The last three contacts go at once

A closed cage of three-bonded atoms with pentagonal and hexagonal faces has exactly twelve pentagons, whatever its size, as twelve pentagons and no way round them showed from the Euler characteristic. The twelve is owed to three faces meeting at every atom, as the twelve belongs to the vertex showed, and on a torus the same count leaves pentagons free as long as heptagons balance them. How close the twelve must be asked where they sit. It counted the contacts, the pairs of pentagons that share a bond, in every cage to forty-four atoms. The fewest contacts fell from thirty at the dodecahedron to eight at forty-four atoms. After the first few sizes they fell by one for each hexagon added, over eight consecutive sizes, and running that rate on reached none at sixty atoms. That is exactly where the truncated icosahedron sits, the first cage whose pentagons are all apart. The essay stated the prediction plainly: 8, 7, 6, 5, 4, 3, 2, 1, 0 from forty-four atoms to sixty, which one more size would begin to test.

Every size to sixty is now counted. The sequence is 8, 8, 7, 5, 5, 4, 4, 3, 0. The endpoint is right and the path is not. The fall pauses at forty-six atoms, drops by two at fifty, pauses again at fifty-two and fifty-six, and loses its last three contacts in a single step from fifty-eight to sixty. The even fall of one a hexagon that the smaller cages showed was not a trend. It was a bound being met. A count round each hexagon proves that no cage with $h$ hexagons can have fewer than $20 - h$ contacts, and the census now says exactly where cages sit on that line.

A faster census #

The earlier census wound up every placement of the twelve pentagons in a spiral of faces and kept the placements that close into a cage. That is $\binom{F}{12}$ attempts for a cage of $F$ faces: 2.7 million at forty-four atoms, and 225 million at sixty. This census winds the spiral up one face at a time inside a depth-first search. A partial spiral that already fails to close is abandoned together with every spiral that begins with it. At sixty atoms the search visits 157 million partial spirals, of which 489,227 close, and each visit costs a handful of operations instead of a full windup.

A cage has many spirals, one from almost every starting face and direction, so every spiral that closes has to be recognised as a cage already found or a new one. The earlier census did that with a canonical code, the least over every starting flag of a breadth-first numbering of the faces. Here the same code is computed more cheaply. The least code begins with the size of its first face, so only the twelve pentagons need be tried as starting faces, and a numbering is abandoned the moment it exceeds the best so far.

The check that the census is complete is the published sequence of fullerene isomer counts. The census finds 116 cages at forty-six atoms, 199 at forty-eight, 271 at fifty, 437 at fifty-two, 580 at fifty-four, 924 at fifty-six, 1,205 at fifty-eight and 1,812 at sixty. Every number matches. A cheaper test of sameness was tried first, a pair of invariants built from distances between pentagons and from repeated relabelling of faces by their neighbours. It merged a few different cages at forty-eight atoms and above, and the published counts caught it, 197 against 199. The canonical code is exact and the invariants were not.

What the census found #

The picture at the head of this essay shows the least contact count at every size from the dodecahedron to sixty atoms. Up to forty-four atoms it is the earlier census’s sequence. From there it runs 8, 8, 7, 5, 5, 4, 4, 3, 0. The cage at sixty atoms with no contacts is unique, and it is the truncated icosahedron. The earlier census built it separately by leapfrogging the dodecahedron and could not reach it by spirals. This one reaches it among 1,812 cages and finds that nothing else at sixty atoms comes close. The next best has two contacts, and three cages have three.

The steps show how far the path is from even. The total fall from forty-four to sixty is eight, as predicted. It comes in steps of none, one, two, none, one, none, one and three. The last step is the largest, three contacts removed by the addition of a single hexagon, from the best cage at fifty-eight atoms to the truncated icosahedron. A cage at sixty atoms can separate all twelve pentagons, and a cage at fifty-eight can do no better than three contacts. There is no intermediate arrangement a size in between could use.

The prediction was a bound #

The earlier essay’s steady fall has a proof, and it explains both the eight sizes where the fall was even and the sizes where it stops being even.

