View source for Triangular number

[[File:First six triangular numbers.svg|thumb|The first six triangular numbers]]

A '''triangular number''' or '''triangle number''' counts objects arranged in an [[equilateral triangle]]. The {{math|''n''}}th triangular number is the number of dots in the triangular arrangement with {{math|''n''}} dots on a side, and is equal to the sum of the {{math|''n''}} [[natural number]]s from 1 to {{math|''n''}}. The sequence of triangular numbers {{OEIS|id=A000217}}, starting at the [[empty sum|0th triangular number]], is
: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...

==Formula==
{{Pascal_triangle_simplex_numbers.svg|2=triangular numbers}}
The triangle 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. --><math> T_n= \sum_{k=1}^n k = 1+2+3+ \dotsb +n = \frac{n(n+1)}{2} = {n+1 \choose 2} ,</math>

where <math>\textstyle {n+1 \choose 2}</math> is 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 "{{math|''n''}} plus one choose two".

The first equation 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-square" 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:

{|cellpadding="7"
|<math> 2T_4 = 4(4+1) = 20 </math> (green plus yellow) implies that <math>T_4 = \frac{4(4+1)}{2} = 10 </math> (green).
|[[File:Illustration of Triangular Number T 4 Leading to a Rectangle.png]]
|&nbsp;&nbsp;&nbsp;
|}

The first equation can also be established using [[mathematical induction]].<ref>Andrews, George E.  ''Number Theory'', Dover, New York, 1971. pp. 3-4.</ref>  Since the sum of the first (one) natural number(s) is clearly equal to one, a basis case is established.  Assuming the inductive hypothesis for some <math>n</math> and adding <math>n+1</math> to both sides immediately gives

:<math> \sum_{k=1}^{n} k + (n+1) = \frac{n(n+1)}{2} + (n+1) \rightarrow \sum_{k=1}^{n+1} k = \frac{(n+1)((n+1)+1)}{2} </math>

In other words, since the [[proposition]] <math>P(n)</math> (that is, the first equation, or inductive hypothesis itself) is true when <math>n=1</math>, and since <math>P(n)</math> being true implies that <math>P(n+1)</math> is also true, then the first equation is true for all natural numbers.  The above argument can be easily modified to start with, and include, zero.

[[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|work=American Scientist|publisher=Computing Science|accessdate=2014-04-16}}</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]] 5th century BC.<ref>{{cite web|last1=Eves|first1=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|accessdate=28 March 2015}}</ref> The two formulae were described by the Irish monk [[Dicuil]] in about 816 in his [[Computus]].<ref>Esposito,M. An unpublished astronomical treatise by the Irish monk Dicuil. Proceedings of the Royal Irish Academy, XXXVI C. Dublin, 1907, 378-446.</ref>

The triangular number {{math|''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 {{math|''n''}} people is {{math|''T''<sub>''n''−1</sub>}}.<ref>https://web.archive.org/web/20160310182700/http://www.mathcircles.org/node/835</ref> The function {{math|''T''}} is the additive analog of the [[factorial]] function, which is the ''products'' of integers from 1 to&nbsp;{{math|''n''}}.

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>L_n = 3 T_{n-1}= 3{n \choose 2};~~~L_n = L_{n-1} + 3(n-1), ~L_1 = 0.</math>

In the limit, the ratio between the two numbers, dots and line segments is
:<math>\lim_{n\to\infty} \frac{T_n}{L_n} = \frac{1}{3}.</math>

==Relations to other figurate numbers==
Triangular numbers have a wide variety of relations to other [[figurate number]]s.

Most simply, the sum of two consecutive triangular numbers is a [[square number]], 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). Algebraically,
:<math>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}{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 fact can be demonstrated graphically by positioning the triangles in opposite directions to create a square:
{| cellpadding="6" style="display: inline-block; height: 150px; margin-top: -10px;"
|style="height: 150px;"|6 + 10 = 16
|style="height: 150px;"|[[File:Square number 16 as sum of two triangular numbers.svg]]
|style="height: 150px;"|&nbsp;&nbsp;&nbsp;
|}
{| cellpadding="6" style="display: inline-block; height: 150px; margin-top: -10px;"
|style="height: 150px;"|10 + 15 = 25
|style="height: 150px;"|[[File:Square number 25 as sum of two triangular numbers.svg]]
|}

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>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>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 {{math|''n''}}th triangular number is equal to the sum of the first {{math|''n''}} [[Cube (algebra)|cube numbers]].]]

