Triangular Numbers

Triangular number sequences are arranged in a series or sequence of equilateral triangles to represent numbers. Each number is in the following sequence: 1, 3, 6, 10, 15, 21, 28, 36, 45, etc. The dots represent the numbers the triangular pattern contains. A sequence of triangular numbers is formed when the previous number is added to the order of the succeeding number. This article explains more about that.

What is the Triangular Number?

Triangular numbers Tn are figurate numbers that can be arranged in several rows of elements in such a way that every subsequent row contains one element more than the preceding row.

Triangular Number List

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, and so on.

Sum of Triangular Numbers

If you pay attention to the pattern of triangular numbers, the next number is added with an extra row. Here's how it works.

• The first number is 1.

• To the first number, two dots are added.

• For number 3, three dots are added to a row of dots.

• In number 4, a row with four dots is added to the third number, and so on.

Therefore, the sequence is as follows:

1, 1 + 2, 1 + 2 + 3, 1 + 2 + 3 + 4, and so on.

How to Find Triangular Numbers

An equilateral triangle contains objects called triangular numbers (also referred to as triangle numbers). Triangular number N is equal to the sum of all "n" natural numbers from "1" to "n" and is the number of black dots in a triangular pattern with n black dots on each side. An arrangement of triangular numbers, starting with the 0th triangular number, is as follows: Numbers 0 through 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,

The following formulas can be used to calculate the triangle numbers: \[T_{n} = \sum_{k=1}^{n}k = 1 + 2 + 3 + . . . + n = \frac{n(n+1)}{2} = \binom{n+1}{2}\]

\[\frac{(n+1)}{2}\] represents the Binomial coefficient. It is written as "n plus one to one choose two" and it represents the number of distinct pairs that can be chosen from (n + 1) items. 

Imagine a "half-square" pattern of objects corresponding to each triangular number Tn. By reversing the pattern and creating a rectangular image, the number of objects doubles, giving a rectangle with dimensions n x (n+1), which is also the number of items in the rectangle. The triangular number is always exactly half of the number of objects in the image or Tn = \[\frac{n(n+1)}{2}\].

Relationship with Other Figurative Numbers

The relationship between triangle numbers and other figures is quite broad. When two triangle numbers are added together, their sum equals a square value. In algebraic terms,

\[T_{n} + T_{n-1} = (\frac{n^{2}}{2} + \frac{n}{2}) + (\frac{(n-1)^{2}}{2} + \frac{n-1}{2}) = (\frac{n^{2}}{2} + \frac{n}{2}) + (\frac{n^{2}}{2} - \frac{n}{2}) = n^{2} = (T_{n} + T_{n-1})^{2}\]

In order to represent this fact graphically, you can create a square by placing triangles in opposite directions: As an example, there are an infinite number of triangular numbers which are also of square value, such as 1, 36, 1225, etc. 

Roots of Triangular Numbers and Tests for Triangular Numbers

By analogy with the square root of x, the (positive) triangular root of x can be defined as the number n such that Tn = x: n = √8x+1−1/2

The quadratic formula yields this result immediately. In this case, a square number 8x + 1 is triangular if and only if x is triangular. A triangular number x is the nth triangular number if n is a positive integer.

Some of them can be calculated by using a simple recursive formula 

\[S_{n+1} = 4S_{n}(8S + 1)\] with \[S_{1}\] = 1

Note: All square triangular numbers are generated from the recursion.

\[S_{n} = 34S_{n-1} - S_{n-2}\] + 2 with \[S_{0}\] = 0 and \[S_{1} = 1\]

Also, the square value of the nth triangle number is the same as the sum of the cubes of the integers from "1" to "n". This can also be represented as: 

\[\int_{k=1}^{n}Tk = \sum_{k=1}^{n} \frac{k(k+1)}{2} = \frac{n(n+1)(n+2)}{6}\]

The difference between the nth m-gon number and the nth (m + 1)-gon number is the (n−1) th triangle number. For example, the difference between the sixth heptagonal number and the sixth hexagonal number is the fifth triangular number. Every other triangle number is a hexagonal number. Knowing the triangle numbers, one can calculate any centered polygonal number; the nth centered k-gon number is obtained by the formula

\[Ck_{n} = kT_{n-1} + 1\]

Where T is a triangle number.

