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Source. Section 4.5, physical pp. 13–14 of the selected author version. This page preserves the ordered template used again by Nielsen and Owens.

Target branch

The deleted holes are $1\pmod6$ and $3\pmod {18}$. Split both by the two odd branches modulo $4$. Equivalently, the prime-$11$ target is

$$x\equiv1\pmod4,\qquad \bigl(x\equiv1\pmod3\ \text{or}\ x\equiv3\pmod9\bigr). \tag{1}$$

The extra $81^\uparrow(1,\_)$ from the initial construction means that a complete package here needs only one input in each of $3$, $27$, and $27^\uparrow$. In the alternative bookkeeping used in two inputs, it needs one input in $3$ and the class $3\pmod9$.

The ten inputs

Define the following packages in the displayed order:

$$\begin{aligned} A_1={}&4,\\ A_2={}&8^\uparrow,\\ A_3={}&3\cdot2+27\cdot1+27^\uparrow\cdot2,\\ A_4={}&3\cdot4+27\cdot4+27^\uparrow\cdot8^\uparrow,\\ A_5={}&3\cdot8^\uparrow+9\cdot8^\uparrow,\\ A_6={}&5^\uparrow(1,2,3\cdot1,4),\\ A_7={}&5^\uparrow(8^\uparrow,3\cdot2+9\cdot2, 3\cdot4,3\cdot8^\uparrow+9\cdot8^\uparrow),\\ A_8={}&3\cdot3(1,2,4)+81^\uparrow(1,4)\\ &\quad+5^\uparrow(27^\uparrow\cdot1,27^\uparrow\cdot2, 27^\uparrow\cdot4,27^\uparrow\cdot8^\uparrow),\\ A_9={}&7^\uparrow(1,2,3\cdot1,5^\uparrow(1,2,x,4),4,8^\uparrow). \tag{2} \end{aligned}$$

For the last input, put

$$B=5^\uparrow(3(3(1,4,\_),\_,\_),\_,\_,\_), \tag{3}$$

and

$$\begin{aligned} C=7^\uparrow\bigl(&A_3,A_4,3\cdot8^\uparrow,\\ &5^\uparrow(3\cdot3(x,x,1)+9\cdot2, 8^\uparrow,x,3\cdot1+9\cdot4),\\ &A_8,\\ &5^\uparrow(3\cdot3(x,x,8^\uparrow),3\cdot2,3\cdot4, 3\cdot8^\uparrow)+9\cdot8^\uparrow\bigr),\\ A_{10}={}&B+C. \tag{4} \end{aligned}$$

The exact prime-$11$ package is therefore

$$\boxed{\mathcal T_{11}=11^\uparrow(A_1,A_2,\ldots,A_{10}).} \tag{5}$$

Complete template verification

The first two packages cover their target children directly. In $A_3$ and $A_4$, the three summands fill respectively the required $3$, $27$, and $27^\uparrow$ inputs. In $A_5$, the $9\cdot8^\uparrow$ summand supplies the class $3\pmod9$ left after the $3\cdot8^\uparrow$ part.

For $A_6$ and $A_7$, the third input of the surrounding $5^\uparrow$ already contains $3\pmod9$ by the prime-$5$ compatibility (7) on the initial page. The other displayed inputs cover the remaining children. Package $A_8$ uses the three available routes separately: $3\cdot3(1,2,4)$ supplies the required class modulo $3$, $81^\uparrow(1,4)$ supplies the required class modulo $27$, and its four $5^\uparrow$ inputs supply the four copies of the remaining $27^\uparrow$ package. Package $A_9$ uses the ordered prime-$7$ compatibility: the third input already has $3\pmod9$, and the $x$ in its nested $5^\uparrow$ is already covered.

In $A_{10}$, $B$ covers the classes $1,4\pmod9$ in the first $5^\uparrow$-input, uniformly over the later $7$-coordinate. The six entries of $C$ then fill all six children of its $7^\uparrow$: the first two are $A_3,A_4$, and the third is direct. In the fourth child, $3\cdot3(x,x,1)$ supplies the missing class $7\pmod9$ in its first $5$-input, while $9\cdot2$ supplies the separate class $3\pmod9$. The fifth child is $A_8$. In the sixth child, $3\cdot3(x,x,8^\uparrow)$ supplies the corresponding $7\pmod9$ classes and $9\cdot8^\uparrow$ supplies $3\pmod9$. The remaining $x$'s are the prime-$5$ and prime-$7$ compatibilities recorded on the initial page. Thus $A_{10}$ has no unresolved child.

All ten $A_i$ are complete packages on the selected branch. Their exact unbounded exponent-region partition is proved in the signature certificate; in particular their regular modulus sets are pairwise disjoint and none contains the prime $11$. Placing them in the ten inputs of $11^\uparrow$ therefore covers the target half and preserves regular injectivity.

The finite meaning of every displayed arrow in (2)–(5) is supplied by arrow finitization.

Used by. The prime-13 template, the prime-17 template, and Owens's construction.

Bears on. Problem 2.