Número triangular

Los números triangulares o números triangulares son la secuencia de enteros positivos que se pueden representar como una red de puntos dispuestos en un triángulo equilátero . La red triangular que representa el${\displaystyle n}$El número triangular contiene${\displaystyle n}$filas: la primera fila contiene un punto, la segunda fila contiene dos, y este patrón continúa hasta la${\displaystyle n}$la fila que contiene${\displaystyle n}$Por lo tanto, los números triangulares también pueden representarse mediante la fórmula $${\displaystyle T_{n}=1+2+3+\cdots +(n-1)+n=\sum _{k=1}^{n}k.}$$

Los números triangulares son el tipo más simple de número figurado ; los números figurados generalizan su concepto a otros polígonos bidimensionales, como los números pentagonales , así como a poliedros de dimensiones superiores, como los números tetraédricos . Tomando${\displaystyle T_{0}=0}$(ver suma vacía ), los primeros términos son

(secuencia A000217 en el OEIS )

Los números triangulares se obtienen mediante las siguientes fórmulas explícitas:

${\displaystyle \displaystyle {\begin{aligned}T_{n}&=\sum _{k=1}^{n}k=1+2+\dotsb +n\\&={\frac {n^{2}+n{\vphantom {(n+1)}}}{2}}={\frac {n(n+1)}{2}}\\&={n+1 \choose 2}\end{aligned}}}$

dónde${\displaystyle \textstyle {n+1 \choose 2}}$es la notación para un coeficiente binomial . Representa el número de pares distintos que se pueden seleccionar de n + 1 objetos, y se lee en voz alta como " n más uno elige dos".

El hecho de que el${\displaystyle n}$El número triangular es igual a${\displaystyle n(n+1)/2}$puede ilustrarse mediante una prueba visual . ^{[ 1 ]} Para cada número triangular${\displaystyle T_{n}}$, imagina una disposición de objetos en forma de "medio rectángulo" que corresponda al número triangular, como en la figura siguiente. Copiar esta disposición y rotarla para crear una figura rectangular duplica el número de objetos, produciendo un rectángulo con dimensiones${\displaystyle n\times (n+1)}$, que también es el número de objetos en el rectángulo. Claramente, el número triangular en sí es siempre exactamente la mitad del número de objetos en dicha figura, o:${\displaystyle T_{n}={\frac {n(n+1)}{2}}}$. El ejemplo${\displaystyle T_{4}}$sigue:

${\displaystyle 2T_{4}=4(4+1)=20}$(verde más amarillo) implica que${\displaystyle T_{4}={\frac {4(4+1)}{2}}=10}$(verde).   

Esta fórmula puede demostrarse formalmente mediante inducción matemática . ^{[ 2 ]} Es claramente cierto para${\displaystyle 1}$:

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

Ahora supongamos que, para algún número natural${\displaystyle m}$,${\displaystyle T_{m}=\sum _{k=1}^{m}k={\frac {m(m+1)}{2}}}$. Luego podemos verificarlo para${\displaystyle m+1}$: $${\displaystyle {\begin{aligned}\sum _{k=1}^{m+1}k&=\sum _{k=1}^{m}k+(m+1)\\&={\frac {m(m+1)}{2}}+m+1\\&={\frac {m^{2}+m}{2}}+{\frac {2m+2}{2}}\\&={\frac {m^{2}+3m+2}{2}}\\&={\frac {(m+1)(m+2)}{2}},\end{aligned}}}$$

entonces si la fórmula es verdadera para${\displaystyle m}$, es cierto para${\displaystyle m+1}$. Dado que es claramente cierto para${\displaystyle 1}$, it is therefore true for ${\displaystyle 2}$, ${\displaystyle 3}$, and ultimately all natural numbers ${\displaystyle n}$ by induction.

