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Triangular numbers

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The \scriptstyle n \,th triangular number is defined as the sum of the first \scriptstyle n \, positive integers

t_n := \sum_{i=1}^{n} i = \frac{n(n+1)}{2} = \binom{n+1}{2},\quad n \ge 0, \,

where \scriptstyle t_0 \,=\, 0 \, since it is the empty sum of positive integers (giving the additive identity, i.e. 0), and \scriptstyle \binom{n}{k} \, is a binomial coefficient. The \scriptstyle n \,th triangular number is thus one half of the \scriptstyle n \,th pronic number (or oblong number).

A000217 Triangular numbers: a(n) = C(n+1,2) = n(n+1)/2 = 0+1+2+...+n, n ≥ 0.

{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, ...}

Contents


This was proved by Euler, by the following trick:

    1,   2,   3, ..., n-2, n-1,   n
 +  n, n-1, n-2, ...,   3,   2,   1
 ----------------------------------
  n+1, n+1, n+1, ..., n+1, n+1, n+1

thus t_n = n*(n+1)/2.

That proof is sometimes also credited to Carl Friedrich Gauß:

Theorem. The sum of the first \scriptstyle n \, positive integers (the \scriptstyle n \,th triangular number \scriptstyle t_n \,) is equal to \scriptstyle \frac{n^2 + n}{2} \,.

Proof. (Gauß) We can write \scriptstyle t_n \,=\, 1 + 2 + 3 + \ldots + (n - 1) + n \,. Since addition is commutative, we can also write \scriptstyle t_n \,=\, n + (n - 1) + \ldots + 3 + 2 + 1 \,. If we add up these expressions term by term, left to right, we obtain \scriptstyle 2 t_n \,=\, (1 + n) + (2 + (n - 1)) + \ldots + ((n - 1) + 2) + (n + 1) \,. Each of these parenthesized addends works out to \scriptstyle n + 1 \, and there are \scriptstyle n \, of these addends. Therefore, \scriptstyle 2 t_n \,=\, n(n + 1) \,=\, n^2 + n \, and dividing both sides by 2 we get \scriptstyle t_n \,=\, \frac{n^2 + n}{2} \, as specified by the theorem. □ [1]

Relations

t_n - t_{n-1} = n,\quad n \ge 1. \,
t_n - 2 t_{n-1} + t_{n-2} = 1,\quad n \ge 2. \,
t_n + t_{n-1} = n^2,\quad n \ge 1. \,
\frac{t_n}{t_{n-1}} = \frac{n+1}{n-1},\quad n \ge 1. \,

Relations involving binomial coefficients

Charles Marion <charliemath@optonline.net> suggested the following two relations:

\sum_{k=0}^{m} (-1)^k \binom{m}{m-k} t_{n-k} = 0,\ n \ge m > 2, \,
\sum_{k=0}^{m} \binom{m}{m-k} t_{n-k} = \sum_{k=0}^{m-1} \binom{m-1}{m-k-1} (n-k)^2,\ n \ge m > 0, \,

where \scriptstyle m \,=\, 1\, gives \scriptstyle t_n + t_{n-1} \,=\, n^2. \,

Sum of reciprocals

The partial sums of the reciprocals of the triangular numbers gives (easily proved by induction)

\sum_{i=1}^n{\frac{1}{t_{i}}} = \frac{n^2}{t_{n}}.\,

The sum of the reciprocals of the triangular numbers converges to 2

\lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{t_{n}} = \lim_{n \to \infty} \frac{n^2}{t_{n}} = \lim_{n \to \infty} \frac{2n^2}{n(n+1)} = 2.\,

Representations of natural numbers as a sum of three triangular numbers

Every natural number may be represented, in at least one way, as a sum of three triangular numbers (with up to three nonzero triangular numbers).

Representations of n as a sum of three triangular numbers
n \, Representations Number of
representations
0 { {0, 0, 0} } 1
1 { {1, 0, 0} } 1
2 { {1, 1, 0} } 1
3 { {3, 0, 0}, {1, 1, 1} } 2
4 { {3, 1, 0} } 1
5 { {3, 1, 1} } 1
6 { {6, 0, 0}, {3, 3, 0} } 2
7 { {6, 1, 0}, {3, 3, 1} } 2
8 { {6, 1, 1} } 1
9 { {6, 3, 0}, {3, 3, 3} } 2
10 { {10, 0, 0}, {6, 3, 1} } 2
11 { {10, 1, 0} } 1
12 { {10, 1, 1}, {6, 6, 0}, {6, 3, 3} } 3
13 { {10, 3, 0}, {6, 6, 1} } 2
14 { {10, 3, 1} } 1
15 { {15, 0, 0}, {6, 6, 3} } 2
16 { {15, 1, 0}, {10, 6, 0}, {10, 3, 3} } 3

A002636 Number of representations of n as a sum of up to three nonzero triangular numbers.

{1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 3, 2, 1, 2, 3, 2, 2, 2, 1, 4, 3, 2, 2, 2, 2, 3, 3, 1, 4, 4, 2, 2, 3, 2, 3, 4, 2, 3, 3, 2, 4, 3, 2, 4, 4, 2, 4, 4, 1, 4, 5, 1, 2, 3, 4, 6, 4, 3, 2, 5, 2, 3, 3, 3, 6, 5, 2, 2, 5, 3, 5, 4, 2, 4, 5, 3, 4, ...}

Even perfect numbers

Every even perfect number is a triangular number, since they are a subset of

2^{n-1} \cdot (2^n - 1) = \frac{(2^n - 1) 2^n}{2} = t_{2^n - 1}, \,

where \scriptstyle 2^n - 1 \, is [necessarily, but not sufficiently] a Mersenne prime.

Every even perfect number is also an hexagonal number, since they are a subset of

2^{n-1} \cdot (2^n - 1) = 2^{n-1} \cdot (2 \cdot 2^{n-1} - 1) = h_{2^{n-1}}, \,

where \scriptstyle 2^n - 1 \, is [necessarily, but not sufficiently] a Mersenne prime.

See also

Notes

  1. Antonella Cupillari, The Nuts and Bolts of Proofs, Belmont, California: Wadsworth Publishing Company (1989): 13–14.

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