Square triangular number

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

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

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

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

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

\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 $n$ is not a square. The first non-trivial solution when $n=8$ is easy to find: it is $(3,1)$ . A solution $(x_{k},y_{k})$ to the Pell equation for $n=8$ yields a square triangular number and its square and triangular roots as follows:

\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 $(3,1)$ , is $1$ , and the next, derived from $6\cdot (3,1)-(1,0)-(17,6)$ , is $36$ .

The sequences $N_{k}$ , $s_{k}$ and $t_{k}$ 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 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 {\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 $s_{k}$ and $t_{k}$ are:^[2]^: 13

\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 {\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 {\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 $b^{2}c^{2}$ , where ${\tfrac {b}{c}}$ is a convergent to the continued fraction expansion of ${\sqrt {2}}$ , the square root of 2.^[4]

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

\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]

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

Notes

^ ^a ^b Dickson, Leonard Eugene (1999) [1920]. History of the Theory of Numbers. Vol. 2. Providence: American Mathematical Society. p. 16. ISBN 978-0-8218-1935-7.
^ ^a ^b ^c Euler, Leonhard (1813). "Regula facilis problemata Diophantea per numeros integros expedite resolvendi (An easy rule for Diophantine problems which are to be resolved quickly by integral numbers)". Mémoires de l'Académie des Sciences de St.-Pétersbourg (in Latin). 4: 3–17. Retrieved 2009-05-11. According to the records, it was presented to the St. Petersburg Academy on May 4, 1778.
^ Weisstein, Eric W. "Square Triangular Number". MathWorld.
^ Ball, W. W. Rouse; Coxeter, H. S. M. (1987). Mathematical Recreations and Essays. New York: Dover Publications. p. 59. ISBN 978-0-486-25357-2.
^ Pietenpol, J. L.; Sylwester, A. V.; Just, Erwin; Warten, R. M. (February 1962). "Elementary Problems and Solutions: E 1473, Square Triangular Numbers". American Mathematical Monthly. 69 (2). Mathematical Association of America: 168–169. doi:10.2307/2312558. ISSN 0002-9890. JSTOR 2312558.
^ Plouffe, Simon (August 1992). "1031 Generating Functions" (PDF). University of Quebec, Laboratoire de combinatoire et d'informatique mathématique. p. A.129. Archived from the original (PDF) on 2012-08-20. Retrieved 2009-05-11.