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Source. Sections 4.13–4.19, physical pp. 20–22 of the selected author version. The phrase “fill $r$ copies” below means that the available ordered package pool is divided among the open inputs without reusing an inner modulus signature at the same outer-prime level.

Prime 41

Restrict to the regular prime-$17$ hole created by deleting the modulus $17$. Besides the restrictions used in the prime-17 template, only one class modulo $17$ remains to be filled.

Let $A_1,\ldots,A_{15}$ be the fifteen complete packages constructed there, including the temporary atomic packages $1,2$, and keep this order. The packages

$$A_1,\ldots,A_{15},\qquad 17A_1,\ldots,17A_{15} \tag{1}$$

give thirty inputs. Put

$$T=(A_1,\ldots,A_{11},17A_1,\ldots,17A_{11}). \tag{2}$$

None of these twenty-two packages contains prime $7$, so $T$ fills one $23^\uparrow$; call the resulting package $P$.

Let $S$ be the partial sixteenth prime-$17$ package. It becomes complete after one remaining $7$-input (equivalently, one appropriate $5$-input) is filled. With $T$ as in (2),

$$S+23^\uparrow(7T) \tag{3}$$

is one further complete input.

Next let $T'$ consist of the first thirteen $A_i$, the corresponding thirteen $17A_i$, and $P$. On the restricted branch one input of a $29^\uparrow$ is already covered, so these $27$ packages give

$$29^\uparrow(x,T'),\qquad 17S+29^\uparrow(x,7T') \tag{4}$$

as two complete inputs. Here the already-covered first regular input is displayed explicitly, and $T'$ is assigned in order to inputs $2,\ldots,28$. The thirty packages in (1), in order, fill one $31^\uparrow$, giving another.

For one final composite package, put the first fifteen $A_i$ into the first fifteen inputs of a $(17^2)^\uparrow$, put $S$ into its last input, and complete the remaining part by a $31^\uparrow(T'')$. Here use the first thirty packages, in multiplier-major order, from

$$(17^2)^\uparrow\!\cdot uA_i,\qquad u\in(1,5,7,35),\quad i=1,\ldots,8. \tag{5}$$

Thus use $u=1$ with $A_1,\ldots,A_8$, then $u=5$, then $u=7$, and finally $u=35$ with $A_1,\ldots,A_6$. The factor $(17^2)^\uparrow$ is the selected last regular prime-$17$ input at all levels $k\ge2$, not another complete sixteen-input arrow. Its $17$-exponent is at least $2$, whereas the packages in (1) have $17$-exponent $0$ or $1$. The source's compatibility restrictions make the partial $(17^2)^\uparrow$ and this $31^\uparrow$ cover complementary children.

At this point there are $36$ complete inputs in the displayed construction order. Split that pool into two consecutive blocks of $18$ to form two $19^\uparrow$ packages, and use the same first-$36$ pool to form one $37^\uparrow$. These raise the count to $39$. Since $41^\uparrow$ has $40$ regular inputs, exactly one input remains empty. Its marked continuation is retained, and the empty regular input is split between primes $97$ and $101$.

Prime 43

Restrict instead to the other prime-$17$ hole, created by deleting modulus $34$. Repeat the prime-$41$ packages with every factor $17$ interpreted on the other compatible class modulo $17$. This gives an ordered pool of $39$ complete packages, none involving prime $41$. In a new $41^\uparrow$, the first regular input is already covered by the global prime-$41$ package because its first inner package is $A_1=1$. Put the $39$ translated packages into inputs $2,\ldots,40$. This supplies one more complete package, for a total of $40$. Thus exactly two of the $42$ regular inputs of $43^\uparrow$ remain empty. Prime $61$ fills this residual hole.

Prime 47

This stage completes the one regular input deliberately left open in the partial prime-19 template. Begin with $1,2$ and its seventeen complete packages, in that order, giving $19$ packages. On this restricted branch the first, fifth, sixth, and tenth inputs of $23^\uparrow$ are already covered. Put the first $18$ packages into the other $18$ inputs in increasing-input order; this produces a twentieth package.

Only one regular input of the surrounding $19^\uparrow$ is open. Multiplying the twenty packages separately by that fixed first-level $19$-condition produces twenty more packages. Fill one $(19^2)^\uparrow$ with the first $18$ packages. There are now $41$ packages. At each of the following steps use the first $q-1$ members of the current ordered pool to fill $q^\uparrow$:

$$29,\quad31,\quad37,\quad41,\quad43. \tag{5}$$

The count is

$$20+20+1+5=46,$$

exactly the number of regular inputs in $47^\uparrow$. Hence the old prime-$19$ hole is now complete.

Prime 53

Return to the modulus-$36$ hole and restrict to the third regular input of its $5^\uparrow$, complementary to the prime-$37$ stage. Start with the sixteen initial packages of that stage. For each $A_i$, apply that selected third input at every prime-$5$ level. This produces sixteen packages $5^\uparrow\!\cdot A_i$ with $v_5\ge1$, for $32$ total. It is the full unbounded family in one selected regular input, rather than a fixed factor $5$; the other three regular inputs belong to the complementary prime-$37$ target.

