Triangular number

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A triangular number is the sum of the n natural numbers from 1 to n.

T_n= 1+2+3+ \dotsb +(n-1)+n = \frac{n(n+1)}{2} = \frac{n^2+n}{2} \overset{\underset{\mathrm{def}}{}} {=} {n+1 \choose 2}

As shown in the rightmost term of this formula, every triangular number is a binomial coefficient: the nth triangular is the number of distinct pairs to be selected from n + 1 objects. In this form it solves the 'handshake problem' of counting the number of handshakes if each person in a room full of n+1 total people shakes hands once with each other person.

The sequence of triangular numbers (sequence A000217 in OEIS) for n = 1, 2, 3... is:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ....

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[edit] 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. 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.

Alternatively, the same fact can be demonstrated graphically:

16 Image:Square number 16 as sum of two triangular numbers.svg 25 Image: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. Some of them can be generated by a simple recursive formula:

S_{n+1} = 4S_n \left( 8S_n + 1\right) with S1 = 1.

All square triangular numbers are found from the recursion

Sn = 34Sn − 1 − Sn − 2 + 2 with S0 = 0 and S1 = 1.

Also, the square of the nth triangular number is the same as the sum of the cubes of the integers 1 to n.

The sum of the n first triangular numbers is the nth tetrahedral number,

\frac {(n)(n+1)(n+2)} {6}.

More generally, the difference between the nth m-gonal number and the nth (m + 1)-gonal number is the (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 nth centered k-gonal number is obtained by the formula

Ckn = kTn − 1 + 1

where T is a triangular number.

[edit] Other properties

Triangular numbers correspond to the 1st-order case of Faulhaber's formula.

Every even perfect number is triangular (given by formula M_n 2^{n-1} = M_n (M_n + 1)/2 = T_{M_n} where Mn is a Mersenne prime), and no odd perfect numbers are known, hence all known perfect numbers are triangular.

In base 10, the digital root of a 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 nine:

6 = 3×2,
10 = 9×1+1,
15 = 3×5,
21 = 3×7,
28 = 9×3+1,
...

The inverse 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.

The sum of the reciprocals of all the 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 interesting formulas regarding triangular numbers are:

Ta + b = Ta + Tb + ab

and

Tab = TaTb + Ta − 1Tb − 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 at most three triangular numbers, writing in his diary his famous words, "Heureka! num= Δ + Δ + Δ." Note that this theorem does not imply that the triangular numbers are different (as in the case of 20=10+10), nor that a solution with three nonzero triangular numbers must exist. This is a special case of Fermat's Polygonal Number Theorem.

The largest triangular number of the form 2k − 1 is 4095, see Ramanujan-Nagell equation.

Wacław Franciszek Sierpiński posed a question if there exist four distinct triangular numbers in geometric progression. It was conjectured by Polish mathematician Szymiczek to be false. This conjecture was proven by Fang and Chen in 2007[1][2].

[edit] Tests for triangular numbers

One can efficiently test whether a positive integer x is a triangular number by computing

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

If n is an integer, then x is the nth triangular number. If n is not an integer, then x is not triangular.

This test is based on identity 8Tn + 1 = S2n + 1.

[edit] References

  1. ^ Chen, Fang: Triangular numbers in geometric progression
  2. ^ Fang: Nonexistence of a geometric progression that contains four triangular numbers

[edit] External links

Retrieved from "http://en.wikipedia.org/wiki/Triangular_number"
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