Abstract
It is easy to assume that whether points can be grouped into triples so that every pair shares exactly one triple can only be settled by trying each number of points in turn. This paper searches for such triple systems from $3$ to $15$ points and counts what decides whether they exist. No new theorem or law is claimed.
First, only five point counts are found: of $3$ to $15$ points, triple systems were found only for $3$, $7$, $9$, $13$ and $15$ (Section 2).
Second, and this is the core. Existence is decided by the number of points modulo six: the five found all have remainder $1$ or $3$, and the eight not found all have remainder $0$, $2$, $4$ or $5$ (Section 2).
Third, every absence is explained by divisibility: for $5$, $8$, $11$ and $14$ points the number of triples is $\mathbf{3.333333}$, $\mathbf{9.333333}$, $\mathbf{18.333333}$ and $\mathbf{30.333333}$, not an integer, and for $4$, $6$, $8$, $10$, $12$ and $14$ points the number of partners of a point is odd (Section 2).
Fourth, when they exist, there are not few: counted with labels, there are $30$ systems for $7$ points and $840$ for $9$ (Section 3).
Fifth, up to isomorphism each is unique: applying every relabelling to one system found gives exactly $30$ and $840$ distinct systems, and the relabellings leaving it unchanged number $168$ and $432$ (Section 3).
Sixth, the separator is whether a system is absent by the divisibility conditions or absent even though they hold: up to $15$, every absence was by the divisibility conditions, and none was absent with the conditions holding (Section 4).
Keywords: Steiner triple system, combinatorial design, pairs of points, remainder modulo six, isomorphism, Kirkman.
Scope (what is not claimed)
No new theorem or law is claimed: that collections in which every pair lies in exactly one triple (called Steiner triple systems) exist exactly when the number of points leaves remainder $1$ or $3$ on division by $6$ is a known theorem since Kirkman ($1847$). That there are $30$ for $7$ points and $840$ for $9$, each unique up to isomorphism, is also known. The theorem is not proved: the construction for every point count with remainder $1$ or $3$ is not reproduced; only $3$ to $15$ points were searched. Sixteen or more is not searched: counts for larger numbers of points are not treated. Uses are not treated: use in experimental design or tournament scheduling is not treated. Relation to earlier papers: Paper 604 counted that no Latin square of order six has an orthogonal mate, and stated as a known result the side on which Euler’s conjecture (no pair for orders leaving remainder $2$ on division by $4$) fails. This paper stands beside it. There, a remainder condition that held within the computed range was known to fail beyond it. Here, the remainder condition holds for every count up to $15$ and is known to hold beyond, as Kirkman’s theorem. What is added is searching whether systems exist from $3$ to $15$ points, laying out that every absence is explained by divisibility, confirming the counts for $7$ and $9$ points in two ways, and placing the separator on absent by the conditions versus absent despite them.
1. Notation and the Question
Setting: there are $v$ points, and some triples of points are chosen. When every pair lies in exactly one triple, the collection is called a triple system. Each triple covers three pairs, so the number of pairs must be divisible by $3$. And the $v-1$ partners of a point are covered two at a time by the triples containing it, so $v-1$ must be even.
$$\text{pairs}=\frac{v(v-1)}{2},\qquad \text{triples}=\frac{v(v-1)}{6},\qquad v-1\ \text{even}$$The search takes the uncovered pair with the smallest labels, tries third points in turn, and backs up when covering fails. For $7$ and $9$ points every system was counted; for other counts the search ran until one was found or the search was exhausted.
Question: can the existence of a triple system only be settled by trying each number of points?
2. Counts With and Without Systems
Computation 2.1: triple systems were searched for from $3$ to $15$ points.
| points | remainder mod $6$ | triples | partners | found |
|---|---|---|---|---|
| $3$ | $3$ | $1.000000$ | $2$ | yes |
| $4$ | $4$ | $2.000000$ | $3$ | no |
| $5$ | $5$ | $3.333333$ | $4$ | no |
| $6$ | $0$ | $5.000000$ | $5$ | no |
| $7$ | $1$ | $7.000000$ | $6$ | yes |
| $8$ | $2$ | $9.333333$ | $7$ | no |
| $9$ | $3$ | $12.000000$ | $8$ | yes |
| $10$ | $4$ | $15.000000$ | $9$ | no |
| $11$ | $5$ | $18.333333$ | $10$ | no |
| $12$ | $0$ | $22.000000$ | $11$ | no |
| $13$ | $1$ | $26.000000$ | $12$ | yes |
| $14$ | $2$ | $30.333333$ | $13$ | no |
| $15$ | $3$ | $35.000000$ | $14$ | yes |
Observation 2.2 this is the core. Existence is decided by the number of points modulo six. Systems were found for five counts, $3$, $7$, $9$, $13$ and $15$, all with remainder $1$ or $3$: the eight without have a non-integer number of triples ($5$, $8$, $11$, $14$ points) or an odd number of partners ($4$, $6$, $8$, $10$, $12$, $14$ points). Both divisibility conditions hold exactly when the remainder is $1$ or $3$. Only for $11$ points did exhausting the search take more than $20$ seconds; the other counts took under $1$ second.