An apocryphal story claims that the German mathematician Gauss found this relationship in his early youth, by multiplying ⁠n/2⁠ pairs of numbers in the sum by the values of each pair n + 1.^[3] In any case, 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.^[4] The two formulas were described by the Irish monk Dicuil in about 816 in his Computus.^[5] An English translation of Dicuil's account is available.^[6]

Occasionally it is necessary to compute large triangular numbers where the standard formula t = n*(n+1)/2 would suffer integer overflow before the final division by 2. For example, T₂₀ = 210 < 256, so will fit into an 8-bit byte, but not the intermediate product 420. This can be solved by dividing either n or n+1 by 2 before the multiplication, whichever is even. This does not require a conditional branch if implemented as t = (n|1) * ((n+1)/2). If n is odd, the binary OR operation n|1 has no effect, so this is equivalent to t = n * ((n+1)/2) and thus correct. If n is even, setting the low bit with n|1 is the same as adding 1, while the 1 added before the division is truncated away, so this is equivalent to t = (n+1) * (n/2) and also correct.

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:^[7][8]

${\displaystyle T_{n-1}+T_{n}}$

${\displaystyle ={\frac {1}{2}}\,n(n-1)+{\frac {1}{2}}\,n(n+1)}$

${\displaystyle ={\frac {1}{2}}\,n{\Bigl (}(n-1)+(n+1){\Bigr )}}$

${\displaystyle =n^{2}}$

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

This property, colloquially known as the theorem of Theon of Smyrna,^[9] is visually demonstrated in the following sum, which represents ${\displaystyle T_{4}+T_{5}=5^{2}}$ as digit sums:

${\displaystyle {\begin{array}{ccccccc}&4&3&2&1&\\+&1&2&3&4&5\\\hline &5&5&5&5&5\end{array}}}$

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

6 + 10 = 16        10 + 15 = 25    

The double of a triangular number, as in the visual proof from the above section § 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: $${\displaystyle S_{n+1}=4S_{n}\left(8S_{n}+1\right)}$$ with ${\displaystyle S_{1}=1.}$

All square triangular numbers are found from the recursion $${\displaystyle S_{n}=34S_{n-1}-S_{n-2}+2}$$ with ${\displaystyle S_{0}=0}$ and ${\displaystyle S_{1}=1.}$

The square of the nth triangular number is also the same as the sum of the cubes of the integers 1 to n. This can also be expressed as $${\displaystyle \sum _{k=1}^{n}k^{3}=\left(\sum _{k=1}^{n}k\right)^{2}.}$$

The sum of the first n triangular numbers is the nth tetrahedral number: $${\displaystyle \sum _{k=1}^{n}T_{k}=\sum _{k=1}^{n}{\frac {k(k+1)}{2}}={\frac {n(n+1)(n+2)}{6}}.}$$

More generally, the difference between the 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 $${\displaystyle Ck_{n}=kT_{n-1}+1}$$

where T is a triangular number.

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

The pattern found for triangular numbers ${\displaystyle \sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{2}+1}{2}}}$ and for tetrahedral numbers ${\displaystyle \sum _{n_{2}=1}^{n_{3}}\sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{3}+2}{3}},}$ which uses binomial coefficients, can be generalized. This leads to the formula:^[11]$${\displaystyle \sum _{n_{k-1}=1}^{n_{k}}\sum _{n_{k-2}=1}^{n_{k-1}}\dots \sum _{n_{2}=1}^{n_{3}}\sum _{n_{1}=1}^{n_{2}}n_{1}={\binom {n_{k}+k-1}{k}}}$$

Triangular numbers correspond to the first-degree case of Faulhaber's formula.

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

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

The sum of the reciprocals of all the nonzero triangular numbers is $${\displaystyle \sum _{n=1}^{\infty }{1 \over {{n^{2}+n} \over 2}}=2\sum _{n=1}^{\infty }{1 \over {n^{2}+n}}=2.}$$

This can be shown by using the basic sum of a telescoping series: $${\displaystyle \sum _{n=1}^{\infty }{1 \over {n(n+1)}}=1.}$$

In addition, the nth partial sum of this series can be written as: $${\displaystyle 2n \over {n+1}.}$$

Two other formulas regarding triangular numbers are $${\displaystyle T_{a+b}=T_{a}+T_{b}+ab}$$ and $${\displaystyle T_{ab}=T_{a}T_{b}+T_{a-1}T_{b-1},}$$ both of which can be established either by looking at dot patterns (see above) or with some simple algebra.

