Graham's number is a large number that arose as an upper bound on the answer of a problem in the mathematical field of Ramsey theory. It is much larger than many other large numbers such as Skewes's number and Moser's number, both of which are in turn much larger than a googolplex. As with these, it is so large that the observable universe is far too small to contain an ordinary digital representation of Graham's number, assuming that each digit occupies one Planck volume, possibly the smallest measurable space. But even the number of digits in this digital representation of Graham's number would itself be a number so large that its digital representation cannot be represented in the observable universe. Nor even can the number of digits of that number—and so forth, for a number of times far exceeding the total number of Planck volumes in the observable universe. Thus Graham's number cannot be expressed even by physical universe-scale power towers of the form ${\displaystyle a^{b^{c^{\cdot ^{\cdot ^{\cdot }}}}}}$.

However, Graham's number can be explicitly given by computable recursive formulas using Knuth's up-arrow notation or equivalent, as was done by Ronald Graham, the number's namesake. As there is a recursive formula to define it, it is much smaller than typical busy beaver numbers. Though too large to ever be computed in full, the sequence of digits of Graham's number can be computed explicitly via simple algorithms; the last 13 digits are ...7262464195387. Using Knuth's up-arrow notation, Graham's number is ${\displaystyle g_{64}}$,^[1] where

$${\displaystyle g_{n}=\left\{{\begin{matrix}3\uparrow \uparrow \uparrow \uparrow 3,&n=1\\3\uparrow ^{g_{n-1}}3,&n\geq 2,n\in \mathbb {N} \end{matrix}}\right.}$$

Graham's number was used by Graham in conversations with popular science writer Martin Gardner as a simplified explanation of the upper bounds of the problem he was working on. In 1977, Gardner described the number in Scientific American, introducing it to the general public. At the time of its introduction, it was the largest specific positive integer ever to have been used in a published mathematical proof. The number was described in the 1980 Guinness Book of World Records, adding to its popular interest. Other specific integers (such as TREE(3)) known to be far larger than Graham's number have since appeared in many serious mathematical proofs, for example in connection with Harvey Friedman's various finite forms of Kruskal's theorem. Additionally, smaller upper bounds on the Ramsey theory problem from which Graham's number derived have since been proven to be valid.

Context

Example of a 2-colored 3-dimensional cube containing one single-coloured 4-vertex coplanar complete subgraph. The subgraph is shown below the cube. Note that this cube would contain no such subgraph if, for example, the bottom edge in the present subgraph were replaced by a blue edge – thus proving by counterexample that N* > 3.

Graham's number is connected to the following problem in Ramsey theory:

Connect each pair of geometric vertices of an n-dimensional hypercube to obtain a complete graph on 2ⁿ vertices. Colour each of the edges of this graph either red or blue. What is the smallest value of n for which every such colouring contains at least one single-coloured complete subgraph on four coplanar vertices?

In 1971, Graham and Rothschild proved the Graham–Rothschild theorem on the Ramsey theory of parameter words, a special case of which shows that this problem has a solution N*. They bounded the value of N* by 6 ≤ N* ≤ N, with N being a large but explicitly defined number

${\displaystyle F^{7}(12)=F(F(F(F(F(F(F(12))))))),}$

where ${\displaystyle F(n)=2\uparrow ^{n}3}$ in Knuth's up-arrow notation; the number is between 4 → 2 → 8 → 2 and 2 → 3 → 9 → 2 in Conway chained arrow notation.^[2] This was reduced in 2014 via upper bounds on the Hales–Jewett number to

${\displaystyle N'=2\uparrow \uparrow 2\uparrow \uparrow (3+2\uparrow \uparrow 8),}$

which contains three tetrations.^[3] In 2019 this was further improved to:^[4]

${\displaystyle N''=(2\uparrow \uparrow 5138)\cdot ((2\uparrow \uparrow 5140)\uparrow \uparrow (2\cdot 2\uparrow \uparrow 5137))\ll 2\uparrow \uparrow (2\uparrow \uparrow 5138)}$

The lower bound of 6 was later improved to 11 by Geoffrey Exoo in 2003,^[5] and to 13 by Jerome Barkley in 2008.^[6] Thus, the best known bounds for N* are 13 ≤ N* ≤ N''.

