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Source. Section 4.6, physical pp. 14–15 of the selected author version.

Statement

Work on the $3\pmod4$ half of the same deleted classes $1\pmod6$ and $3\pmod {18}$ used by [[covering_systems/nielsen_2009_covering_system_smallest_modulus_40/prime_11_template|the prime-$11$ template]]. Thus the target is

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

There is an ordered list of twelve pairwise modulus-disjoint complete packages

$$\mathcal T_{13}=13^\uparrow(D_1,\ldots,D_{12}) \tag{1}$$

that covers this half.

Exact construction

For $1\le i\le10$, let $D_i$ be the package $A_i$ on the prime-$11$ page, with every contextual $4$ or $8^\uparrow$ interpreted on $3\pmod4$ rather than on $1\pmod4$. This changes residue positions but not modulus signatures.

For $D_{11}$, take an $11^\uparrow$ whose ten ordered inputs are the same prime-$11$ recipes, now on $3\pmod4$, and replace by $x$ every atomic entry ending in $2^0$ or $2^1$. Those children were covered on the first half, so only the complementary higher-$2$ pieces remain.

For $D_{12}$, start with the same modified $11^\uparrow$: again replace every entry ending in $2^0$ or $2^1$ by $x$, then replace each contextual $4$ by $1$ and each contextual $8^\uparrow$ by $2$. This uses the other half of each prime-$11$ input and supplies the twelfth package.

Complete proof

The ten packages $D_1,\ldots,D_{10}$ cover the first ten children because the prime-$11$ verification depends only on the local child relations, all of which are preserved by translating from $1\pmod4$ to $3\pmod4$.

On this translated branch, exactly half of each relevant $11^\uparrow$ was already covered by the prime-$11$ stage. Deleting the $2^0$- and $2^1$-ending entries removes precisely those previously used classes. The remaining higher-$2$ packages fill the uncovered half, giving $D_{11}$. Changing $4,8^\uparrow$ to $1,2$ moves to the complementary unused $2$-profiles, so the same child check gives $D_{12}$ without reusing a regular modulus.

Thus all twelve children in (1) are covered. The signature certificate shows that $D_{11}$ contains exactly the transformed prime-$11$ profiles with $v_2\ge2$, while $D_{12}$ contains exactly those with $v_2\in\{0,1\}$. The first ten inputs contain no prime $11$. Consequently the twelve regular signature sets are disjoint. Applying the finite-arrow lemma realizes (1) as a finite package.

Used by. The prime-17 template, the prime-19 template, and Owens's construction.

Bears on. Problem 2.