In 1796, Gauss discovered that every positive integer is representable as a sum of three triangular numbers, writing in his diary his famous words, "ΕΥΡΗΚΑ!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 2^k −1 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.^[12][13]

Formulas involving expressing an integer as the sum of triangular numbers are connected to theta functions, in particular the Ramanujan theta function.^[14][15]

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

In the limit, the ratio between the two numbers, dots and line segments is $${\displaystyle \lim _{n\to \infty }{\frac {T_{n}}{L_{n}}}={\frac {1}{3}}.}$$

The triangular number T_n solves the handshake problem of counting the number of handshakes if each person in a room with n + 1 people shakes hands once with each person. In other words, the solution to the handshake problem of n people is T_n−1.^[16]

Equivalently, a fully connected network of n computing devices requires the presence of T_{n −1} cables or other connections.

A triangular number ${\displaystyle T_{n}}$ is equivalent to the number of principal rotations in dimension ${\displaystyle n+1}$. For example, in five dimensions the number of principal rotations is 10 which is ${\displaystyle T_{4}}$.^[17]

In a tournament format that uses a round-robin group stage, the number of matches that need to be played between n teams is equal to the triangular number T_{n −1}. 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.

Una forma de calcular la depreciación de un activo es el método de la suma de los dígitos de los años , que consiste en hallar T _n , donde n es la duración en años de la vida útil del activo. Cada año, el artículo pierde ( b − s ) × ⁠ n − y / T _n ⁠ , donde b es el valor inicial del artículo (en unidades monetarias), s es su valor residual final, n es el número total de años que el artículo es utilizable, e y es el año actual en el plan de depreciación. Según este método, un artículo con una vida útil de n = 4 años perdería ⁠ 4 / 10 ⁠ de su valor "perdible" en el primer año, ⁠ 3 / 10 ⁠ en el segundo, ⁠ 2 / 10 ⁠ en el tercero y ⁠ 1 / 10 ⁠ en el cuarto, acumulando una depreciación total de ⁠ 10 / 10 ⁠ (la totalidad) del valor perdible.

Los diseñadores de juegos de mesa Geoffrey Engelstein e Isaac Shalev describen los números triangulares como algo que ha alcanzado "casi el estatus de un mantra o koan entre los diseñadores de juegos ", describiéndolos como "profundamente intuitivos" y "presentes en una enorme cantidad de juegos, [demostrando] una increíble versatilidad para proporcionar recompensas crecientes para conjuntos más grandes sin incentivar excesivamente la especialización excluyendo todas las demás estrategias". ^{[ 18 ]}

Por analogía con la raíz cuadrada de x , se puede definir la raíz triangular (positiva) de x como el número n tal que T _n = x : ^{[ 19 ]}$${\displaystyle n={\frac {{\sqrt {8x+1}}-1}{2}}}$$

which follows immediately from the quadratic formula. So an integer x is triangular if and only if8x + 1 is a square. Equivalently, if the positive triangular root n of x is an integer, then x is the nth triangular number.^[19]

By analogy with the factorial function, a product whose factors are the integers from 1 to n, Donald Knuth proposed the name Termial function,^[20] with the notation n? for the sum whose terms are the integers from 1 to n (the nth triangular number). Although some other sources use this name and notation,^[21] they are not in wide use. Thus, the termial function can be defined in notation as:^[20]

${\displaystyle n?=\sum _{k=1}^{n}{k}{\text{ for }}n\in \mathbb {N} }$

• 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
• Šindel sequence

• "Arithmetic series", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Triangular numbers at cut-the-knot
• There exist triangular numbers that are also square at cut-the-knot
• Weisstein, Eric W."Triangular Number". MathWorld.
• Hypertetrahedral Polytopic Roots by Rob Hubbard, including the generalisation to triangular cube roots, some higher dimensions, and some approximate formulas