In mathematics, a square triangular number (or triangular square number) is a number which is both a triangular number and a square number. There are infinitely many square triangular numbers; the first few are:

0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in the OEIS)

## Explicit formulas

Write ${\displaystyle N_{k))$ for the ${\displaystyle k}$th square triangular number, and write ${\displaystyle s_{k))$ and ${\displaystyle t_{k))$ for the sides of the corresponding square and triangle, so that

${\displaystyle \displaystyle N_{k}=s_{k}^{2}={\frac {t_{k}(t_{k}+1)}{2)).}$

Define the triangular root of a triangular number ${\displaystyle N={\tfrac {n(n+1)}{2))}$ to be ${\displaystyle n}$. From this definition and the quadratic formula,

${\displaystyle \displaystyle n={\frac ((\sqrt {8N+1))-1}{2)).}$

Therefore, ${\displaystyle N}$ is triangular (${\displaystyle n}$ is an integer) if and only if ${\displaystyle 8N+1}$ is square. Consequently, a square number ${\displaystyle M^{2))$ is also triangular if and only if ${\displaystyle 8M^{2}+1}$ is square, that is, there are numbers ${\displaystyle x}$ and ${\displaystyle y}$ such that ${\displaystyle x^{2}-8y^{2}=1}$. This is an instance of the Pell equation ${\displaystyle x^{2}-ny^{2}=1}$ with ${\displaystyle n=8}$. All Pell equations have the trivial solution ${\displaystyle x=1,y=0}$ for any ${\displaystyle n}$; this is called the zeroth solution, and indexed as ${\displaystyle (x_{0},y_{0})=(1,0)}$. If ${\displaystyle (x_{k},y_{k})}$ denotes the ${\displaystyle k}$th nontrivial solution to any Pell equation for a particular ${\displaystyle n}$, it can be shown by the method of descent that the next solution is

{\displaystyle \displaystyle {\begin{aligned}x_{k+1}&=2x_{k}x_{1}-x_{k-1},\\y_{k+1}&=2y_{k}x_{1}-y_{k-1}.\end{aligned))}

Hence there are infinitely many solutions to any Pell equation for which there is one non-trivial one, which is true whenever ${\displaystyle n}$ is not a square. The first non-trivial solution when ${\displaystyle n=8}$ is easy to find: it is ${\displaystyle (3,1)}$. A solution ${\displaystyle (x_{k},y_{k})}$ to the Pell equation for ${\displaystyle n=8}$ yields a square triangular number and its square and triangular roots as follows:

${\displaystyle \displaystyle s_{k}=y_{k},\quad t_{k}={\frac {x_{k}-1}{2)),\quad N_{k}=y_{k}^{2}.}$

Hence, the first square triangular number, derived from ${\displaystyle (3,1)}$, is ${\displaystyle 1}$, and the next, derived from ${\displaystyle 6\cdot (3,1)-(1,0)-(17,6)}$, is ${\displaystyle 36}$.

The sequences ${\displaystyle N_{k))$, ${\displaystyle s_{k))$ and ${\displaystyle t_{k))$ are the OEIS sequences , , and respectively.

In 1778 Leonhard Euler determined the explicit formula[1][2]: 12–13

${\displaystyle \displaystyle N_{k}=\left({\frac {\left(3+2{\sqrt {2))\right)^{k}-\left(3-2{\sqrt {2))\right)^{k)){4{\sqrt {2))))\right)^{2}.}$

Other equivalent formulas (obtained by expanding this formula) that may be convenient include

{\displaystyle \displaystyle {\begin{aligned}N_{k}&={\tfrac {1}{32))\left(\left(1+{\sqrt {2))\right)^{2k}-\left(1-{\sqrt {2))\right)^{2k}\right)^{2}\\&={\tfrac {1}{32))\left(\left(1+{\sqrt {2))\right)^{4k}-2+\left(1-{\sqrt {2))\right)^{4k}\right)\\&={\tfrac {1}{32))\left(\left(17+12{\sqrt {2))\right)^{k}-2+\left(17-12{\sqrt {2))\right)^{k}\right).\end{aligned))}

The corresponding explicit formulas for ${\displaystyle s_{k))$ and ${\displaystyle t_{k))$ are:[2]: 13

{\displaystyle \displaystyle {\begin{aligned}s_{k}&={\frac {\left(3+2{\sqrt {2))\right)^{k}-\left(3-2{\sqrt {2))\right)^{k)){4{\sqrt {2)))),\\t_{k}&={\frac {\left(3+2{\sqrt {2))\right)^{k}+\left(3-2{\sqrt {2))\right)^{k}-2}{4)).\end{aligned))}

## Recurrence relations

There are recurrence relations for the square triangular numbers, as well as for the sides of the square and triangle involved. We have[3]: (12)

{\displaystyle \displaystyle {\begin{aligned}N_{k}&=34N_{k-1}-N_{k-2}+2,&{\text{with ))N_{0}&=0{\text{ and ))N_{1}=1;\\N_{k}&=\left(6{\sqrt {N_{k-1))}-{\sqrt {N_{k-2))}\right)^{2},&{\text{with ))N_{0}&=0{\text{ and ))N_{1}=1.\end{aligned))}

We have[1][2]: 13

{\displaystyle \displaystyle {\begin{aligned}s_{k}&=6s_{k-1}-s_{k-2},&{\text{with ))s_{0}&=0{\text{ and ))s_{1}=1;\\t_{k}&=6t_{k-1}-t_{k-2}+2,&{\text{with ))t_{0}&=0{\text{ and ))t_{1}=1.\end{aligned))}

## Other characterizations

All square triangular numbers have the form ${\displaystyle b^{2}c^{2))$, where ${\displaystyle {\tfrac {b}{c))}$ is a convergent to the continued fraction expansion of ${\displaystyle {\sqrt {2))}$, the square root of 2.[4]

A. V. Sylwester gave a short proof that there are infinitely many square triangular numbers: If the ${\displaystyle n}$th triangular number ${\displaystyle {\tfrac {n(n+1)}{2))}$ is square, then so is the larger ${\displaystyle 4n(n+1)}$th triangular number, since:

${\displaystyle \displaystyle {\frac ((\bigl (}4n(n+1){\bigr )}{\bigl (}4n(n+1)+1{\bigr ))){2))=4\,{\frac {n(n+1)}{2))\,\left(2n+1\right)^{2}.}$

The left hand side of this equation is in the form of a triangular number, and as the product of three squares, the right hand side is square.[5]

The generating function for the square triangular numbers is:[6]

${\displaystyle {\frac {1+z}{(1-z)\left(z^{2}-34z+1\right)))=1+36z+1225z^{2}+\cdots }$