Sequentially fill five $7^\uparrow$, three $13^\uparrow$, one each of $31^\uparrow,37^\uparrow,41^\uparrow,43^\uparrow$, two $23^\uparrow$, and one $47^\uparrow$. For several copies of one arrow, split the shortest required prefix of the current pool into consecutive blocks of $q-1$ inputs. This adds $15$ packages and gives $47$. On the present branch the first two inputs of $11^\uparrow$ are already covered. Split the first $40$ packages into five consecutive blocks of eight and put them, in order, in inputs $3,\ldots,10$. These add five more packages. All $52$ regular inputs of $53^\uparrow$ are filled.

Primes 59 and 61

For prime $59$, assume the interface left open on the [[covering_systems/nielsen_2009_covering_system_smallest_modulus_40/primes_29_37|prime-$37$ page]]: an ordered pool of $35$ complete, pairwise signature-disjoint packages when the temporary atomic $1$ is retained. In each new $37^\uparrow$, inputs $1,\ldots,34$ are already covered and only inputs $35,36$ are open. Split the first $34$ packages into $17$ consecutive pairs; they fill $17$ copies of $37^\uparrow$. Together with the original $35$, this gives $52$ packages. One package for each of

$$41^\uparrow,\quad43^\uparrow,\quad47^\uparrow,\quad53^\uparrow$$

raises the count to $56$; at each step use the first $q-1$ packages of the current pool. Two of the $58$ inputs of $59^\uparrow$ remain open; prime $89$ completes them. This entire prime-$59$ paragraph is a valid conditional deduction from the stated interface; it does not construct that interface.

For prime $61$, return to the two-input hole left at prime $43$. The prime-$43$ stage supplied $40$ complete packages. In each new $43^\uparrow$, inputs $1,\ldots,40$ are already covered and only two inputs are open. Split the ordered pool into twenty consecutive pairs; they fill $20$ copies of $43^\uparrow$. The resulting $40+20=60$ packages fill every regular input of $61^\uparrow$.

Prime 67

This stage closes the partially filled gray hole left by the third prime-$5$ input, rather than the $20\pmod {25}$ class. Its exact ideal regular target is the union, over $k\ge2$, of

$$x\equiv4\pmod8,\qquad x\equiv3\pmod5,\qquad x\equiv2\cdot3^{k-1}-1\pmod {3^k}.$$

It is the second regular input of the $9^\uparrow$ rooted at $2\pmod3$. On this restricted branch it is enough to fill one input in an $16^\uparrow$, a $9^\uparrow$, or the relevant $5$-node.

Start with

$$1,\quad2,\quad4,\quad8,\quad16^\uparrow,$$

then take the required $3$-multiple of each and the required $9^\uparrow$-multiple of each. These are $15$ complete packages. Putting each separately into the open $5$-input gives $15$ more.

The original fifteen packages fill three complete $25^\uparrow$ packages: use packages $1$–$4$, $5$–$8$, and $9$–$12$ in the four regular inputs. Packages $13$–$15$ fill the first three regular inputs of a fourth. Its last input is completed by the cross-piece

$$7^\uparrow\bigl((5^2)^\uparrow\!\cdot A_1,\ldots, (5^2)^\uparrow\!\cdot A_6\bigr). \tag{6}$$

Here $(5^2)^\uparrow\!\cdot A_i$ is the selected missing prime-$5$ input at levels at least $2$. It supplies precisely the rectangle left by the partial fourth $25^\uparrow$. The first thirty packages also fill five $7^\uparrow$ packages in consecutive blocks of six. The running count is therefore

$$30+3+1+5=39. \tag{7}$$

Next extend the ordered pool by filling, in order,

$$\begin{gathered} 37^\uparrow,\ 41^\uparrow,\ 4(11^\uparrow),\ 43^\uparrow,\ 47^\uparrow,\ 2(23^\uparrow),\\ 4(13^\uparrow),\ 53^\uparrow,\ 3(19^\uparrow). \end{gathered} \tag{8}$$

At each step, consecutive blocks are taken from the shortest required prefix of the current pool. This adds $18$, giving $57$. On this branch the third, sixth, ninth, tenth, and eleventh inputs of $17^\uparrow$ are already covered. Split the first $55$ packages into five consecutive blocks of eleven and put them, in order, in the other regular inputs. Finally use the first $56$ packages in two blocks of $28$ for two $29^\uparrow$, and the first $60$ in two blocks of $30$ for two $31^\uparrow$. This gives

$$57+5+2+2=66,$$

exactly the number of regular inputs of $67^\uparrow$. The prime-$5$ gray hole is complete.

Verification

Every count above uses a complete input package or a union whose missing children are explicitly complementary. The ordered choices are part of the construction. Their exact unbounded exponent regions and fresh-prime block induction are checked on the later signature-certificate page. The prime-$59$ row, and the later prime-$89$ row that imports it, have the conditional scope stated above. All other rows on this page are independent of the missing second-$13$ allocation at prime $37$; in particular, prime $53$ uses only the first sixteen packages constructed before that step. In particular, changing a residue or child label is never used as evidence for a different modulus. The two translated prime-$17$ contexts are not identified: their regular holes are completed separately at $41,43,61,97,101$. The global coverage and regular-signature ledger is given on the construction ledger.

Bears on. Problem 2.