\uparrow$.

Printed p. 14 calls this total $41$ and says that deleting $1$ leaves the inputs of $41^\uparrow$. The displayed arithmetic instead gives $42$. Deleting atomic $1$ leaves $41$ available packages; selecting any $40$ of them is enough at the capacity level, so one further surplus package is unused. Which package may be omitted while preserving every downstream ordered reuse belongs to the unresolved allocation interface. The corrected arithmetic does not assert an erratum to the thesis and does not by itself prove the distinct-modulus condition.

Scope

Formulas (1)–(10), the target correction (6), and all package arithmetic have been checked against the source. Where the source displays every input, notably in the cross-packages (3)–(5) and (9), their relative coverage follows from the shown $x$ masks and the stated earlier precoverage. The compressed prime-$37$ and prime-$41$ completions still require explicit residue maps. A full proof from prime $29$ onward also needs the source-omitted ordered allocation and a cross-stage unbounded signature check; that interface is stated on the construction ledger.