3. How Many When They Exist
Computation 3.1: all labelled triple systems were counted for $7$ and $9$ points. As a separate method, one system was found by a different search, every relabelling of the points ($5040$ for $7$ points, $362880$ for $9$) was applied, and the distinct systems produced were counted.
| points | systems counted | systems from relabelling | relabellings leaving it unchanged |
|---|---|---|---|
| $7$ | $30$ | $30$ | $168$ |
| $9$ | $840$ | $840$ | $432$ |
Observation 3.2 up to isomorphism each is unique. The number of distinct systems produced by relabelling one system equalled the number counted by search. That is, every system is a relabelling of the one found. $5040=30\times168$ and $362880=840\times432$.
4. The Separator
Observation 4.1 the separator is whether a system is absent by the divisibility conditions or absent even though they hold. An absence by the divisibility conditions is known without trying a single triple: an absence with the conditions holding could only be known by trying. Up to $15$ points, every absence was of the first kind, and there was none of the second.
Observation 4.2 placed beside Paper 604. 604 counted that no Latin square of order six has an orthogonal mate, and stated as a known result that Euler’s conjecture, “no pair for orders leaving remainder $2$ on division by $4$”, fails beyond the computed range. Here the remainder condition holds for every count up to $15$ points and is known to hold beyond, as Kirkman’s theorem. Of two guesses stated by a remainder, one failed and one held.
Observation 4.3 what is not said. A proof of the theorem, a construction for every count with remainder $1$ or $3$, counts for $16$ or more points, and use in experimental design or tournament scheduling are not treated. What is shown is that, up to $15$ points, the existence of a triple system is decided by the remainder modulo six, and every absence is explained by divisibility.
5. Census and Closing
- triples $v(v-1)/6$ and partners $v-1$ (Section 1): two divisibility conditions.
- found for $3$, $7$, $9$, $13$, $15$ (Computation 2.1): only five.
- remainder $1$ or $3$ (Observation 2.2): the core.
- $\mathbf{3.333333}$, $\mathbf{9.333333}$ (Computation 2.1): not integers for $5$ and $8$ points.
- $\mathbf{18.333333}$, $\mathbf{30.333333}$ (Computation 2.1): nor for $11$ and $14$.
- odd number of partners (Observation 2.2): $4$, $6$, $8$, $10$, $12$, $14$ points.
- $30$ for $7$ points (Computation 3.1): $30$ by relabelling too.
- $840$ for $9$ points (Computation 3.1): $840$ by relabelling too.
- $168$ and $432$ (Observation 3.2): unique up to isomorphism.
- absent by the conditions, or despite them (Observation 4.1): the separator; Paper 604.
Closing: Whether triples can cover every pair exactly once is decided by the number of points modulo six: up to fifteen, they exist only for remainders one and three. Searching from $3$ to $15$ points found systems only for $3$, $7$, $9$, $13$ and $15$: the eight without had a non-integer number of triples or an odd number of partners per point, all known to be absent before trying any triple. There were $30$ systems for $7$ points and $840$ for $9$, relabelling gave the same numbers, and each is unique up to isomorphism. Here the remainder condition did not just predict; it was the reason for every absence. Placed among the earlier papers: beside the Latin squares of Paper 604, where a guess stated by a remainder failed beyond the computed range, we placed triples for which the remainder condition holds. To be honest: the search went only up to $15$ points, and that the condition holds for every count rests on Kirkman’s known theorem.
References
- Colbourn, C. J., & Dinitz, J. H. (Eds.) (2007). Handbook of Combinatorial Designs (2nd ed.). Chapman & Hall/CRC.
- Kirkman, T. P. (1847). On a problem in combinations. Cambridge and Dublin Mathematical Journal.
- Yamagishi, Y. (2026). No Latin Square of Order Six Has an Orthogonal Mate [Paper 604]. YOI: https://www.yamagishiyuuki.com/yoi/604; DOI: 10.5281/zenodo.23014438
Reference
Yamagishi, Y. (2026). Triples Covering Every Pair Exactly Once Exist, up to Fifteen Points, Only When the Count Leaves Remainder One or Three Mod Six [Paper 898]. YOI: https://www.yamagishiyuuki.com/yoi/898
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 898. This paper is a preprint and has not been peer reviewed.