Also, the [[squared triangular number|square of the {{math|''n''}}th triangular number]] is the same as the sum of the cubes of the integers 1 to {{math|''n''}}. This can also be expressed as
:<math> \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 {{math|''n''}}th [[tetrahedral number]]:
:<math> \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 {{math|''n''}}th [[polygonal number|{{math|''m''}}-gonal number]] and the {{math|''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 {{math|''n''}}th centered {{math|''k''}}-gonal number is obtained by the formula
:<math>Ck_n = kT_{n-1}+1</math>

where {{math|''T''}} is a triangular number.

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

==Other properties==
Triangular numbers correspond to the first-degree case of [[Faulhaber's formula]].

Alternating triangular numbers (1, 6, 15, 28, ...) are also [[hexagonal number]]s.

Every even [[perfect number]] is triangular (as well as hexagonal), given by the formula
:<math>M_p 2^{p-1} = \frac{M_p (M_p + 1)}2 = T_{M_p}</math>
where {{math|''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.

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:
: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
:…
:There is a more specific property to the triangular numbers that aren't divisible by 3, that is, they either have a remainder 1 or 10 when divided by 27.  Those that are equal to 10 mod 27, are also equal to 10 mod 81.
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 {{math|''x''}} is a triangular number, then {{math|''ax'' + ''b''}} is also a triangular number, given {{math|''a''}} is an odd square and {{math|''b'' {{=}} {{sfrac|''a'' − 1|8}}}}

{{math|''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 {{math|''b''}} given {{math|''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 {{math|''x''}} is equal to {{math|''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> \!\ \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> \!\ \sum_{n=1}^{\infty}{1 \over {n(n+1)}} = 1 .</math>

Two other formulas regarding triangular numbers are
:<math>T_{a+b} = T_a + T_b + ab</math>

and
:<math>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.

In 1796, German mathematician and scientist [[Carl Friedrich Gauss]] discovered that every positive integer is representable as a sum of three triangular numbers (possibly including {{math|''T''<sub>0</sub>}} = 0), writing in his diary his famous words, "[[Eureka (word)|ΕΥΡΗΚΑ!]] {{nowrap|1=num = Δ + Δ + Δ}}". This theorem does not imply that the triangular numbers are different (as in the case of 20 = 10 + 10 + 0), nor that a solution with exactly three nonzero triangular numbers must exist. This is a special case of the [[Fermat polygonal number theorem]].

The largest triangular number of the form {{math|2<sup>''k''</sup> − 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|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>

==Applications==
A [[fully connected network]] of {{math|''n''}} computing devices requires the presence of {{math|''T''<sub>''n'' − 1</sub>}} cables or other connections; this is equivalent to the handshake problem mentioned above.

In a tournament format that uses a round-robin [[group stage]], the number of matches that need to be played between {{math|''n''}} teams is equal to the triangular number {{math|''T''<sub>''n'' − 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.

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 {{math|''T''<sub>''n''</sub>}}, where {{math|''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 {{math|''b''}} is the item's beginning value (in units of currency), {{math|''s''}} is its final salvage value, {{math|''n''}} is the total number of years the item is usable, and {{math|''y''}} the current year in the depreciation schedule. Under this method, an item with a usable life of {{math|''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.

==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 {{math|''n''}} such that {{math|1=''T''<sub>''n''</sub> = ''x''}}:<ref name="EulerRoots">{{Citation |last=Euler |first=Leonhard |authorlink=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>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" />

==See also==
* [[Centered triangular number]]
* [[Polygonal number]]
* [[Simplex]]
* [[Square triangular number]]
* [[Tetractys]]

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

==External links==
{{Commons category|triangular numbers}}
* {{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 formulae

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

[[Category:Integer sequences]]
[[Category:Proof without words]]
[[Category:Squares in number theory]]
[[Category:Triangles]]
[[Category:Simplex numbers]]