Round any hexagon, two neighbouring faces that share one of its corners also share an edge with each other, because three faces meet at every atom. So if a hexagon is bordered by $k$ pentagons, placed somewhere round its six sides, consecutive pentagons among them touch. With $k$ pentagons and $6 - k$ other faces in a ring of six, at least $2k - 6$ pairs of pentagons are consecutive, and each such pair is a contact. A single contact, two pentagons sharing an edge, can be seen this way at most twice, from the face at each end of the shared edge. Summing over all $h$ hexagons,

$$\sum_h (2k_h - 6) \le 2c.$$

The pentagons have sixty edges between them, and each contact uses two, so $60 - 2c$ pentagon edges face hexagons and $\sum_h k_h = 60 - 2c$. Putting the two together,

$$2(60 - 2c) - 6h \le 2c, \qquad\text{that is}\qquad c \ge 20 - h.$$

The prediction of one fewer contact a hexagon is exactly this bound. At $h = 20$ it gives none, which is the earlier essay’s alternate-edge argument that isolated pentagons need sixty atoms, now as the endpoint of a line. For small $h$ the older edge bound $30 - 3h$ is stronger, and the two cross at five hexagons.

The census says where cages meet the bound. They meet the edge bound at twenty to twenty-eight atoms, the new bound at thirty-six to forty-four atoms, and the new bound again at fifty and at sixty. Everywhere else they sit above it. The eight consecutive sizes of steady fall the earlier census saw were the bound being met size after size. Past forty-four atoms the cages meet it only twice more. The pauses at forty-six, fifty-two and fifty-six are sizes where no arrangement of the pentagons can achieve the bound. The jump at fifty and the collapse at sixty are the cages catching up with it.

Meeting the bound has a precise meaning. Every hexagon must have its pentagons in as few runs as possible, and every contact must be seen from both ends. At fifty atoms the single best cage does this, and at sixty the truncated icosahedron does it trivially, with every hexagon bordered by exactly three pentagons in alternate positions and no contacts at all.

The excess is a sum of local defects #

The two inequalities in the argument can be made exact, and doing so turns the bound into an identity. The first inequality was that a hexagon with $k$ pentagons has at least $2k - 6$ touching pairs among them. Call the difference, the actual number of touching pairs round the hexagon less $2k - 6$, that hexagon’s defect, and let $S$ be its sum over all hexagons. A hexagon contributes nothing to $S$ when its pentagons sit in as few runs as possible. It contributes something when it has fewer than three pentagons, or more runs than it needs. The second inequality was that each contact is seen from at most two hexagons. It is seen from fewer exactly when the face at an end of the shared edge is a third pentagon. Let $M$ count those ends. Then the counting is exact:

$$6c = 120 - 6h + S + M.$$

Every cage’s contacts are the bound plus a sixth of its local defects. The census checks the identity cage by cage, on every one of the 5,770 cages from twenty to sixty atoms, and it holds for all of them. The dodecahedron has no hexagons, so $S = 0$, and every one of its sixty contact ends meets a third pentagon, so $M = 60$ and $c = 30$.

The identity turns the bound into a statement about when it is met. Both defects are counts, so neither is negative, and the bound follows at once. It is met exactly when both vanish, when every hexagon carries at least three pentagons in as few runs as possible and no three pentagons meet at a point. Those are conditions a chemist can check on a drawing, one hexagon and one vertex at a time. The count of contacts, which looks global, is the sum of a constant set by the number of hexagons and local charges set by the neighbourhood of each face. A rule that charges strain to each pair of touching pentagons can equally be written as a rule charging hexagons whose pentagons are badly placed and atoms where three pentagons meet. The two rules must agree on every cage, since they differ only by the constant $20 - h$.

The picture sets the two defects of the best cage at each size side by side, and they divide the sizes into three regimes. From twenty to thirty-four atoms the excess is all $M$. The small cages cannot keep their pentagons out of triples, and every vertex where three pentagons meet costs contacts. From thirty-six to forty-four atoms the best cages have no defect of either kind and sit exactly on the bound. From forty-six atoms onwards $M$ is zero, since no best cage has three pentagons at a point, and the whole excess is $S$. Some hexagon has its pentagons arranged worse than it could. The pauses in the fall are the sizes where some $S$ cannot be avoided, and the census shows which hexagons carry it.

What the best cages do round each hexagon #

At the sizes where the bound is met, the hexagons of the best cage come in only two kinds. Some are bordered by three pentagons on alternate sides, none touching. The others are bordered by four pentagons in two touching pairs, the fewest runs four pentagons can make among six sides. The forty-four-atom cage has four hexagons of the first kind and eight of the second. The fifty-atom cage, with its five-fold axis, has ten of the first and five of the second, and its five contacts are the five touching pairs that ring its equator. The truncated icosahedron has twenty hexagons all of the first kind.

Where the bound is missed, the census shows why. The best cage at fifty-two atoms has six hexagons with three pentagons of which two touch, one run too many, and that adds six to $S$ and one contact to the bound. The best cage at fifty-eight atoms has three hexagons bordered by only two pentagons and six with three pentagons in two runs. They add twelve to $S$ and two contacts to the bound, and that is why fifty-eight atoms cannot do better than three contacts when the bound allows one. The last step to sixty is large because the truncated icosahedron removes all the defects at once. Every hexagon gets exactly three alternating pentagons, which no smaller cage can arrange.

