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Source. Sections 4.20–4.23, physical pp. 22–23 of the selected author version.

Primes 71, 73, 79, and 83

On $2\pmod4$, the only remaining part of the prime-$7$ branch $6\pmod7$ consists of

$$2\pmod5,\qquad (3\pmod5)\cap(2\pmod3). \tag{1}$$

Split the first class in (1) into its three classes modulo $3$. The four resulting holes are filled separately by $71,73,79,83$.

For any one of the four holes, begin with the ten packages

$$\begin{gathered} 1,\ 2,\ 4,\ 8^\uparrow,\ 3\cdot1,\ 3\cdot2,\ 3\cdot4,\ 3\cdot8^\uparrow,\\ 9^\uparrow(1,2),\qquad9^\uparrow(4,8^\uparrow). \end{gathered} \tag{2}$$

The relevant class modulo $3$ is the one fixed by that hole. Multiplying the ten packages separately by its fixed $5$-condition gives ten more. Use the first eight original packages in two consecutive blocks of four to fill two $(5^2)^\uparrow$ packages. This gives an ordered pool of $22$.

Apply the fixed $7$-condition separately to those $22$ packages. Use the first $18$ packages of the $7$-free pool, in three consecutive blocks of six, to fill three $(7^2)^\uparrow$ packages. The total is then

$$22+22+3=47. \tag{3}$$

From this pool, fill in order

$$\begin{gathered} 47^\uparrow,\quad4(13^\uparrow),\quad5(11^\uparrow), \quad53^\uparrow,\quad2(29^\uparrow),\quad2(31^\uparrow),\\ 59^\uparrow,\quad61^\uparrow,\quad4(17^\uparrow), \quad3(23^\uparrow),\quad41^\uparrow,\quad43^\uparrow,\\ 2(37^\uparrow),\quad4(19^\uparrow). \end{gathered} \tag{4}$$

For every repeated arrow, split the shortest required prefix of the current pool into consecutive blocks of $q-1$ packages. These operations add $32$ complete packages, giving $79$. Select the first $70,72,78$ packages to fill respectively all regular inputs of $71^\uparrow,73^\uparrow,79^\uparrow$. For the fourth hole, add one already completed package for each of

$$71^\uparrow,\qquad73^\uparrow,\qquad79^\uparrow. \tag{5}$$

The $79+3=82$ available packages fill all regular inputs of $83^\uparrow$. The optional $67^\uparrow$ mentioned by the source is not needed.

Prime 89

Conditional on the prime-$59$ interface stated on the preceding page, return to its two-input hole and reuse the ordered pool of $56$ complete packages. In a new $59^\uparrow$, the first $56$ inputs are already covered and the last two are open. Partition the pool into $28$ consecutive pairs and put one pair in those two inputs of each copy. This produces $28$ completed $59$-packages. Add one package for each

$$61^\uparrow,\quad67^\uparrow,\quad71^\uparrow,\quad 73^\uparrow,\quad79^\uparrow. \tag{6}$$

using the first $q-1$ members of the current pool at each step. There are $56+28+5=89$ packages. The first $88$ fill the regular inputs of $89^\uparrow$; the one-package surplus is immaterial. This verifies the local implication from the $56$-package prime-$59$ pool; it does not close the missing prime-$37$ allocation from which that pool depends.

Primes 97 and 101

Return to the one-input hole left at prime $41$. The prime-$41$ stage supplied $39$ complete packages. For each $A_i$, apply the selected empty regular prime-$41$ input at every level. This gives another $39$ packages $41^\uparrow\!\cdot A_i$, with $v_{41}\ge1$. It is the full unbounded family in that selected input rather than one fixed prime-$41$ level. Add one package for each of

$$53^\uparrow,\ 59^\uparrow,\ 61^\uparrow,\ 67^\uparrow,\ 71^\uparrow,\ 73^\uparrow,\ 79^\uparrow, \tag{7}$$

then two $43^\uparrow$, one $83^\uparrow$, one $89^\uparrow$, and one $47^\uparrow$. Use the shortest required prefix at each step, splitting the first $84$ packages into the two $43$-blocks. The total is

$$39+39+7+2+1+1+1=90. \tag{8}$$

Place these in the first $90$ inputs of a $97^\uparrow$, leaving six inputs open.

Reuse the same ordered $90$-package pool and partition it into fifteen consecutive blocks of six. On the present branch, the first $90$ inputs of each copy of $97^\uparrow$ are contextual $x$'s covered by the first partial $97$-package; one block fills its six open inputs. Only those six new pieces contribute leaves to each completed copy. Thus the unshifted $90$ packages and the fifteen new positive-$97$ packages give $90+15=105$ pairwise disjoint complete packages. Select $101$ and put them in the $101$ children of an ordinary $101$-node. The source explicitly uses $101$, without an arrow, so there is no further $101$-adic tail. This ordinary node, together with the partial $97^\uparrow$ coverage used in producing its inputs, closes the prime-$41$ hole.

Prime 103

The only remaining hole lies inside the partial sixteenth package from the prime-$17$ stage. In this context it suffices to fill two inputs of one $25^\uparrow$, one input of $7$, or one input of $17^\uparrow$.

Take the first eight packages of the prime-$17$ construction, including $1,2$. They use only primes $2,3$. Apply the needed fixed $5$-condition separately to obtain eight more. Group the original eight into four consecutive pairs and put each pair in the two open inputs of a $(5^2)^\uparrow$; the other two inputs are contextual $x$'s. There are $20$ packages. Apply the needed fixed $7$-condition separately to all $20$, and use the first $18$ packages in three consecutive blocks of six to fill three $(7^2)^\uparrow$ packages. This produces $23$ more, for $43$.

Sequentially fill four $11^\uparrow$, one $41^\uparrow$, four $13^\uparrow$, one $43^\uparrow$, and one $47^\uparrow$, always using consecutive blocks from the shortest required prefix. This adds $11$, giving $54$. The one-open-input condition at prime $17$ allows each package to fill that selected regular input at every level, producing $54$ further $17^\uparrow$ packages. There are $108$ available packages, so any $102$ fill all regular inputs of $103^\uparrow$. This closes the final hole.

Verification

The arithmetic checks are

$$\begin{array}{c|c|c} \text{outer prime or node}&\text{packages available}&\text{needed}\\ \hline 71&79&70\\ 73&79&72\\ 79&79&78\\ 83&82&82\\ 89&89&88\\ 97&90&90\text{ of }96\\ 101&105&101\\ 103&108&102. \end{array} \tag{9}$$

Surplus packages are omitted in the specified prefix order. The exact base pools, all repeated-arrow blocks, and the absence of each newly adjoined prime from its input pool are recorded on the later signature-certificate page. The four holes in (1) have different fixed $3$- and $5$-conditions, but that residue distinction is used only for coverage; modulus injectivity follows from the exponent-region and block checks. The prime-$101$ and prime-$103$ stages unconditionally close the residual holes recorded at primes $41$ and $17$. Prime $89$ closes the residual prime-$59$ hole conditionally on its input interface. Thus the local schedules on this page leave no new hole, but the full construction still inherits the prime-$37$ reconstruction boundary.

Bears on. Problem 2.