The difference between the two triangular numbers is a trapezoidal number. 

Other Properties

• Triangular numbers correlate to the first-degree case of Faulhaber’s formula.

• It’s interesting to notice that every alternate triangular number (1, 6, 15, 28, 45….) is also a hexagonal number.

• Every even perfect number is triangular (as well as hexagonal), according to the given formula.

\[M_{p} 2^{p-1} = \frac{M_{p}(M_{p} + 1)}{2} = TM_{p}\]

Here, ‘Mp’ is a Mersenne prime. No odd perfect numbers are known, therefore, all known perfect numbers are triangular.

For example, the third triangle 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 the case of 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 3 or it leaves 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

• Triangular numbers have one more specific property that they aren't divisible by 3, that is, when divided by 27, they either have to leave a remainder of 1 or 10. Those equal to 10 mod 27, are equal to 10 mod 81 too.

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

• The converse of the statement above is, not necessarily’ always holds true. For example, the digital root of 12(not a triangular number), is 3 and divisible by three. 

Suppose, x is a triangular number, then ax + b is also a triangular number, given a, which is an odd-square, and b = (a – 1)/8.

Where, b will always be a triangular number because Tn + 1 = (2n + 1)2/8, which gives all the odd squares are revealed by multiplication of 8 to the triangle number and adding 1, and the process for b given a is an odd square is the inverse of this operation.

The first few pairs of this form (excluding 1x + 0) are: 9x + 1, 25x + 3, 49x + 6, 81x + 10, 121x + 15, 169x + 21 … and so on. Provided, x is equal to Tn, these formulas yield T3n + 1, T5n + 2, T7n + 3, T9n + 4, so on and so forth.

The summation of the reciprocals of all the non-zero triangle numbers is:  

\[\sum_{n=1}^{\infty} \frac{1}{\frac{n^{2} + n}{2}} = 2\sum_{n=1}^{\infty} \frac{1}{n^{2} + n} = 2\]

This can be shown through a sum of telescopic series:

\[\sum_{n=1}^{\infty} \frac{1}{n(n+1)} = 1\]

Two other formulas relating to triangle numbers are:

\[T_{a+b} = T_{a} + T_{b} + ab\]

\[T_{ab} = T_{a}T_{b} + T_{a-1}T_{b-1}\]

Both of these can be established either by looking at dot patterns (shown above) or with some simple algebra. 

In 1796, a German mathematician and scientist Carl Fredrich Gauss discovered that every positive integer can be represented as a sum of three triangular numbers (possibly including T0= 0), num = Δ + Δ + Δ".  

Note: This theorem does not say that the triangular numbers are different (as in the case of 20 = 10 + 10 + 0), nor does it say that a solution with exactly three non-zero triangle numbers has to exist: a special case of the Fermat Polygonal number Theorem.

The largest triangular number of the form 2k − 1 is 4095(reference Ramanujan Nagell equation).

Wacław Franciszek Sierpiński raised the question as to the existence of four distinct triangular numbers in geometric progression. It was concluded by Polish mathematician Kazimierz Szymiczek that it was impossible and was later proven by Fang and Chen also in 2007.

Applications  

A fully connected network of n computing devices needs the presence of Tn − 1 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 are required to be played between n teams is equal to the triangular number Tn − 1. For example, a group stage with 4 teams needs 6 matches, and a group stage with 8 teams requires 28 matches. This is also similar to the handshake problem and fully connected network problems. 

One way of calculating the depreciation of an asset is the sum of years' digit method, which involves finding Tn, where n is the length in years of the asset's useful life. Every year, the item loses (b − s) × n − y/ Tn, where b is the item's starting value (in units of currency), ‘s’ is its final salvage values’’ is the total number of years for which the item is usable, and ‘y’ the current year in the depreciation schedule. In this method, an item with a usable life of n = 4 years would lose 4/10 of its "losable" value in the first year, 3/10 in the second, 2/10in the third, and 1/10 in the fourth, accumulating a total depreciation of 10/10 (the whole) of the losable value.