Triangular numbers

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

$t_{n}:=\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}={\binom {n+1}{2}},\quad n\geq 0,\,$

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, ...}

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 $\scriptstyle 2t_{n}\,=\,(1+n)+(2+(n-1))+\ldots +((n-1)+2)+(n+1)\,$. Each of these parenthesized addends works out to $\scriptstyle n+1\,$ and there are $\scriptstyle n\,$ of these addends. Therefore, $\scriptstyle 2t_{n}\,=\,n(n+1)\,=\,n^{2}+n\,$ and dividing both sides by 2 we get $\scriptstyle t_{n}\,=\,{\frac {n^{2}+n}{2}}\,$ as specified by the theorem. □

Relations

$t_{n}-t_{n-1}=n,\quad n\geq 1.\,$

$t_{n}-2t_{n-1}+t_{n-2}=1,\quad n\geq 2.\,$

$t_{n}+t_{n-1}=n^{2},\quad n\geq 1.\,$

${\frac {t_{n}}{t_{n-1}}}={\frac {n+1}{n-1}},\quad n\geq 1.\,$

Relations involving binomial coefficients

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

$\sum _{k=0}^{m}(-1)^{k}{\binom {m}{m-k}}t_{n-k}=0,\ n\geq m>2,\,$

$\sum _{k=0}^{m}{\binom {m}{m-k}}t_{n-k}=\sum _{k=0}^{m-1}{\binom {m-1}{m-k-1}}(n-k)^{2},\ n\geq m>0,\,$

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

• 3-sided polygonal numbers.

• 2-dimensional simplicial polytopic numbers.

External links

• Ken Ono, Sinai Robins, and Patrick T. Wahl, ON THE REPRESENTATION OF INTEGERS AS SUMS OF TRIANGULAR NUMBERS.