Triangular number

The first six triangular numbers

A triangular number or triangle number counts objects arranged in an equilateral triangle. The Template:Mathth triangular number is the number of dots in the triangular arrangement with Template:Math dots on a side, and is equal to the sum of the Template:Math natural numbers from 1 to Template:Math. The sequence of triangular numbers Template:OEIS, starting at the 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

Template:Pascal triangle simplex numbers.svg The triangle numbers are given by the following explicit formulas:

$T_{n}=\sum _{k=1}^{n}k=1+2+3+\dotsb +n={\frac {n(n+1)}{2}}={n+1 \choose 2},$

where $\textstyle {n+1 \choose 2}$ is a binomial coefficient. It represents the number of distinct pairs that can be selected from Template:Math objects, and it is read aloud as "Template:Math plus one choose two".

The first equation can be illustrated using a visual proof. For every triangular number $T_{n}$, 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 $n\times (n+1)$, 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: $T_{n}={\frac {n(n+1)}{2}}$. The example $T_{4}$ follows:

$2T_{4}=4(4+1)=20$ (green plus yellow) implies that $T_{4}={\frac {4(4+1)}{2}}=10$ (green).

The first equation can also be established using mathematical induction. 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 $n$ and adding $n+1$ to both sides immediately gives

$\sum _{k=1}^{n}k+(n+1)={\frac {n(n+1)}{2}}+(n+1)\rightarrow \sum _{k=1}^{n+1}k={\frac {(n+1)((n+1)+1)}{2}}$

In other words, since the proposition $P(n)$ (that is, the first equation, or inductive hypothesis itself) is true when $n=1$, and since $P(n)$ being true implies that $P(n+1)$ 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 Template:Math pairs of numbers in the sum by the values of each pair Template:Math. 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. The two formulae were described by the Irish monk Dicuil in about 816 in his Computus.

The triangular number Template:Math solves the handshake problem of counting the number of handshakes if each person in a room with Template:Math people shakes hands once with each person. In other words, the solution to the handshake problem of Template:Math people is Template:Math. The function Template:Math is the additive analog of the factorial function, which is the products of integers from 1 to Template:Math.

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:

$L_{n}=3T_{n-1}=3{n \choose 2};~~~L_{n}=L_{n-1}+3(n-1),~L_{1}=0.$

In the limit, the ratio between the two numbers, dots and line segments is

$\lim _{n\to \infty }{\frac {T_{n}}{L_{n}}}={\frac {1}{3}}.$

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, 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,

$T_{n}+T_{n-1}=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {\left(n-1\right)^{2}}{2}}+{\frac {n-1}{2}}\right)=\left({\frac {n^{2}}{2}}+{\frac {n}{2}}\right)+\left({\frac {n^{2}}{2}}-{\frac {n}{2}}\right)=n^{2}=(T_{n}-T_{n-1})^{2}.$

This fact can be demonstrated graphically by positioning the triangles in opposite directions to create a square:

6 + 10 = 16

10 + 15 = 25

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:

$S_{n+1}=4S_{n}\left(8S_{n}+1\right)$ with $S_{1}=1.$

All square triangular numbers are found from the recursion

$S_{n}=34S_{n-1}-S_{n-2}+2$ with $S_{0}=0$ and $S_{1}=1.$

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 Template:Mathth triangular number is equal to the sum of the first Template:Math cube numbers.

Also, the [[squared triangular number|square of the Template:Mathth triangular number]] is the same as the sum of the cubes of the integers 1 to Template:Math. This can also be expressed as

$\sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.$

The sum of the first Template:Mvar triangular numbers is the Template:Mathth tetrahedral number:

$\sum _{k=1}^{n}T_{k}=\sum _{k=1}^{n}{\frac {k(k+1)}{2}}={\frac {n(n+1)(n+2)}{6}}.$

More generally, the difference between the Template:Mathth [[polygonal number|Template:Math-gonal number]] and the Template:Mathth Template:Math-gonal number is the Template:Mathth 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 Template:Mathth centered Template:Math-gonal number is obtained by the formula

$Ck_{n}=kT_{n-1}+1$

where Template:Math 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 numbers.

Every even perfect number is triangular (as well as hexagonal), given by the formula

$M_{p}2^{p-1}={\frac {M_{p}(M_{p}+1)}{2}}=T_{M_{p}}$

where Template:Math 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 Template:Math is a triangular number, then Template:Math is also a triangular number, given Template:Math is an odd square and Template:Math

Template:Math will always be a triangular number, because Template:Math, which yields all the odd squares are revealed by multiplying a triangular number by 8 and adding 1, and the process for Template:Math given Template:Math is an odd square is the inverse of this operation.

The first several pairs of this form (not counting Template:Math) are: Template:Math, Template:Math, Template:Math, Template:Math, Template:Math, Template:Math, … etc. Given Template:Math is equal to Template:Math, these formulas yield Template:Math, Template:Math, Template:Math, Template:Math, and so on.

The sum of the reciprocals of all the nonzero triangular numbers is

$\!\ \sum _{n=1}^{\infty }{1 \over {{n^{2}+n} \over 2}}=2\sum _{n=1}^{\infty }{1 \over {n^{2}+n}}=2.$

This can be shown by using the basic sum of a telescoping series:

$\!\ \sum _{n=1}^{\infty }{1 \over {n(n+1)}}=1.$

Two other formulas regarding triangular numbers are

$T_{a+b}=T_{a}+T_{b}+ab$

and

$T_{ab}=T_{a}T_{b}+T_{a-1}T_{b-1},$

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 Template:Math = 0), writing in his diary his famous words, "ΕΥΡΗΚΑ! Template:Nowrap". 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 Template:Math is 4095 (see Ramanujan–Nagell equation).

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.

Formulas involving expressing an integer as the sum of triangular numbers are connected to theta functions, in particular the Ramanujan theta function.

Applications

A fully connected network of Template:Math computing devices requires the presence of Template:Math 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 Template:Math teams is equal to the triangular number Template:Math. 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 sum-of-years' digits method, which involves finding Template:Math, where Template:Math is the length in years of the asset's useful life. Each year, the item loses Template:Math, where Template:Math is the item's beginning value (in units of currency), Template:Math is its final salvage value, Template:Math is the total number of years the item is usable, and Template:Math the current year in the depreciation schedule. Under this method, an item with a usable life of Template:Math = 4 years would lose Template:Sfrac of its "losable" value in the first year, Template:Sfrac in the second, Template:Sfrac in the third, and Template:Sfrac in the fourth, accumulating a total depreciation of Template:Sfrac (the whole) of the losable value.

Triangular roots and tests for triangular numbersTemplate:Anchor

By analogy with the square root of Template:Mvar, one can define the (positive) triangular root of Template:Mvar as the number Template:Math such that Template:Math:

$n={\frac {{\sqrt {8x+1}}-1}{2}}$

which follows immediately from the quadratic formula. So an integer Template:Mvar is triangular if and only if Template:Math is a square. Equivalently, if the positive triangular root Template:Mvar of Template:Mvar is an integer, then Template:Mvar is the Template:Mvarth triangular number.

See also

• Centered triangular number
• Polygonal number
• Simplex
• Square triangular number
• Tetractys

External links

• Template:SpringerEOM
• Triangular numbers at cut-the-knot
• There exist triangular numbers that are also square at cut-the-knot
• Template:MathWorld
• Hypertetrahedral Polytopic Roots by Rob Hubbard, including the generalisation to triangular cube roots, some higher dimensions, and some approximate formulae