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Source. Sections 3.14–3.20, printed pp. 14–18, physical pp. 20–24 of the selected thesis.

Owens describes these stages by naming a target branch, counting available complete packages, and using earlier precoverage to reduce the number of inputs needed by selected arrows. The following is the exact arithmetic of that schedule. A count of $r\cdot q^\uparrow$ means $r$ new complete packages, not $r(q-1)$ new leaves: earlier packages serve as its regular inputs and stay in the pool.

Primes 43, 47, and 53

For prime $43$, the target is the middle prime-$3$ input in the fourth prime-$5$ input of the $4$-hole. The progression is

$$4\xrightarrow{5,25^\uparrow}9\xrightarrow{11^\uparrow}10 \xrightarrow{3,9^\uparrow}25\xrightarrow{5\cdot7^\uparrow}30 \xrightarrow{31,29,2\cdot17}34 \xrightarrow{23,37}36\xrightarrow{3\cdot13}39 \xrightarrow{3\cdot19}42. \tag{1}$$

For prime $47$, work in the first prime-$3$ input of that same prime-$5$ branch. Start from the first $25$ packages of (1), with the prime-$3$ and prime-$9$ packages now taken in the first class modulo $3$. Seven partially precovered $7^\uparrow$ give $32$; two $17^\uparrow$, one each of $29^\uparrow,31^\uparrow$, and three $13^\uparrow$ give $39$; three thirteen-input $19^\uparrow$ give $42$; and $41^\uparrow,43^\uparrow$, and two $23^\uparrow$ give $46=47-1$.

For prime $53$, the target is the last prime-$5$ input of the $4$-hole, where only one $25\cdot3^\uparrow\cdot4$ branch remains. Four atomic/two-adic packages and their prime-$3$ uses give $8$. Prime-$5$, prime-$25$, and two $125^\uparrow$ completions give $26$. Two thirteen-input $19^\uparrow$ give $28$, followed by $29^\uparrow$ and the partially precovered $31^\uparrow$ to give $30$. Six five-input $7^\uparrow$ give $36$. Then three $13^\uparrow$, one $37^\uparrow$, four $11^\uparrow$, and one each of $41^\uparrow,43^\uparrow$ give $46$; one $47^\uparrow$, two $23^\uparrow$, and three $17^\uparrow$ give $52=53-1$.

Primes 59, 61, 67, and 89

For prime $59$, the target is the last prime-$3$ input in the first prime-$5$ input of the $8$-hole. The five starting packages $1,2,4,8,16^\uparrow$ become $12$ after the prime-$3$ and prime-$9$ steps, then $25$ after the prime-$5$ step. The last of those is the cross-package

$$5\cdot9^\uparrow(16^\uparrow,\_) +9^\uparrow(\_,16^\uparrow). \tag{2}$$

Three $25^\uparrow$ give $28$; $29^\uparrow$ and $23^\uparrow$ give $30$; ten three-input $7^\uparrow$ give $40$; and the source's remaining one $41$, four $11$, one $43$, one $47$, one $37$, three $17$, four $13$, and three $19$ arrows add $18$, giving $58=59-1$.

The prime-$61$ branch begins with $28$ packages built as for prime $59$, adds $29^\uparrow$ and the partially precovered $31^\uparrow$, and then nine four-open-input $7^\uparrow$ to reach $39$. The subsequent schedule adds

$$3+1+1+1+2+1+3+4+1+6+1=24 \tag{3}$$

packages, in the order

$$3\cdot19, 37, 41, 43, 2\cdot23, 47, 3\cdot17, 4\cdot13, 53, 6\cdot11, 59.$$

Thus the source constructs $63$ available packages. A $61^\uparrow$ needs only $60$ regular inputs, so three are surplus at this stage.

On the complementary prime-$2$ target, repeat the same construction to obtain another $63$-package pool. Two doubled $61^\uparrow$ packages raise the count to $65$, and one $89^\uparrow$ package supplies the last of the $66$ regular inputs of $67^\uparrow$. This is the only use of regular prime $89$.

Primes 71, 73, 79, and 83

For prime $71$, begin with $1,2,4,8,16^\uparrow$ and one partially precovered $7^\uparrow$. Prime-$3$ and prime-$9$ packages bring the count to $18$. One each of $17^\uparrow,19^\uparrow$, two $11^\uparrow$, and two partially precovered $13^\uparrow$ give $24$. Prime-$5$ and prime-$25$ packages give $54$. The remaining arrows add

$$1+1+1+1+1+1+2+1+3+3+1=16, \tag{4}$$

in the order $53,47,43,41,59,37,2\cdot31,61,3\cdot23, 3\cdot29,67$, giving $70$. Repeating this on the complementary half and adding two $71^\uparrow$ packages gives the $72$ inputs for $73^\uparrow$.

For prime $79$, the seven packages $1,2,4,8,16,32,64^\uparrow$ become $14$ after the prime-$3$ step, $21$ after seven two-input $7^\uparrow$, and $47$ after the prime-$5$ and prime-$25$ steps. The later list

$$47, 3\cdot17, 5\cdot11, 2\cdot29, 59, 53, 61, 2\cdot31, 5\cdot13, 67, 3\cdot23, 4\cdot19, 71, 73 \tag{5}$$

adds $31$ packages, giving $78$. Finally one $79^\uparrow$, one $43^\uparrow$, and two $41^\uparrow$ added to a translated $78$-package pool give the $82$ inputs of $83^\uparrow$.

Scope

Equations (1)–(5) and all intermediate totals reproduce the source exactly. They prove that enough packages are claimed at every stage. The source does not list the ordered packages used in most of these arrows or the exact earlier residue inputs denoted by its prose precoverage assertions. In particular, the counts do not exclude a repeated unbounded prime exponent when a package is nested under a prime it already contains. The entire page therefore remains conditional on the ordered-allocation interface stated on the construction ledger.