Graham's number, G, is much larger than N: ${\displaystyle f^{64}(4)}$, where ${\displaystyle f(n)=3\uparrow ^{n}3}$. This weaker upper bound for the problem, attributed to an unpublished work of Graham, was eventually published and named by Martin Gardner in Scientific American in November 1977.^[7]

Publication

The number gained a degree of popular attention when Martin Gardner described it in the "Mathematical Games" section of Scientific American in November 1977, writing that Graham had recently established, in an unpublished proof, "a bound so vast that it holds the record for the largest number ever used in a serious mathematical proof." The 1980 Guinness Book of World Records repeated Gardner's claim, adding to the popular interest in this number. According to physicist John Baez, Graham invented the quantity now known as Graham's number in conversation with Gardner. While Graham was trying to explain a result in Ramsey theory which he had derived with his collaborator Bruce Lee Rothschild, Graham found that the said quantity was easier to explain than the actual number appearing in the proof. Because the number which Graham described to Gardner is larger than the number in the paper itself, both are valid upper bounds for the solution to the problem studied by Graham and Rothschild.^[8]

Definition

Using Knuth's up-arrow notation, Graham's number G (as defined in Gardner's Scientific American article) is

$${\displaystyle \left.{\begin{matrix}G&=&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \cdots \uparrow } 3\\&&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \uparrow } 3\\&&\underbrace {\qquad \quad \vdots \qquad \quad } \\&&3\underbrace {\uparrow \uparrow \cdots \cdots \uparrow } 3\\&&3\uparrow \uparrow \uparrow \uparrow 3\end{matrix}}\right\}{\text{64 layers}}}$$

where the number of arrows in each layer is specified by the value of the next layer below it; that is,

$${\displaystyle G=g_{64},}$$

where ${\displaystyle g_{1}=3\uparrow \uparrow \uparrow \uparrow 3,}$ ${\displaystyle g_{n}=3\uparrow ^{g_{n-1}}3,}$

and where a superscript on an up-arrow indicates how many arrows there are. In other words, G is calculated in 64 steps: the first step is to calculate g₁ with four up-arrows between 3s; the second step is to calculate g₂ with g₁ up-arrows between 3s; the third step is to calculate g₃ with g₂ up-arrows between 3s; and so on, until finally calculating G = g₆₄ with g₆₃ up-arrows between 3s.

Equivalently,

$${\displaystyle G=f^{64}(4),{\text{ where }}f(n)=3\uparrow ^{n}3,}$$

and the superscript on f indicates an iteration of the function, e.g., ${\displaystyle f^{4}(n)=f(f(f(f(n))))}$. Expressed in terms of the family of hyperoperations ${\displaystyle {\text{H}}_{0},{\text{H}}_{1},{\text{H}}_{2},\cdots }$, the function f is the particular sequence ${\displaystyle f(n)={\text{H}}_{n+2}(3,3)}$, which is a version of the rapidly growing Ackermann function A(n, n). (In fact, ${\displaystyle f(n)>A(n,n)}$ for all n.) The function f can also be expressed in Conway chained arrow notation as ${\displaystyle f(n)=3\rightarrow 3\rightarrow n}$, and this notation also provides the following bounds on G:

$${\displaystyle 3\rightarrow 3\rightarrow 64\rightarrow 2<G<3\rightarrow 3\rightarrow 65\rightarrow 2.}$$

Magnitude

To convey the difficulty of appreciating the enormous size of Graham's number, it may be helpful to express—in terms of exponentiation alone—just the first term (g₁) of the rapidly growing 64-term sequence. First, in terms of tetration (${\displaystyle \uparrow \uparrow }$) alone:

$${\displaystyle g_{1}=3\uparrow \uparrow \uparrow \uparrow 3=3\uparrow \uparrow \uparrow (3\uparrow \uparrow \uparrow 3)=3\uparrow \uparrow (3\uparrow \uparrow (3\uparrow \uparrow \ \dots \ (3\uparrow \uparrow 3)\dots ))}$$

where the number of 3s in the expression on the right is

$${\displaystyle 3\uparrow \uparrow \uparrow 3=3\uparrow \uparrow (3\uparrow \uparrow 3).}$$

Now each tetration (${\displaystyle \uparrow \uparrow }$) operation reduces to a power tower (${\displaystyle \uparrow }$) according to the definition

$${\displaystyle 3\uparrow \uparrow X=3\uparrow (3\uparrow (3\uparrow \dots (3\uparrow 3)\dots ))=3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}$$

where there are X 3s.

Thus,

$${\displaystyle g_{1}=3\uparrow \uparrow (3\uparrow \uparrow (3\uparrow \uparrow \ \dots \ (3\uparrow \uparrow 3)\dots ))\quad {\text{where the number of 3s is}}\quad 3\uparrow \uparrow (3\uparrow \uparrow 3)}$$

becomes, solely in terms of repeated "exponentiation towers",

$${\displaystyle g_{1}=\underbrace {\left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{\cdot ^{3}}}}}}\end{matrix}}\right\}\left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}\end{matrix}}\right\}\dots \left.{\begin{matrix}3^{3^{3}}\end{matrix}}\right\}3} _{\left.{\begin{matrix}3^{3^{\cdot ^{\cdot ^{\cdot ^{3}}}}}\end{matrix}}\right\}\left.{\begin{matrix}3^{3^{3}}\end{matrix}}\right\}3}}$$

and where the number of 3s in each tower, starting from the leftmost tower, is specified by the value of the next tower to the right.

In other words, g₁ is computed by first calculating the number of towers, ${\displaystyle n=3\uparrow (3\uparrow (3\ \dots \ \uparrow 3))}$ (where the number of 3s is ${\displaystyle 3\uparrow (3\uparrow 3)=7625597484987}$), and then computing the n^th tower in the following sequence:

      1st tower:  3
     
      2nd tower:  3↑3↑3 (number of 3s is 3) = 7625597484987
     
      3rd tower:  3↑3↑3↑3↑...↑3 (number of 3s is 7625597484987) = …
     
      ⋮
     
 g1 = nth tower:  3↑3↑3↑3↑3↑3↑3↑...↑3 (number of 3s is given by the n − 1th tower)

where the number of 3s in each successive tower is given by the tower just before it. Note that the result of calculating the third tower is the value of n, the number of towers for g₁.

The magnitude of this first term, g₁, is so large that it is practically incomprehensible, even though the above display is relatively easy to comprehend. Even n, the mere number of towers in this formula for g₁, is far greater than the number of Planck volumes (roughly 10¹⁸⁵ of them) into which one can imagine subdividing the observable universe. And after this first term, still another 63 terms remain in the rapidly growing g sequence before Graham's number G = g₆₄ is reached. To illustrate just how fast this sequence grows, while g₁ is equal to ${\displaystyle 3\uparrow \uparrow \uparrow \uparrow 3}$ with only four up arrows, the number of up arrows in g₂ is this incomprehensibly large number g₁.

Rightmost decimal digits

Graham's number, ${\displaystyle G}$, is a "power tower" of the form ${\displaystyle 3\uparrow \uparrow n}$ (with an extremely large value of ${\displaystyle n}$), so its ${\displaystyle n-1}$ rightmost decimal digits are stable digits (for a proof, see Lemma 1 of Reference,^[9] and thus they are the same of the rightmost ${\displaystyle n-1}$ digits of the ${\displaystyle 10}$-adic fixed point of ${\displaystyle x\mapsto 3^{x}}$. This follows from a peculiar property of tetration which is called "the constancy of the congruence speed"^[10] so that all such towers of height greater than ${\displaystyle d:d>1}$ (say) are characterized by the same sequence of ${\displaystyle d-2}$ rightmost decimal digits (i.e., ${\displaystyle 2\uparrow \uparrow 5}$ has exactly ${\displaystyle 3}$ stable digits since ${\displaystyle 2\uparrow \uparrow 5\equiv 2\uparrow \uparrow 6{\pmod {10}}^{3}}$, but ${\displaystyle 2\uparrow \uparrow 5}$ is not congruent modulo ${\displaystyle 10^{4}}$ to ${\displaystyle 2\uparrow \uparrow 6}$).

This is a special case of the above-mentioned general property.^[11]

The following table illustrates, for a few values of ${\displaystyle d}$, how this happens. For a given height of tower and number of digits ${\displaystyle d}$, the full range of ${\displaystyle d}$-digit numbers (${\displaystyle 10^{d}}$ of them) does not occur; instead, a certain smaller subset of values repeats itself in a cycle. The length of the cycle and some of the values (in parentheses) are shown in each cell of this table:

Number of different possible values of 3↑3↑${\displaystyle \cdots }$3↑x when all but the rightmost d decimal digits are ignored

Number of digits (d)	3↑x	3↑3↑x	3↑3↑3↑x	3↑3↑3↑3↑x	3↑3↑3↑3↑3↑x
1	4 (1,3,9,7)	2 (3,7)	1 (7)
2	20 (01,03,…,87,…,67)	4 (03,27,83,87)	2 (27,87)	1 (87)
3	100 (001,003,…,387,…,667)	20 (003,027,…387,…,587)	4 (027,987,227,387)	2 (987,387)	1 (387)

