For example,

(6) The sequence of primes is endless.

For, if $p$ is any prime, the number ${\begin{array}{|c}\color{red}p\\\hline\end{array} + 1}$ is greater than $p$ and is not divisible by $p$ or by any smaller prime. If then ${\begin{array}{|c}p\\\hline\end{array} + 1}$ is not a prime, it must have a prime divisor greater than $p$, and in either case a prime greater than $p$ exists.

And here is another one:

9. Theorems.

(1) The product of any $n$ consecutive integers is divisible by $\begin{array}{|c}n\\\hline\end{array}$

For $(m+1)(m+2)...(m+n)/\begin{array}{|c}n\\\hline\end{array} = \begin{array}{|c}\color{red}{m + n}\\\hline\end{array}/\begin{array}{|c}\color{red}m\\\hline\end{array}\begin{array}{|c}\color{red}n\\\hline\end{array}$, and to show that the last expression is an integer it is sufficient to show that any prime $p$ which occurs in $\begin{array}{|c}\color{red}m\\\hline\end{array}\begin{array}{|c}\color{red}n\\\hline\end{array}$ occurs to at least as high a power in $\begin{array}{|c}\color{red}{m+n}\\\hline\end{array}$. Thus we have to show that

$$I[(m+n)/p]+I[(m+n)/p^2]+I[(m+n)/p^3]+... \\ \geq I[m/p]+I[m/p^2]+I[m/p^3]+... \\ +I[n/p] +I[n/p^2]+I[n/p^3]+...$$

  • 12
    $\begingroup$ I guess it denotes the factorial: $\begin{array}{|c}n\\\hline\end{array}=n!$ $\endgroup$
    – egreg
    Commented Apr 2, 2016 at 10:26
  • 10
    $\begingroup$ It seems to denote the factorial. It is a quite unusual notation, though. Never seen before. $\endgroup$
    – Crostul
    Commented Apr 2, 2016 at 10:26
  • 6
    $\begingroup$ @egreg You beat me to it cause I was still thinking how to TeX that symbol :) $\endgroup$ Commented Apr 2, 2016 at 10:26
  • 17
    $\begingroup$ It would be good to cite the work in which you found that notation. More than likely the author defined it earlier in the written text. $\endgroup$
    – hardmath
    Commented Apr 2, 2016 at 10:29
  • 4
    $\begingroup$ It's used in Ramanujan's notebooks, to denote factorials. I don't suppose he invented it, so I'd guess it was common a century ago. Ramanujan wasn't constrained by TeX, so it didn't cause him any trouble :-) $\endgroup$
    – bubba
    Commented Apr 2, 2016 at 12:20

5 Answers 5


In Cajori's "A History of Mathematical Notations", this symbol is attributed to Thomas Jarrett and means $n!$. See article 447 of Cajori's book for the attribution and articles 448 and 449 for the history of its use, mainly in the 19th century.


This is the notation once used for factorials. I have a copy of Hall and Knight's Higher Algebra (1964 reprint, first published in 1887) that uses this notation. A scan from two pages of the book is at the top of the page https://kconrad.math.uconn.edu/factorials/ shows that corner notation defined and also n! is mentioned as another notation that is "sometimes used." Lower down on that webpage are screenshots of the corner factorial notation in work by Eisenhart and Hilbert and at the bottom of the page is a photo at Colorado State from long ago that shows someone using the corner factorial notation at a blackboard.

The Wikipedia page for factorials says the use of ! for factorial was introduced in the early 1800s.


Adding to @Bernard Masse, the history of notations for products of terms in arithmetical progression is quite long. The following is taken from Christian Kramp, 1808, Elémens d'arithmétique universelle (pdf), p. XI-XII:

Image of text from a book, transcribed below

Pour désigner les produits dont les facteurs constituent entr’eux une progression arithmétique, tels que $a(a+r)(a+2r)....(a+nr-r)$, j’ai conservé la notation $a^{n|r}$, déjà proposée dans mon analyse des refractions; je leur avois donné le nom de facultés. Arbogast lui avoit substitué la dénomination plus nette et plus françoise de factorielles; j’ai reconnu l’avantage de cette nouvelle dénomination; et en adoptant son idée, je me suis félicité de pouvoir rendre hommage à la mémoire de mon ami.

where $1^{p|1}$ would denote the (nowadays) factorial.

He used the term "faculties (facultées en French)", acknowledged Louis Arbogast for the term "factorial (factorielles)", and on page 348, he introduces the notation $p!$ for $1^{p|1}$ (additional details in History of notation: "!"). You can find this from Article 445 (Volume 2, Page 66) in Florian Cajori, 1993, A History of Mathematical Notations (Dover Publications).

From Article 447, Cajori mentions Thomas Jarrett (1805–1882) for his extensive study of algebraic notations, An essay on algebraic development: containing the principal expansions in common algebra, in the differential and integral calculus, and in the calculus of finite differences, 1831, Page 15, Article 38, you find:

$$\begin{array}{|c}p\\\hline\end{array}=p(p-1)\ldots 1$$

Image of text from a book, transcribed below

The symbol $\underset{n,m}{\begin{array}{|c}a\\\hline\end{array}}$ denotes the product of $n$ factors forming an arithmetical progression, of which the first term is $a$, and the common difference $m$; if $m = -1$, the $m$ may be omitted; and if, in the same case, $n = a$ the $n$ also may be omitted: thus

$\underset{n,m}{\begin{array}{|c}a\\\hline\end{array}} = a(a+m)(a+2m)...(a+\overline{n-1}.m)$,
$\underset{n\phantom{,m}}{\begin{array}{|c}a\\\hline\end{array}} = a(a-1)(a-2)...(a-n+1)$, and
${\begin{array}{|c}a\\\hline\end{array}} = a(a-1)(a-2)...2.1$.

  • 4
    $\begingroup$ Why are old books yellow $\endgroup$
    – Olórin
    Commented Apr 2, 2016 at 20:27
  • 2
    $\begingroup$ @MSE is a dating site : yellow to books is what grey is to hair. Kramp's 1808 book is older and the scan is whiter $\endgroup$ Commented Apr 2, 2016 at 20:30
  • 3
    $\begingroup$ @LaurentDuval (Most) paper made before about 1850 is more durable than (most) paper made between 1850 and 1980, due to different industrial processes. The yellowing is a symptom of this. See loc.gov/preservation/care/deterioratebrochure.html $\endgroup$
    – zwol
    Commented Apr 4, 2016 at 0:36
  • 1
    $\begingroup$ I'm not sure how I made it to this old answer but if you'd like to put $n,m$ underneath, one option is \underset{n,m}{\begin{array}{|c}p\\\hline\end{array}}=p(p-1)\dots 1 $$\underset{n,m}{\begin{array}{|c}p\\\hline\end{array}}=p(p-1)\dots 1$$ $\endgroup$ Commented Feb 2, 2021 at 8:00

From the context, it seems to mean factorial ($n!=1\cdot 2 \cdot 3 \cdot \ldots \cdot n$). But I've never seen it before.


The first example is a well known proof of Euclid's result about infiniteness of primes; one takes a prime number $p$, then does $$ N=1+p! $$ and shows that $N$ is not divisible by $p$ nor by any smaller prime, because each one divides $p!$. So there exists a bigger prime than $p$, because $N>1$ is divisible by a prime. This means $\begin{array}{|c}p\\\hline\end{array}=p!$ (sorry for the bad emulation of the symbol); it could mean the primorial, that is, the product of all prime numbers from $2$ up to $p$, but this interpretation would contradict the usage in the second example.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .