The residues as a sequence

Take a hot position and pile up n copies of it. The pile is worth about n times the mean, and the difference between the pile and that number is the residue — a game, not a number, and the thing the mean-and-temperature reading throws away.

What is left when the copies pair off found the rule that produces one residue from the last. It is $r(n+1) = r(n) + r(1)$, so $r(n) = n \cdot r(1)$, and the arithmetic is settled. It closed on what a recurrence is not:

four of the eight sequences do not repeat, and a sequence of games that never repeats can still have a description … what is wanted is a closed form for the $n$-th term rather than a recurrence.

There is one. It is not a closed form for the game, and the reason it is not is the interesting half.

What repeats and what does not #

Two questions can be asked of a sequence of residues and they have different answers.

Does the sequence of games repeat? On five of the eight, yes: $\{2 \mid 0\}$ alternates between $\{1 \mid -1\}$ and $0$ for ever, and $\{1 \mid \ast\}$ runs a cycle of four. On three of them, no — $\{5 \mid \{4 \mid 0\}\}$, $\{2 \mid \{1 \mid 0\}\}$ and $\{\{4 \mid 2\} \mid 0\}$ produce a different game at every $n$ as far as the sweep goes, thirteen distinct residues each.

The five that repeat are worth a moment, because they are where the answer would have stopped if the pool had been chosen differently. All five are positions whose options are numbers or nearly so: a plain switch, a switch about nothing, a switch with a star in it. Pile up copies of a plain switch and the odd piles are all worth $\{1 \mid -1\}$ and the even ones nought, for ever. Nothing is accumulating because there is nothing below the first level to accumulate.

The three that do not repeat are exactly the three with a follow-up — an option that is itself a fight rather than a number. That is not a coincidence and the last section of this page is about why.

Does the reading repeat? Take the two stops and the temperature, which is what a player has of a hot position, and the answer is yes on all eight. The periods are one, two and four, and the three sequences whose forms never repeat have reading periods of one, four and four.

So the closed form asked for exists, and it is a closed form for what the position is worth to read rather than for what it is. Knowing $n$ modulo four gives the residue’s stops and its temperature, on every sequence in this pool. Knowing $n$ exactly does not give the game, and no amount of arithmetic on $r(1)$ will, because the game is different every time.

That is a smaller answer than the rung below was hoping for and it is worth saying why it is nevertheless the answer to the question. The reason anybody wants a closed form for a residue is to know what a pile of copies is worth without building it — and what it is worth is a reading. A player holding fifteen copies of a position wants to know the pile’s stops and its temperature; the canonical form of a sum of fifteen games is not a thing anybody wants, and would not fit on a page if they did.

The sharpest case #

One of the three is worth its own table.

$\{5 \mid \{4 \mid 0\}\}$ has a reading period of one. Every residue in its sequence has left stop 1, right stop 0 and temperature 1 — the same three numbers at every $n$, from the first copy to the thirteenth. And every residue is a different game: thirteen terms, thirteen distinct forms, no two equal, with the form six levels deeper every two copies.

The depth column is the one to watch alongside. It runs 6, 6, 12, 12, 18, 18 — so the twelfth residue is a form seven levels deeper than the second, and every level is a real branching that a canonical form has kept because nothing dominates it. These are not the same game written at different sizes; they are different games that happen to present the same face.

That is a sequence of games a player cannot tell apart. Whatever n is, the pile is worth $4n$ plus something whose stops are 1 and 0 and whose temperature is 1, and the difference between the n-th such something and the m-th is invisible to every quantity this anchor has defined.

It is worth being exact about what “invisible” means, since it is a strong claim. The residues differ as games: they are not equal, they are not even comparable in some cases, and a sum containing one behaves differently from a sum containing another. What is identical is the triple a player reads — and that triple is the whole of what the mean and the temperature give.

So a reader who wanted to distinguish the third residue from the ninth would have to do something the anchor has no vocabulary for: put both in the same sum and see which wins, or compute both forms and compare them. Every quantity this ladder has defined says they are the same, and they are not.

That is the honest size of the finding. It is not that the reading is wrong — the residues really do all have stops 1 and 0 and temperature 1, and a player using that will play correctly against any of them in isolation. It is that the reading is not injective on this sequence, and the sequence is one a player meets by doing the most ordinary thing there is: making another copy.

How the ones that never repeat grow #

The forms are not arbitrary either. They grow, and they grow predictably.

$\{5 \mid \{4 \mid 0\}\}$ gains six levels every two copies, at every step. $\{\{4 \mid 2\} \mid 0\}$ gains five, at every step. $\{2 \mid \{1 \mid 0\}\}$ alternates: three levels, then four, then three, then four.

The rates themselves are worth a glance because they are not the same number and there is no obvious reason they should not have been. All three positions have a single follow-up on one side and all three are built the same way, and they grow at six, five and three-and-a-half levels per two copies. Whatever sets the rate is reading something finer about the position than has a follow-up, and this page has not found what.

That third one is the reason the statement is the gains are periodic rather than the gains are constant. Writing it the second way is what a reader would guess from the other two, and the sweep refuses to draw if the growing sequences all gain the same amount at every step — which is the check that turned a guess into the correct statement.

The alternation lines up with the reading period, and the alignment is worth noting even though this page cannot explain it. $\{2 \mid \{1 \mid 0\}\}$ is the one whose reading period is four and whose gains alternate; $\{\{4 \mid 2\} \mid 0\}$ has reading period four and constant gains; $\{5 \mid \{4 \mid 0\}\}$ has reading period one and constant gains. So the two periodicities are not the same phenomenon and are not independent either, and three cases is not enough to say which.