The particular rightmost ${\displaystyle d}$ digits that are ultimately shared by all sufficiently tall towers of ${\displaystyle 3}$s are in bold text, and can be seen developing as the tower height increases. For any fixed number of digits ${\displaystyle d}$ (row in the table), the number of values possible for ${\displaystyle 3\uparrow 3\uparrow \cdots \uparrow x{\pmod {10}}^{d}}$, as ${\displaystyle x}$ ranges over all nonnegative integers, is seen to decrease steadily as the height increases, until eventually reducing the "possibility set" to a single number (colored cells) when the height exceeds ${\displaystyle d+2}$.

A simple algorithm^[12] for computing these digits may be described as follows: let ${\displaystyle x=3}$, then iterate, ${\displaystyle d}$ times, the assignment ${\displaystyle x=3^{x}{\pmod {10}}^{d}}$. Except for omitting any leading ${\displaystyle 0}$s, the final value assigned to ${\displaystyle x}$ (as a base-ten numeral) is then composed of the ${\displaystyle d}$ rightmost decimal digits of ${\displaystyle 3\uparrow \uparrow n}$, for all ${\displaystyle n>d}$ (if the final value of ${\displaystyle x}$ has fewer than ${\displaystyle d}$ digits, then the required number of leading ${\displaystyle 0}$s must be added).

The numerousness of these stable digits is ${\displaystyle b-1}$ since ${\displaystyle G:=3\uparrow \uparrow b}$, which satisfy the congruence relation

$${\displaystyle G{\pmod {10}}^{b-1}\equiv G^{G}{\pmod {10}}^{b-1}}$$

(whereas ${\displaystyle G{\pmod {10}}^{b}}$ is not congruent modulo ${\displaystyle 10^{b}}$ to ${\displaystyle G^{G}}$).^[13] It follows that, ${\displaystyle g_{63}<<k<<g_{64}}$.

The algorithm above produces the following ${\displaystyle 500}$ rightmost decimal digits of Graham's number (OEIS: A133613):

...02425950695064738395657479136519351798334535362521
   43003540126026771622672160419810652263169355188780
   38814483140652526168785095552646051071172000997092
   91249544378887496062882911725063001303622934916080
   25459461494578871427832350829242102091825896753560
   43086993801689249889268099510169055919951195027887
   17830837018340236474548882222161573228010132974509
   27344594504343300901096928025352751833289884461508
   94042482650181938515625357963996189939679054966380
   03222348723967018485186439059104575627262464195387

$${\displaystyle \left.{\begin{matrix}g_{n}&=&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \cdots \uparrow } 3\\&&3\underbrace {\uparrow \uparrow \cdots \cdots \cdots \cdots \uparrow } 3\\&&\underbrace {\qquad \quad \vdots \qquad \quad } \\&&3\underbrace {\uparrow \uparrow \cdots \cdots \uparrow } 3\\&&3\uparrow \uparrow \uparrow \uparrow 3\end{matrix}}\right\}n\ \mathrm {layers} =...02425950695064738395657479136519351798334535362521430035401260267716226721604198106522631693551887803881448314065252616878509555264605107117200099709291249544378887496062882911725063001303622934916080254594614945788714278323508292421020918258967535604308699380168924988926809951016905591995119502788717830837018340236474548882222161573228010132974509273445945043433009010969280253527518332898844615089404248265018193851562535796399618993967905496638003222348723967018485186439059104575627262464195387}$$

References

External links

• OEIS sequence A133613 (Graham's number)
• Sbiis Saibian's article on Graham's number
• "A Ramsey Problem on Hypercubes" by Geoff Exoo
• Weisstein, Eric W. "Graham's Number". MathWorld.
• How to calculate Graham's number
• Conceptualizing Graham's number
• Some Ramsey results for the n-cube prepublication mentions Graham's number
• Padilla, Tony; Parker, Matt. "Graham's Number". Numberphile. Brady Haran. Archived from the original on 2014-05-27. Retrieved 2013-04-08.
• Archived at Ghostarchive and the Wayback Machine: Ron Graham. "What is Graham's Number? (feat Ron Graham)" (video). Numberphile. Brady Haran.
• Archived at Ghostarchive and the Wayback Machine: Ron Graham. "How Big is Graham's Number? (feat Ron Graham)" (video). Numberphile. Brady Haran.
• The last 16 million digits of Graham's number by the Darkside communication group