View source for Triangular number

{{Short description|Figurate number}}
[[File:First six triangular numbers.svg|thumb|The first six triangular numbers (not starting with ''T''<sub>0</sub>, but rather, ''T''<sub>1</sub>)]]
[[File:Triangular Numbers Plot.svg|thumb|Triangular Numbers Plot]]
A '''triangular number''' or '''triangle number''' counts objects arranged in an [[equilateral triangle]]. Triangular numbers are a type of [[figurate number]], other examples being [[square number]]s and [[Cube (algebra)#In integers|cube numbers]]. The {{mvar|n}}th triangular number is the number of dots in the triangular arrangement with {{mvar|n}} dots on each side, and is equal to the sum of the {{mvar|n}} [[natural number]]s from 1 to {{mvar|n}}. The first 100 terms [[Integer sequence|sequence]] of triangular numbers, starting with the [[Empty sum|0th triangular number]], are
{{block indent|left=1.6|0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035, 1081, 1128, 1176, 1225, 1275, 1326, 1378, 1431, 1485, 1540, 1596, 1653, 1711, 1770, 1830, 1891, 1953, 2016, 2080, 2145, 2211, 2278, 2346, 2415, 2485, 2556, 2628, 2701, 2775, 2850, 2926, 3003, 3081, 3160, 3240, 3321, 3403, 3486, 3570, 3655, 3741, 3828, 3916, 4005, 4095, 4186, 4278, 4371, 4465, 4560, 4656, 4753, 4851, 4950, 5050...}}
{{OEIS|id=A000217}}

==Formula==
{{Pascal triangle simplex numbers.svg|2=triangular numbers}}
The triangular numbers are given by the following explicit formulas:
<!-- THE END OF THE NEXT LINE HAS BINOMIAL NOTATION. PLEASE DO NOT CHANGE IF YOU ARE UNFAMILIAR. -->
{{bi|left=1.6|1=<math>\displaystyle
\begin{align} T_n &= \sum_{k=1}^n k = 1+2+ \dotsb +n \\
&= \frac{n^2+n \vphantom{(n+1)}}{2} = \frac{n(n+1)}{2} \\
&= {n+1 \choose 2}
\end{align}</math>}}
where <math>\textstyle {n+1 \choose 2}</math> is notation for a [[binomial coefficient]]. It represents the number of distinct pairs that can be selected from {{math|''n'' + 1}} objects, and it is read aloud as "{{mvar|n}} plus one choose two".

The fact that the <math>n</math>th triangular number equals <math>n(n+1)/2</math> can be illustrated using a [[Proof without words|visual proof]].<ref>{{cite web|url=https://www.mathsisfun.com/algebra/triangular-numbers.html|title=Triangular Number Sequence|website=Math Is Fun}}</ref> For every triangular number <math>T_n</math>, imagine a "half-rectangle" arrangement of objects corresponding to the triangular number, as in the figure below.  Copying this arrangement and rotating it to create a rectangular figure doubles the number of objects, producing a rectangle with dimensions <math>n \times (n+1)</math>, which is also the number of objects in the rectangle.  Clearly, the triangular number itself is always exactly half of the number of objects in such a figure, or: <math>T_n = \frac{n(n+1)}{2} </math>.  The example <math>T_4</math> follows:

{{bi|left=1.6
|<math> 2T_4 = 4(4+1) = 20 </math> (green plus yellow) implies that <math>T_4 = \frac{4(4+1)}{2} = 10 </math> (green).&nbsp;&nbsp;&nbsp; [[File:Illustration of Triangular Number T 4 Leading to a Rectangle (yellow-green).svg]]
}}

This formula can be proven formally using [[mathematical induction]].<ref>{{cite book |last=Spivak |first=Michael |author-link=Michael Spivak |date=2008 |title=Calculus |edition=4th |url=https://books.google.com/books?id=7JKVu_9InRUC&pg=PA21 |location=Houston, Texas |publisher=Publish or Perish |pages=21–22 |isbn=978-0-914098-91-1}}</ref> It is clearly true for <math>1</math>:

<math display=block>T_1 = \sum_{k=1}^{1}k = \frac{1(1 + 1)}{2} = \frac{2}{2} = 1.</math>

Now assume that, for some natural number <math>m</math>, <math>T_m = \sum_{k=1}^{m}k = \frac{m(m + 1)}{2}</math>.  We can then verify it for <math>m+1</math>:
<math display=block>
\begin{align}
  \sum_{k=1}^{m+1}k &= \sum_{k=1}^{m}k + (m + 1) \\
                    &= \frac{m(m + 1)}{2} + m + 1\\
                    &= \frac{m^2 + m}{2} + \frac{2m + 2}{2}\\
                    &= \frac{m^2 + 3m + 2}{2}\\
                    &= \frac{(m + 1)(m + 2)}{2},
\end{align}
</math>

so if the formula is true for <math>m</math>, it is true for <math>m+1</math>. Since it is clearly true for <math>1</math>, it is therefore true for <math>2</math>, <math>3</math>, and ultimately all natural numbers <math>n</math> by induction.

The German mathematician and scientist, [[Carl Friedrich Gauss]], is said to have found this relationship in his early youth, by multiplying {{math|{{sfrac|''n''|2}}}} pairs of numbers in the sum by the values of each pair {{math|''n'' + 1}}.<ref name=Gauss>{{cite web |last=Hayes |first=Brian |title=Gauss's Day of Reckoning |url=http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning/1 |website=American Scientist |publisher=Computing Science |access-date=2014-04-16 |archive-date=2015-04-02 |archive-url=https://web.archive.org/web/20150402141244/http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning/1 }}</ref> However, regardless of the truth of this story, Gauss was not the first to discover this formula, and some find it likely that its origin goes back to the [[Pythagoreans]] in the 5th century BC.<ref>{{cite web|last=Eves|first=Howard|title=Webpage cites AN INTRODUCTION TO THE HISTORY OF MATHEMATICS|url=http://mathcentral.uregina.ca/QQ/database/QQ.09.98/matt1.html|publisher=Mathcentral|access-date=28 March 2015}}</ref> The two formulas were described by the Irish monk [[Dicuil]] in about 816 in his [[Computus]].<ref>{{cite journal |last=Esposito |first=Mario |author-link=Mario Esposito (scholar) |title=An unpublished astronomical treatise by the Irish monk Dicuil |journal=[[Proceedings of the Royal Irish Academy, Section C]] |volume=26 |location=Dublin |date=August 1907 |pages=378–446+i (PDF pages 704–773) |lang=en,la |url=https://archive.org/details/proceedingsofroy2619roya}}</ref> An English translation of Dicuil's account is available.<ref>{{cite journal |last1=Ross |first1=H.E. |last2=Knott |first2=B.I. |title=Dicuil (9th century) on triangular and square numbers |journal=British Journal for the History of Mathematics |year=2019 |volume=34 |issue=2 |pages=79–94 |doi=10.1080/26375451.2019.1598687 |url=https://dspace.stir.ac.uk/handle/1893/29437|hdl=1893/29437 |hdl-access=free }}</ref>

Occasionally it is necessary to compute large triangular numbers where the standard formula <code>t = n*(n+1)/2</code> would suffer [[integer overflow]] before the final division by 2.  For example, {{math|''T''<sub>20</sub>}} = 210 < 256, so will fit into an [[8-bit byte]], but not the intermediate product 420.  This can be solved by dividing either {{mvar|n}} or {{mvar|n+1}} by 2 before the multiplication, whichever is even.  This does not require a [[conditional branch]] if implemented as <code>t = (n|1) * ((n+1)/2)</code>.  If <code>n</code> is odd, the [[binary OR]] operation <code>n|1</code> has no effect, so this is equivalent to <code>t = n * ((n+1)/2)</code> and thus correct.  If <code>n</code> is even, setting the low bit with <code>n|1</code> is the same as adding 1, while the 1 added before the division is [[Division (mathematics)#Of integers|truncated away]], so this is equivalent to <code>t = (n+1) * (n/2)</code> and also correct.

==Relations to other figurate numbers==
Triangular numbers have a wide variety of relations to other figurate numbers.

Most simply, the sum of two consecutive triangular numbers is a square number, since:<ref>{{cite journal
| last1 = Beldon
| first1 = Tom
| last2 = Gardiner
| first2 = Tony
| title = Triangular Numbers and Perfect Squares
| journal = The Mathematical Gazette
| volume = 86
| issue = 507
| year = 2002
| pages = 423–431
| doi = 10.2307/3621134
| url = https://doi.org/10.2307/3621134
| access-date = 25 April 2024
| jstor = 3621134
}}</ref><ref>{{cite web |title=Triangular Number |url=https://mathworld.wolfram.com/TriangularNumber.html |author=Eric W. Weisstein |publisher=Wolfram MathWorld |accessdate=2024-04-14}} See equations 18 - 20.</ref>

:<math>T_{n - 1} + T_{n}</math>
:<math>= \frac{1}{2} \, n(n-1) + \frac{1}{2} \, n (n + 1)</math>
:<math>= \frac{1}{2} \, n\Bigl((n - 1) + (n + 1)\Bigr)</math>
:<math>= n^2</math> 

with the sum being the square of the difference between the two (and thus the difference of the two being the square root of the sum):
<math display="block">T_n + T_{n-1} = \left (\frac{n^2}{2} + \frac{n}{2}\right) + \left(\frac{\left(n-1\right)^2}{2} + \frac{n-1 \vphantom{\left(n-1\right)^2}}{2} \right ) = \left (\frac{n^2}{2} + \frac{n}{2}\right) + \left(\frac{n^2}{2} - \frac{n}{2} \right ) = n^2 = (T_n - T_{n-1})^2.</math>

This property, colloquially known as the theorem of [[Theon of Smyrna]],<ref>{{cite book
    | last1 = Shell-Gellasch
    | first1 = Amy
    | last2 = Thoo
    | first2 = John
    | author-link = 
    | date = October 15, 2015
    | title = Algebra in Context: Introductory Algebra from Origins to Applications
    | url = https://doi.org/10.1353/book.49475
    | location = 
    | publisher = Johns Hopkins University Press
    | page = 210
    | doi = 10.1353/book.49475
 | isbn = 9781421417288
}}</ref> is visually demonstrated in the following sum, which represents <math>T_{4} + T_{5} = 5^2</math> as [[Digit sum|digit sums]]:

<math>\begin{array}{ccccccc}
    & 4 & 3 & 2 & 1 & \\
+   & 1 & 2 & 3 & 4 & 5 \\
\hline
    & 5 & 5 & 5 & 5 & 5
\end{array}</math>

This fact can also be demonstrated graphically by positioning the triangles in opposite directions to create a square:
{{bi|left=1.6
|6 + 10 {{=}} 16 &nbsp;&nbsp;&nbsp; [[File:Square number 16 as sum of two triangular numbers.svg]] &nbsp;&nbsp;&nbsp; 10 + 15 {{=}} 25 &nbsp;&nbsp;&nbsp; [[File:Square number 25 as sum of two triangular numbers.svg]]
}}

The double of a triangular number, as in the visual proof from the above section {{section link||Formula}}, is called a [[pronic number]].

There are infinitely many triangular numbers that are also square numbers; e.g., 1, 36, 1225. Some of them can be generated by a simple recursive formula:
<math display=block>S_{n+1} = 4S_n \left( 8S_n + 1\right)</math> with <math>S_1 = 1.</math>

''All'' [[square triangular number]]s are found from the recursion
<math display=block>S_n = 34S_{n-1} - S_{n-2} + 2</math> with <math>S_0 = 0</math> and <math>S_1 = 1.</math>

[[File:Nicomachus_theorem_3D.svg|thumb|A square whose side length is a triangular number can be partitioned into squares and half-squares whose areas add to cubes. This shows that the square of the {{mvar|n}}th triangular number is equal to the sum of the first {{mvar|n}} cube numbers.]]

Also, the [[squared triangular number|square of the {{mvar|n}}th triangular number]] is the same as the sum of the cubes of the integers 1 to {{mvar|n}}. This can also be expressed as
<math display="block"> \sum_{k=1}^n k^3 = \left(\sum_{k=1}^n k \right)^2.</math>

The sum of the first {{mvar|n}} triangular numbers is the {{mvar|n}}th [[tetrahedral number]]:
<math display=block> \sum_{k=1}^n T_k = \sum_{k=1}^n \frac{k(k+1)}{2} = \frac {n(n+1)(n+2)} {6}.</math>

More generally, the difference between the {{mvar|n}}th [[polygonal number|{{mvar|m}}-gonal number]] and the {{mvar|n}}th {{math|(''m'' + 1)}}-gonal number is the {{math|(''n'' − 1)}}th triangular number. For example, the sixth [[heptagonal number]] (81) minus the sixth [[hexagonal number]] (66) equals the fifth triangular number, 15. Every other triangular number is a hexagonal number. Knowing the triangular numbers, one can reckon any [[centered polygonal number]]; the {{mvar|n}}th centered {{mvar|k}}-gonal number is obtained by the formula
<math display=block>Ck_n = kT_{n-1}+1</math>

where {{mvar|T}} is a triangular number.

The positive difference of two triangular numbers is a [[trapezoidal number]].

The pattern found for triangular numbers <math> \sum_{n_1=1}^{n_2}n_1=\binom{n_2+1}{2}</math> and for tetrahedral numbers <math> \sum_{n_2=1}^{n_3}\sum_{n_1=1}^{n_2} n_1=\binom{n_3+2}{3},</math> which uses [[binomial coefficient]]s, can be generalized. This leads to the formula:<ref>{{Cite journal| last=Baumann|first=Michael Heinrich |date=2018-12-12|title=Die {{mvar|k}}-dimensionale Champagnerpyramide |journal=Mathematische Semesterberichte |language=de|volume=66|pages=89–100 |doi=10.1007/s00591-018-00236-x |s2cid=125426184 |issn=1432-1815 |url=https://epub.uni-bayreuth.de/3850/1/Baumann_Champagnerpyramide.pdf}}</ref>
<math display=block> \sum_{n_{k-1}=1}^{n_k}\sum_{n_{k-2}=1}^{n_{k-1}} \dots \sum_{n_2=1}^{n_3}\sum_{n_1=1}^{n_2}n_1 =\binom{n_k+k-1}{k}</math>
[[File:tetrahedral_triangular_number_10.svg|thumb|The fourth triangular number equals the third tetrahedral number as the ''n''th ''k''-simplex number equals the ''k''th ''n''-simplex number due to the symmetry of [[Pascal's triangle]], and its diagonals being simplex numbers; similarly, the fifth triangular number (15) equals the third [[pentatope number]], and so forth]]

==Other properties==
Triangular numbers correspond to the first-degree case of [[Faulhaber's formula]].
	
{{annotated image
|image      =hexagonal_number_visual_proof.svg
|height     =200
|width      =200
|image-width=350
|image-left =-135
|image-top  =-5
|caption    =[[Proof without words]] that all hexagonal numbers are odd-sided triangular numbers
}}
Alternating triangular numbers (1, 6, 15, 28, ...) are also hexagonal numbers.

Every even [[perfect number]] is triangular (as well as hexagonal), given by the formula
<math display=block>M_p 2^{p-1} = \frac{M_p (M_p + 1)}2 = T_{M_p}</math>
where {{mvar|M<sub>p</sub>}} is a [[Mersenne prime]]. No odd perfect numbers are known; hence, all known perfect numbers are triangular.

For example, the third triangular number is (3 × 2 =) 6, the seventh is (7 × 4 =) 28, the 31st is (31 × 16 =) 496, and the 127th is (127 × 64 =) 8128.

The final digit of a triangular number is 0, 1, 3, 5, 6, or 8, and thus such numbers never end in 2, 4, 7, or 9. A final 3 must be preceded by a 0 or 5; a final 8 must be preceded by a 2 or 7.

In [[base 10]], the [[digital root]] of a nonzero triangular number is always 1, 3, 6, or 9. Hence, every triangular number is either divisible by three or has a remainder of 1 when divided by 9:
{{Block indent|left=1.6|1={{Break lines|1=
0 = 9 × 0
1 = 9 × 0 + 1
3 = 9 × 0 + 3
6 = 9 × 0 + 6
10 = 9 × 1 + 1
15 = 9 × 1 + 6
21 = 9 × 2 + 3
28 = 9 × 3 + 1
36 = 9 × 4
45 = 9 × 5
55 = 9 × 6 + 1
66 = 9 × 7 + 3
78 = 9 × 8 + 6
91 = 9 × 10 + 1
...}}}}

The digital root pattern for triangular numbers, repeating every nine terms, as shown above, is "1, 3, 6, 1, 6, 3, 1, 9, 9".

The converse of the statement above is, however, not always true. For example, the digital root of 12, which is not a triangular number, is 3 and divisible by three.

If {{mvar|x}} is a triangular number, {{mvar|a}} is an odd square, and {{math|''b'' {{=}} {{sfrac|''a'' − 1|8}}}}, then {{math|''ax'' + ''b''}} is also a triangular number. Note that {{mvar|b}} will always be a triangular number, because {{math|1=8''T<sub>n</sub>'' + 1 = (2''n'' + 1)<sup>2</sup>}}, which yields all the odd squares are revealed by multiplying a triangular number by 8 and adding 1, and the process for {{mvar|b}} given {{mvar|a}} is an odd square is the inverse of this operation.
The first several pairs of this form (not counting {{math|1''x'' + 0}}) are: {{math|9''x'' + 1}}, {{math|25''x'' + 3}}, {{math|49''x'' + 6}}, {{math|81''x'' + 10}}, {{math|121''x'' + 15}}, {{math|169''x'' + 21}}, ... etc. Given {{mvar|x}} is equal to {{mvar|T<sub>n</sub>}}, these formulas yield {{math|''T''<sub>3''n'' + 1</sub>}}, {{math|''T''<sub>5''n'' + 2</sub>}}, {{math|''T''<sub>7''n'' + 3</sub>}}, {{math|''T''<sub>9''n'' + 4</sub>}}, and so on.

The sum of the [[multiplicative inverse|reciprocals]] of all the nonzero triangular numbers is
<math display=block> \sum_{n=1}^\infty{1 \over {{n^2 + n} \over 2}} = 2\sum_{n=1}^\infty{1 \over {n^2 + n}} = 2 .</math>

This can be shown by using the basic sum of a [[telescoping series]]:
<math display=block> \sum_{n=1}^\infty{1 \over {n(n+1)}} = 1 .</math>

In addition, the ''n''th partial sum of this series can be written as {{sfrac|''2n''|''n'' + 1}}

Two other formulas regarding triangular numbers are
<math display=block>T_{a+b} = T_a + T_b + ab</math>
and
<math display=block>T_{ab} = T_aT_b + T_{a-1}T_{b-1},</math>
both of which can easily be established either by looking at dot patterns (see above) or with some simple algebra. The first formula are relevant to [[multiplication algorithm#Quarter square multiplication]].

In 1796, Gauss discovered that every positive integer is representable as a sum of three triangular numbers, writing in his diary his famous words, "[[Eureka (word)|ΕΥΡΗΚΑ!]] {{nowrap|1=num = Δ + Δ + Δ}}". The three triangular numbers are not necessarily distinct, or nonzero; for example 20 = 10 + 10 + 0. This is a special case of the [[Fermat polygonal number theorem]].

The largest triangular number of the form {{math|2<sup>''k''</sup> −&thinsp;1}} is [[4000 (number)#4001 to 4099|4095]] (see [[Ramanujan–Nagell equation]]).

[[Wacław Sierpiński|Wacław Franciszek Sierpiński]] posed the question as to the existence of four distinct triangular numbers in [[geometric progression]]. It was conjectured by Polish mathematician [[Kazimierz Szymiczek]] to be impossible and was later proven by Fang and Chen in 2007.<ref>[http://www.emis.de/journals/INTEGERS/papers/h19/h19.pdf Chen, Fang: Triangular numbers in geometric progression]</ref><ref>[http://www.emis.de/journals/INTEGERS/papers/h57/h57.pdf Fang: Nonexistence of a geometric progression that contains four triangular numbers]</ref>

Formulas involving expressing an integer as the sum of triangular numbers are connected to [[theta function]]s, in particular the [[Ramanujan theta function]].<ref>{{Cite journal|last=Liu|first=Zhi-Guo|date=2003-12-01|title=An Identity of Ramanujan and the Representation of Integers as Sums of Triangular Numbers|journal=The Ramanujan Journal|language=en|volume=7|issue=4|pages=407–434|doi=10.1023/B:RAMA.0000012425.42327.ae|s2cid=122221070|issn=1382-4090}}</ref><ref>{{cite arXiv|last=Sun|first=Zhi-Hong|date=2016-01-24|title=Ramanujan's theta functions and sums of triangular numbers|eprint=1601.06378|class=math.NT}}</ref>

The number of line segments between closest pairs of dots in the triangle can be represented in terms of the number of dots or with a [[recurrence relation]]:
<math display=block>L_n = 3 T_{n-1} = 3{n \choose 2};\qquad L_n = L_{n-1} + 3(n-1), ~L_1 = 0.</math>

In the [[Limit of a sequence|limit]], the ratio between the two numbers, dots and line segments is
<math display=block>\lim_{n\to\infty} \frac{T_n}{L_n} = \frac{1}{3}.</math>

==Applications==
{{Anchor|Handshake problem}}<!-- "Handshake problem" redirects here. -->
[[File:handshake_problem_visual_proof.svg|thumb|upright|[[Proof without words]] that the number of possible handshakes between n people is the (n−1)th triangular number]]
The triangular number {{mvar|T<sub>n</sub>}} solves the '''handshake problem''' of counting the number of handshakes if each person in a room with {{math|''n'' + 1}} people shakes hands once with each person. In other words, the solution to the handshake problem of {{mvar|n}} people is {{math|''T''<sub>''n''−1</sub>}}.<ref>{{cite web |url=http://www.mathcircles.org/node/835 |title=The Handshake Problem &#124; National Association of Math Circles |website=MathCircles.org |access-date=12 January 2022 |archive-url=https://web.archive.org/web/20160310182700/http://www.mathcircles.org/node/835 |archive-date=10 March 2016 }}</ref> 

Equivalently, a [[fully connected network]] of {{mvar|n}} computing devices requires the presence of {{math|''T''<sub>''n'' −&thinsp;1</sub>}} cables or other connections.

In a tournament format that uses a round-robin [[group stage]], the number of matches that need to be played between {{mvar|n}} teams is equal to the triangular number {{math|''T''<sub>''n'' −&thinsp;1</sub>}}. For example, a group stage with 4 teams requires 6 matches, and a group stage with 8 teams requires 28 matches. This is also equivalent to the handshake problem and fully connected network problems.

{{central_polygonal_numbers.svg}}
One way of calculating the [[depreciation]] of an asset is the [[depreciation#Sum-of-years-digits method|sum-of-years' digits method]], which involves finding {{mvar|T<sub>n</sub>}}, where {{mvar|n}} is the length in years of the asset's useful life. Each year, the item loses {{math|(''b'' − ''s'') × {{sfrac|''n'' − ''y''|''T<sub>n</sub>''}}}}, where {{mvar|b}} is the item's beginning value (in units of currency), {{mvar|s}} is its final salvage value, {{mvar|n}} is the total number of years the item is usable, and {{mvar|y}} the current year in the depreciation schedule. Under this method, an item with a usable life of {{mvar|n}} = 4 years would lose {{sfrac|4|10}} of its "losable" value in the first year, {{sfrac|3|10}} in the second, {{sfrac|2|10}} in the third, and {{sfrac|1|10}} in the fourth, accumulating a total depreciation of {{sfrac|10|10}} (the whole) of the losable value.

[[Board game]] designers Geoffrey Engelstein and Isaac Shalev describe triangular numbers as having achieved "nearly the status of a mantra or koan among [[Game design|game designers]]", describing them as "deeply intuitive" and "featured in an enormous number of games, [proving] incredibly versatile at providing escalating rewards for larger sets without overly incentivizing specialization to the exclusion of all other strategies".<ref>{{Cite book |last1=Engelstein |first1=Geoffrey |last2=Shalev |first2=Isaac |date=2019-06-25 |title=Building Blocks of Tabletop Game Design |url=http://dx.doi.org/10.1201/9780429430701 |doi=10.1201/9780429430701|isbn=978-0-429-43070-1 |s2cid=198342061 }}</ref>
{| class="wikitable" style="text-align:center; margin:0 auto;"
|+ Relationship between the maximum number of pips on an end of a [[domino]] and the number of dominoes in its set<br />(values in bold are common)
! Max. pips
| 0 || 1 || 2 || 3 || 4 || 5
! 6
| 7 || 8
! 9
| 10 || 11
! 12
| 13 || 14
! 15
| 16 || 17
! 18
| 19 || 20 || 21
|-
! ''n''
| 1 || 2 || 3 || 4 || 5 || 6
! 7
| 8 || 9
! 10
| 11 || 12
! 13
| 14 || 15
! 16
| 17 || 18
! 19
| 20 || 21 || 22
|-
! ''T<sub>n</sub>''
| 1 || 3 || 6 || 10 || 15 || 21
! 28
| 36 || 45
! 55
| 66 || 78
! 91
| 105 || 120
! 136
| 153 || 161
! 190
| 210 || 231 || 253
|}

==Triangular roots and tests for triangular numbers{{Anchor|Triangular root}}==
By analogy with the [[square root]] of {{mvar|x}}, one can define the (positive) triangular root of {{mvar|x}} as the number {{mvar|n}} such that {{math|1=''T<sub>n</sub>'' = ''x''}}:<ref name="EulerRoots">{{Citation |last1=Euler |first1=Leonhard |author-link=Leonhard Euler |last2=Lagrange |first2=Joseph Louis |author2-link=Joseph Louis Lagrange |year=1810 |title=Elements of Algebra |edition=2nd |volume=1 |publisher=J. Johnson and Co. |pages=332–335|title-link=Elements of Algebra }}</ref>
<math display=block>n = \frac{\sqrt{8x+1}-1}{2}</math>

which follows immediately from the [[quadratic formula]]. So an integer {{mvar|x}} is triangular [[if and only if]] {{math|8''x'' + 1}} is a square. Equivalently, if the positive triangular root {{mvar|n}} of {{mvar|x}} is an integer, then {{mvar|x}} is the {{mvar|n}}th triangular number.<ref name="EulerRoots" />

==Alternative name==
By analogy with the [[factorial]] function, a product whose factors are the integers from 1 to&nbsp;{{mvar|n}}, [[Donald Knuth]] proposed the name '''Termial function''',<ref>{{TAOCP|volume=1|edition=3|pages=48}}</ref> with the notation {{mvar|n}}? for the sum whose terms are the integers from 1 to&nbsp;{{mvar|n}} (the {{mvar|n}}th triangular number).  Although some other sources use this name and notation,<ref>{{citation | last = Stone | first = John David | doi = 10.1007/978-3-662-57970-1 | page = 282 | publisher = Springer | title = Algorithms for Functional Programming | year = 2018| isbn = 978-3-662-57968-8 | s2cid = 53079729 }}</ref> they are not in wide use.

==See also==
* [[1 + 2 + 3 + 4 + ⋯]]
* [[Doubly triangular number]], a triangular number whose position in the sequence of triangular numbers is also a triangular number
* [[Tetractys]], an arrangement of ten points in a triangle, important in Pythagoreanism
* [[Factoriangular number]]

==References==
{{Reflist|30em}}

==External links==
{{Commons category|Triangular numbers|lcfirst=yes}}
* {{SpringerEOM|title=Arithmetic series|id=p/a013370}}
* [http://www.cut-the-knot.org/do_you_know/numbers.shtml#square Triangular numbers] at [[cut-the-knot]]
* [http://www.cut-the-knot.org/do_you_know/triSquare.shtml There exist triangular numbers that are also square] at [[cut-the-knot]]
* {{MathWorld|urlname=TriangularNumber|title=Triangular Number}}
* [http://rhubbarb.wordpress.com/2009/04/23/hypertetrahedral-polytopic-roots/ Hypertetrahedral Polytopic Roots] by Rob Hubbard, including the generalisation to ''triangular cube roots'', some higher dimensions, and some approximate formulas

{{Figurate numbers}}
{{Series (mathematics)}}
{{Classes of natural numbers}}

[[Category:Figurate numbers]]
[[Category:Factorial and binomial topics]]
[[Category:Integer sequences]]
[[Category:Proof without words]]
[[Category:Squares in number theory]]
[[Category:Triangles]]
[[Category:Simplex numbers]]