Abstract
Heawood’s formula $\chi(g)=\lfloor(7+\sqrt{1+48g})/2\rfloor$ gives the chromatic number of a surface of genus $g$. Put in $g=0$ and out comes $\mathbf{4}$: the very value of the four-colour theorem. This paper asks whether that is then a proof of the four-colour theorem: the answer is no. No new mathematical theorem and no new law is claimed.
First, the formula returns the right value at $g=0$. $(7+\sqrt1)/2=4.000000$, and the floor is $\mathbf{4}$ (Section 2).
Second, this is the core of the paper. It is a theorem for $g\ge1$, and at $g=0$ the proof does not reach: Heawood’s argument fails at that one point (Section 3).
Third, a gap of $78$ years opened. $g=1$ was settled in $1890$, all $g\ge1$ in $1968$, and $g=0$ in $1976$ (Section 4).
Fourth, the simplest surface was left for last. And it fell only after $1936$ reducible configurations were checked by computer (Section 4).
Fifth, the non-orientable side has one exception too. For the Klein bottle the formula returns $7$, and the truth is $\mathbf{6}$ (Section 5).
Sixth, the separator is the value being right against the proof reaching it (Section 6).
Scope (what is not claimed)
No new mathematical theorem and no new law is claimed: Heawood’s formula, the Ringel--Youngs theorem, the four-colour theorem, and Franklin’s exception are all standard. We do not build topology: all we use is one square root and one floor. We do not prove the four-colour theorem: we do not enter the Appel--Haken proof nor its later simplifications. We do not reproduce Heawood’s argument: the derivation of the bound from Euler’s formula is merely quoted. We assert no count of reducible configurations: $1936$ is the number in the $1976$ version, reduced to $633$ in Robertson et al. ($1997$). The claim is that the quantity is beyond human reading, not the count itself. We do not argue about computer proof: whether it should be accepted is not treated. We record only that it was formalised in Coq in $2005$. Non-orientable surfaces are not a principal object: the Klein bottle of Section 5 is placed to show that an exception exists. Colouring of non-orientable surfaces is not treated systematically. Relation to earlier papers: Paper 302 treated how “it must appear” does not say when. This paper treats the case where the value is known and the proof does not reach it, the gap running the other way. Paper 163 separated “a good code exists” from “here is a good code”. This paper separates “the value is this” from “why”. Paper 179 showed that “cannot be constructed” has distinct roots: the $g=0$ here is only one argument failing to reach, and another argument did reach. Paper 112 counted “six distinct roots sharing one rhyme”. Here $g=0$ and $g\ge1$ share the value of one formula while their grounds differ. What is added is confirming that the formula returns $4.000000$ at $g=0$, measuring the $78$-year gap between the settlement of $g\ge1$ and of $g=0$, recording the inversion by which the simplest surface was left for last, and placing the separator at the value’s correctness against the proof’s reach.
1. Notation and the Question
Setting: Any map drawn on a closed orientable surface of genus $g$ can be coloured with $\chi(g)$ colours. Heawood ($1890$) gave this.
$$\chi(g)=\left\lfloor\frac{7+\sqrt{1+48g}}{2}\right\rfloor$$Question: What comes out at $g=0$ (the sphere, hence the plane)? And is that a proof of the four-colour theorem?
Observation 1.1 The first half of the answer is “$4$ comes out.”The second half is “it is not”. This paper treats that gap.
2. The Formula Returns $4$ at $g=0$
Check 2.1:
| Genus $g$ | $(7+\sqrt{1+48g})/2$ | $\chi(g)$ | Surface |
|---|---|---|---|
| $\mathbf{0}$ | $\mathbf{4.000000}$ | $\mathbf{4}$ | sphere |
| $1$ | $7.000000$ | $7$ | torus |
| $2$ | $8.424429$ | $8$ | double torus |
| $3$ | $9.520797$ | $9$ | triple torus |
| $4$ | $10.446222$ | $10$ | |
| $5$ | $11.262087$ | $11$ | |
| $6$ | $12.000000$ | $12$ | |
| $7$ | $12.678780$ | $12$ | |
| $10$ | $14.465856$ | $14$ |
Observation 2.2 At $g=0$, $1+48\cdot0=1$, $\sqrt1=1$, and $(7+1)/2=4.000000$. No floor is even needed: exactly $4$: the very value of the four-colour theorem.
Observation 2.3 Note that $g=6$ and $g=7$ both give $12$. The floor collapses them to one value. There are steps where the genus rises and the chromatic number does not.