So there are two closed forms and between them they describe the n-th residue completely except for its identity. $n$ modulo the reading period gives what it is worth. $n$ gives how deep it is. What neither gives is the form.

What a player actually holds #

It is worth putting the pile back beside its residue, because the residue is a difference and differences are easy to lose the scale of.

A pile of eight copies of $\{5 \mid \{4 \mid 0\}\}$ is worth about thirty-two. Its residue has stops of 1 and 0 — so the pile’s own stops sit within one move of $8 \times 4$, and that is true of every pile in the sequence, because the residue’s stops never move. The thing that grows without bound as the pile grows is not the error; it is the description of the error.

That is the practical content of the whole page. A player piling up copies of a hot position accumulates no additional uncertainty about what the pile is worth. What accumulates is structure inside a bounded interval, and structure inside a bounded interval is exactly what a reading is designed not to see.

Why the form is the wrong thing to ask for #

Standing back, the shape of this answer was available from the recurrence and nobody had looked at it that way.

$r(n) = n \cdot r(1)$ means the n-th residue is a sum of $n$ copies of one mean-zero game. A sum of $n$ copies is a game whose form is built out of $n$ things, and there is no reason for it to stabilise: a sum’s canonical form is not bounded by its parts’, and adding another copy adds another layer of options for both players to choose among. The forms growing linearly is what a sum of $n$ things ought to do.

What is not obvious in advance is that the readings stabilise. The mean of $n \cdot r(1)$ is $n$ times nought, which is nought, so means are trivially constant — but the stops and the temperature of a sum are not functions of the parts’, which is the whole content of temperatures do not add. So the reading of $n \cdot G$ settling into a cycle of length at most four, on every position in the pool, is a genuine fact and not a consequence of the mean being zero.

The mechanism, as far as this page can see it, is parity. A pile of an even number of copies of a mean-zero game is a sum that pairs off: whatever Left gains from one copy Right gains from another, and the residue collapses. A pile of an odd number leaves one copy unpaired. Two-cycles are that and nothing else. Four-cycles are the same thing at one more level down, where the leftover copy’s own follow-ups pair off across two more copies — which is why the positions with a follow-up are the ones with the longer period.

The story has the right shape for the split observed above. A position whose options are numbers has nothing below the first level, so the pairing is finished after one level and the period is two. A position with a follow-up has a second level for the leftover copy to be answered at, so the pairing takes two more copies and the period is four. And if that is right, a follow-up to a follow-up should give period eight — which is the prediction the next rung is for.

That is a description and not a proof, and the sweep does not test it. What the sweep tests is that the periods are one, two and four, and none of the eight is anything else.

What is settled and what the pool leaves open #

The rung’s question is answered. There is a closed form for the n-th residue, in the sense that matters for a player: two numbers determine what it is worth, and one more determines its size.

It is worth writing the form out for the sharpest case, since it is short enough to state. For $G = \{5 \mid \{4 \mid 0\}\}$, the residue of $n$ copies has left stop 1, right stop 0 and temperature 1 for every $n \ge 1$, and its form has $6\lceil n/2 \rceil$ levels. Three constants and one linear function, and between them they say everything about $r(n)$ except which game it is.

For the two with period four the statement is one line longer: the stops and the temperature cycle through four values as $n$ runs through the residues modulo four, and the depth grows at five levels per two copies in one case and alternately three and four in the other. Still four numbers and a rule, and still nothing about the form.

The five that repeat need no closed form at all, which is worth saying so the answer is not made to sound harder than it is. A sequence with period two is described by two games, written out; a sequence with period four by four. The closed form is only interesting where the list would be infinite, and that is the three.

And the answer explains why nobody found a closed form for the game. They were looking for the wrong object. A sequence of sums of $n$ things has no reason to have a formula for its $n$-th term as a form, and the recurrence $r(n) = n \cdot r(1)$ is already the best statement of that kind available.

The pool is eight positions and every claim here is about them. The reading periods being one, two and four is a fact about eight sequences, and a ninth position with a deeper follow-up might well have period eight — indeed the parity story above predicts it should. Nothing here has tried, because the pool is the same eight the rung below used and changing it mid-ladder would make the two pages incomparable.

The three that never repeat are the three with a follow-up, and that is the only structural fact here. Five positions in the pool have options that are numbers or numbers with a star, and all five have residue sequences that repeat as forms. Three have an option that is itself a fight, and none of those three repeats. Five against three is a small sample and the split is clean, which is worth recording as an observation rather than a rule — but it is the observation that the parity story below has to explain.

And thirteen copies is where the sweep stops. A residue’s form is the canonical form of a sum of thirteen games, and the twenty-seventh would be deeper than anything on this site. The reading periods are established over three full cycles of the longest of them, which is enough to be a period and is not a proof of one — the usual caution, and it applies here as it does to any period found by search.

Where the ladder goes next #

The temperature anchor reaches eight rungs: what is at stake, cooling, playing the hottest, the same fight eight times over, the first time it told somebody something, big is not the same as hot, what is left when the copies pair off, and now what the leavings are as a sequence.

The rung above is the parity story, made into a claim. This page offers a mechanism for why the periods are one, two and four — copies pairing off, and the leftover copy’s follow-ups pairing off one level down — and does not test it. The test is direct: build positions with follow-ups nested two and three deep and predict period eight and period sixteen before running them. If the prediction holds the periods are a fact about the depth of the position’s own form; if it fails, the periods are something else, and the eight sequences here are a coincidence of a small pool.

Two neighbours are worth the trip. What is at stake is where the mean and the temperature are defined, and reading it beside this page shows exactly how much a pair of numbers throws away: thirteen distinct games with one reading between them. And temperatures do not add is why the readings settling is a result rather than an arithmetic identity.