The same accounting explains the regimes of the whole range of sizes. For small cages the pentagons outnumber the hexagons’ capacity to separate them. Three pentagons must meet at points, and $M$ is forced. Between thirty-six and forty-four atoms there are enough hexagons to keep every triple apart and few enough that each one can be given its pentagons efficiently, and the bound is met. Past forty-four atoms there are hexagons to spare, and the difficulty changes. The pentagons can no longer be packed so that every hexagon has at least three of them, and a hexagon left with two, or with three awkwardly placed, charges $S$. The truncated icosahedron is the one arrangement at which twenty hexagons exactly suffice to give every hexagon three alternating pentagons, and its defects are zero.

Most counts are nowhere near the least #

The least is an extreme, and the distribution above it is broad. At sixty atoms one cage has no contacts, one has two, three have three and seventeen have four. The great majority of the 1,812 have far more. At every size the least is attained by very few cages, most often just one: at fifty, fifty-two, fifty-four, fifty-eight and sixty atoms the best arrangement is unique. The picture also shows a gap. No cage up to sixty atoms has exactly one contact. At sixty atoms the counts jump from none to two. At every smaller size the least is at least three. A single pair of touching pentagons with the other ten isolated is not an arrangement any cage in the census achieves.

What the census does and does not settle #

The counts are complete to sixty atoms. Every cage the spiral search can reach is found, and the isomer counts match the published ones at every size. That also confirms, for these sizes, that every fullerene has a face spiral. An exhaustive search elsewhere found no counterexample below 380 atoms.

The bound is proved and its attainment is measured. That $c \ge 20 - h$ holds for every cage follows from the counting argument, with no appeal to the census. Where it is attained is read from the census and is known only to sixty atoms. That is the same separation everything except the hexagons found in Eberhard’s theorem: a count says what a cage must carry, and only a construction says that a cage carrying it exists.

The gap at one contact is observed, not proved. No cage in the census has exactly one contact. Whether some larger cage has one, a single fused pair of pentagons among ten isolated ones, is not settled here. The bound permits it from nineteen hexagons on, so the question is one of arrangement, not of counting. A count the arithmetic permits and no cage realises is the kind of gap the projective plane’s cages leave in their hexagon counts.

Who worked out the pieces #

The spiral description of fullerenes and the algorithm that winds spirals into cages are due to David Manolopoulos and Patrick Fowler, whose An Atlas of Fullerenes of 1995 lists the isomers used here as a check. The isolated-pentagon rule, the empirical observation that stable fullerenes avoid pentagon contacts, was stated by Harold Kroto in 1987, and its combinatorial consequence, that sixty atoms is the smallest cage satisfying it, is the endpoint of the bound here. The pentagon adjacency count as a measure of strain in fullerenes is standard in the chemical literature, where the fewest adjacencies at each size have been tabulated before; the counting argument for the bound is given here in full so that it can be checked.

What the census must refuse #

The first two tests carry the essay. The bound is a proof, and the identity is the same counting made exact, so a census finding a cage below the bound, or one whose defects did not add up, would mean an error in the census or in the counting. The identity is the stronger check, since it has to hold with equality on all 5,770 cages rather than as an inequality on the best ones. The refusal is the isolated-pentagon threshold. A cage below sixty atoms with its pentagons apart would violate the bound at $h < 20$, and none is found.

Still open: why the pauses fall where they do #

The bound $20 - h$ is met at fifty atoms and not at forty-six, forty-eight or fifty-two, and that pattern has no explanation here. Meeting the bound requires every hexagon to carry its pentagons in the fewest possible runs and every contact to be seen from both ends. Those are local conditions, and whether a cage can satisfy all of them at once depends on how the faces can be arranged on a sphere.

Symmetry is the obvious suspect, and the census rules it out as the whole story. The unique best cage at fifty atoms has twenty symmetries, with a five-fold axis, and the truncated icosahedron at sixty has a hundred and twenty. But at forty-four atoms, where the bound is also met, the two best cages have only four symmetries each. Which cages have a centre of symmetry at all is a separate question, and a centre at every other ring found it decided, for capped tubes, by the parity of the tube’s length. At fifty-eight, where the bound is missed by two, the unique best cage has six. What decides attainment is something about how pentagon runs round different hexagons can be fitted together, and the census gives, for every size to sixty, the cages at which the question can be examined one hexagon at a time. That examination has not been made.