Observation 2.4 By value alone, $g=0$ looks like the natural first term of the sequence. $4,7,8,9,10,11,12,\dots$ runs on without a break: the break is not on the value’s side but on the proof’s.
3. The Core: A Theorem for $g\ge1$; at $g=0$ the Proof Does Not Reach
Observation 3.1 What Heawood showed in $1890$ is an upper bound: “$\chi(g)$ colours suffice.”The argument bounds the average degree from Euler’s formula, and goes through for $g\ge1$.
Observation 3.2 The lower bound (that a map actually needing that many colours exists) was completed by Ringel and Youngs in $1968$. With that, the formula became a theorem for $g\ge1$.
Observation 3.3 This is the core of the paper. At $g=0$ alone, Heawood’s argument does not give the bound $4$: the same argument gets only down to $5$ colours (the five-colour theorem).
Check 3.4: We line up how far each reached.
| Bound from Heawood’s argument | True value | |
|---|---|---|
| $g\ge1$ | $\chi(g)$ (as the formula) | $\chi(g)$ (agrees) |
| $g=0$ | $\mathbf{5}$ (five-colour theorem) | $\mathbf{4}$ (does not agree) |
Observation 3.5 The difference is just $1$. The same argument reaches $5$, and the last $1$ will not fall, and that $1$ took $86$ years ($1890$ to $1976$).
Observation 3.6 So the formula’s standing splits in two.
| Standing of $\chi(g)$ | |
|---|---|
| $g\ge1$ | a theorem (upper and lower bounds proved) |
| $g=0$ | an expression returning the right value (verified by another proof) |
Observation 3.7 One and the same expression changes standing at a single point: the figure by which Paper 16 wrote that the equals sign too has distinct roots, occurring here within the domain of one function.
4. $78$ Years: The Simplest Surface Was Left for Last
Check 4.1:
| Surface | Chromatic number | Settled | How |
|---|---|---|---|
| Torus $g=1$ | $7$ | $\mathbf{1890}$ (Heawood) | by hand |
| $g=2$ | $8$ | $1968$ (Ringel--Youngs) | by hand |
| All $g\ge1$ | the formula | $\mathbf{1968}$ | by hand |
| Sphere $g=0$ | $\mathbf{4}$ | $\mathbf{1976}$ (Appel--Haken) | by computer |
Observation 4.2 The simplest surface fell last. From the torus ($1890$) to the sphere ($1976$) is $\mathbf{86}$ years; from the settlement of all $g\ge1$ ($1968$) it is still $8$: the order of difficulty runs opposite to the order of simplicity.
Observation 4.3 The reason is how much slack there is. The larger the genus, the more slack in the Euler characteristic, and the better the degree bound works: the sphere has the least slack and so the argument works least.
Check 4.4: And the manner of settling differed too.
| Quantity | Value |
|---|---|
| Reducible configurations checked by Appel--Haken ($1976$) | $\mathbf{1936}$ |
| Simplification by Robertson et al. ($1997$) | $633$ |
| Formalisation in Coq (Gonthier) | $2005$ |
Observation 4.5 Reading $1936$ configurations through is effectively impossible for a person. Even reduced to $633$ in $1997$, it still cannot be read through: only in $2005$, when a machine verified it, did it become true that someone had seen all of it.
Observation 4.6 A point to write honestly. This paper does not argue about computer proof. What is recorded is only that $g\ge1$ fell by hand and $g=0$ did not.
5. The Non-Orientable Side Has One Exception Too
Setting: There is a corresponding formula for non-orientable surfaces.
Check 5.1:
| Surface | Formula returns | Truth |
|---|---|---|
| Projective plane | $6$ | $6$ (agrees) |
| Klein bottle | $\mathbf{7}$ | $\mathbf{6}$ (does not agree) |
| All others | as the formula | agrees |
Observation 5.2 The Klein bottle is the only exception (Franklin, $1934$). The formula returns $7$; six colours suffice. Here the value itself is wrong.
Observation 5.3 So there are two kinds of gap.
| Value | Proof | |
|---|---|---|
| $g=0$ (sphere) | right | does not reach |
| Klein bottle | wrong | — |
Observation 5.4 They are different failures. On the sphere the formula is right and cannot say why; on the Klein bottle it is simply wrong: the single phrase “the formula fails” covers both.
6. Separator: The Value Being Right against the Proof Reaching It
Observation 6.1 Separator. What must be moved for $g=0$ to take the same standing as $g\ge1$? The argument. Neither the surface nor the formula needs changing: bring another argument (checking reducible configurations) and the value stays while the standing changes.
