Square triangular number

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 OEIS: A001110, OEIS: A001109, and OEIS: A001108 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

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

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

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

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]

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

• Cannonball problem, on numbers that are simultaneously square and square pyramidal
• Sixth power, numbers that are simultaneously square and cubical

• Triangular numbers that are also square at cut-the-knot
• Weisstein, Eric W. "Square Triangular Number". MathWorld.
• Michael Dummett's solution