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

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

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

Therefore, $N$ is triangular ( $n$ is an integer) if and only if $8 N + 1$ is square. Consequently, a square number $M 2$ is also triangular if and only if $8 M 2 + 1$ is square, that is, there are numbers $x$ and $y$ such that $x 2 - 8 y 2 = 1$ . This is an instance of the Pell equation 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

{\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 an infinity of solutions to any Pell equation for which there is one non-trivial one, which holds 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:

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 × (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

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

{\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

{\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}}

Pell's equation

The problem of finding square triangular numbers reduces to Pell's equation in the following way.^[3]

Every triangular number is of the form $⁠ t (t + 1) / 2 ⁠$ . Therefore we seek integers $t$ , $s$ such that

{\frac {t(t+1)}{2}}=s^{2}.

Rearranging, this becomes

\left(2t+1\right)^{2}=8s^{2}+1,

and then letting $x = 2 t + 1$ and $y = 2 s$ , we get the Diophantine equation

x^{2}-2y^{2}=1,

which is an instance of Pell's equation. This particular equation is solved by the Pell numbers $P k$ as^[4]

x=P_{2k}+P_{2k-1},\quad y=P_{2k};

and therefore all solutions are given by

s_{k}={\frac {P_{2k}}{2}},\quad t_{k}={\frac {P_{2k}+P_{2k-1}-1}{2}},\quad N_{k}=\left({\frac {P_{2k}}{2}}\right)^{2}.

There are many identities about the Pell numbers, and these translate into identities about the square triangular numbers.

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^[5]^: (12)

{\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

{\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 $⁠ b / c ⁠$ is a convergent to the continued fraction expansion of √2.^[6]

A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers:^[7] If the $n$ th triangular number $⁠ n (n + 1) / 2 ⁠$ is square, then so is the larger $4 n (n + 1)$ th triangular number, since:

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

As the product of three squares, the right hand side is square. The triangular roots $t k$ are alternately simultaneously one less than a square and twice a square if $k$ is even, and simultaneously a square and one less than twice a square if $k$ is odd. Thus,

49 = 7² = 2 × 5² − 1,

288 = 17² − 1 = 2 × 12², and

1681 = 41² = 2 × 29² − 1.

In each case, the two square roots involved multiply to give $s k$ : 5 × 7 = 35, 12 × 17 = 204, and 29 × 41 = 1189.^{[citation needed]}

Additionally:

N_{k}-N_{k-1}=s_{2k-1};

36 − 1 = 35, 1225 − 36 = 1189, and 41616 − 1225 = 40391. In other words, the difference between two consecutive square triangular numbers is the square root of another square triangular number.^{[citation needed]}

The generating function for the square triangular numbers is:^[8]

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

Numerical data

As $k$ becomes larger, the ratio $⁠ t k / s k ⁠$ approaches √2 ≈ 1.41421356, and the ratio of successive square triangular numbers approaches (1 + √2)⁴ = 17 + 12√2 ≈ 33.970562748. The table below shows values of $k$ between 0 and 11, which comprehend all square triangular numbers up to 10¹⁶.

$k$	$N k$	$s k$	$t k$	$⁠ t k / s k ⁠$	$⁠ N k / N k - 1 ⁠$
0	0	0	0	$⁠ t k / s k ⁠$
1	1	1	1	1
2	36	6	8	1.33333333	36
3	1225	35	49	1.4	34.027777778
4	41616	204	288	1.41176471	33.972244898
5	1413721	1189	1681	1.41379310	33.970612265
6	48024900	6930	9800	1.41414141	33.970564206
7	1631432881	40391	57121	1.41420118	33.970562791
8	55420693056	235416	332928	1.41421144	33.970562750
9	1882672131025	1372105	1940449	1.41421320	33.970562749
10	63955431761796	7997214	11309768	1.41421350	33.970562748
11	2172602007770041	46611179	65918161	1.41421355	33.970562748

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.
^ Barbeau, Edward (2003). Pell's Equation. Problem Books in Mathematics. New York: Springer. pp. 16–17. ISBN 978-0-387-95529-2. Retrieved 2009-05-10.
^ Hardy, G. H.; Wright, E. M. (1979). An Introduction to the Theory of Numbers (5th ed.). Oxford University Press. p. 210. ISBN 0-19-853171-0. Theorem 244
^ 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.

External links

[Dickson-1] 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.

[Euler-2] 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.

[3] Barbeau, Edward (2003). Pell's Equation. Problem Books in Mathematics. New York: Springer. pp. 16–17. ISBN 978-0-387-95529-2. Retrieved 2009-05-10.

[4] Hardy, G. H.; Wright, E. M. (1979). An Introduction to the Theory of Numbers (5th ed.). Oxford University Press. p. 210. ISBN 0-19-853171-0. Theorem 244

[5] Weisstein, Eric W. "Square Triangular Number". MathWorld.

[Ball-6] 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.

[Sylwester-7] 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.

[8] 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.

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