Observation 6.2 Paper 302 treated how “it must appear” does not say when. Here the gap runs the other way. There the value was unknown; here the value was known and the reason could not be given.
| Existence | Value | |
|---|---|---|
| Paper 302 (Ramsey) | known | unknown |
| This paper ($g=0$, $1890$--$1976$) | known | known (but unexplained) |
Observation 6.3 So a third column is needed: the ground. Paper 163 separated existence from construction; Paper 302 separated existence from quantity. This paper separates the value from its ground.
Observation 6.4 And a value can be right while its ground comes from elsewhere. The $4$ at $g=0$ is not a correct prediction by Heawood’s formula but another proof arriving at the same number: at $g=0$ the formula merely “happens to be right”.
Observation 6.5 A point to write honestly: we do not say the formula is right at $g=0$ by accident. The $4$ comes out from the same estimate of the Euler characteristic, and is not unrelated. What can be said is only that at $g=0$ that estimate does not reach the last step.
7. Roll Call and Closing
- $4.000000$ at $g=0$ (Check 2.1): No floor even needed.
- $g=6$ and $g=7$ both $12$ (Observation 2.3): The floor collapses them.
- The bound at $g=0$ stops at $5$ (Check 3.4): A difference of just $1$.
- That $1$ took $86$ years (Observation 3.5): $1890$ to $1976$.
- $g\ge1$ settled in $1968$ (Check 4.1): By hand.
- $g=0$ settled in $1976$ (Check 4.1): By computer.
- $1936$ reducible configurations (Check 4.4): $633$ by $1997$.
- Formalised in $2005$ (Check 4.4): A machine saw all of it.
- Klein bottle: $7$ against $6$ (Check 5.1): The only exception.
- Two kinds of failure (Observation 5.4): The value, or the ground.
Closing: Put $g=0$ into Heawood’s formula and $\mathbf{4.000000}$ comes out: the very value of the four-colour theorem. And it is not a proof of that theorem. The same argument gives both bounds for $g\ge1$ and makes the formula a theorem, yet at $g=0$ it gets only down to $5$ colours: a difference of just $\mathbf{1}$, and that $1$ took $\mathbf{86}$ years. And the order inverted: the torus fell by hand in $1890$, all $g\ge1$ by hand in $1968$, and only the simplest surface, the sphere, fell in $1976$ by computer. The $\mathbf{1936}$ reducible configurations could not be read through by a person, nor could the $633$ of $1997$, and only in $2005$, when a machine verified them, did it become true that someone had seen all of it. One thing separates them: a value being right, and a proof reaching that value. The two are independent. The $4$ the formula returns at $g=0$ is correct, and its correctness came from elsewhere. One and the same expression changes standing at a single point of its domain.
Revision Record
Second edition (October 2026): the dashes used as separators in the text were replaced by punctuation; places where LaTeX commands had been displayed as raw text were repaired; no claim, number or reference was changed.
References
- Heawood, P. J. (1890). Map-Colour Theorem. Quarterly Journal of Pure and Applied Mathematics, 24.
- Franklin, P. (1934). A Six Color Problem. Journal of Mathematics and Physics, 13(1-4).
- Ringel, G. and Youngs, J. W. T. (1968). Solution of the Heawood Map-Coloring Problem. Proceedings of the National Academy of Sciences, 60(2).
- Appel, K. and Haken, W. (1977). Every Planar Map Is Four Colorable. Illinois Journal of Mathematics, 21(3).
- Robertson, N., Sanders, D., Seymour, P. and Thomas, R. (1997). The Four-Colour Theorem. Journal of Combinatorial Theory, Series B, 70(1).
- Gonthier, G. (2008). Formal Proof — The Four-Color Theorem. Notices of the American Mathematical Society, 55(11).
- Yamagishi, Y. (2026). The Equals Sign Has Distinct Roots Too [Paper 16]. Zenodo. 10.5281/zenodo.21600471
- Yamagishi, Y. (2026). The Allotted Region and the Freedom Within [Paper 112]. Zenodo. 10.5281/zenodo.21962834
- Yamagishi, Y. (2026). “A Good Code Exists” and “Here Is a Good Code” Are Different Sentences [Paper 163]. Zenodo. 10.5281/zenodo.22092677
- Yamagishi, Y. (2026). “Cannot Be Constructed” Has Distinct Roots [Paper 179]. Zenodo. 10.5281/zenodo.22098419
Reference
Yamagishi, Y. (2026). The Formula Returns $4$, and at That One Point the Proof Does Not Reach [Paper 303]. YOI: https://www.yamagishiyuuki.com/yoi/303
DOI: 10.5281/zenodo.22172026
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 303. This paper is a preprint and has not been peer reviewed.