Square triangular number

For squares of triangular numbers, see squared triangular number.

In mathematics, a square triangular number (or triangular square number) is a number which is both a triangular number and a perfect square. There are an infinite number of square triangular numbers; the first few are 0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025 (sequence A001110 in OEIS).

Explicit formulas

Write N_k for the kth square triangular number, and write s_k and t_k for the sides of the corresponding square and triangle, so that

$N_k = s_k^2 = \frac{t_k(t_k+1)}{2}.$

The sequences N_k, s_k and t_k are the OEIS sequences  A001110,  A001109, and  A001108 respectively.

In 1778 Leonhard Euler determined the explicit formula

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

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

$\begin{align} N_k &= {1 \over 32} \left( ( 1 + \sqrt{2} )^{2k} - ( 1 - \sqrt{2} )^{2k} \right)^2 = {1 \over 32} \left( ( 1 + \sqrt{2} )^{4k}-2 + ( 1 - \sqrt{2} )^{4k} \right) \\ &= {1 \over 32} \left( ( 17 + 12\sqrt{2} )^k -2 + ( 17 - 12\sqrt{2} )^k \right). \end{align}$

The corresponding explicit formulas for s_k and t_k are

$s_k = \frac{(3 + 2\sqrt{2})^k - (3 - 2\sqrt{2})^k}{4\sqrt{2}}$

and

$t_k = \frac{(3 + 2\sqrt{2})^k + (3 - 2\sqrt{2})^k - 2}{4}.$

Pell's equation

The problem of finding square triangular numbers reduces to Pell's equation in the following way. Every triangular number is of the form t(t + 1)/2. Therefore we seek integers t, s such that

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

With a bit of algebra this becomes

(2t + 1)² = 8s² + 1,

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

x² − 2y² = 1

which is an instance of Pell's equation. This particular equation is solved by the Pell numbers P_k as

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

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

N_k = 34N_{k − 1} − N_{k − 2} + 2, with N₀ = 0 and N₁ = 1.

$N_k = \left(6\sqrt{N_{k-1}} - \sqrt{N_{k-2}}\right)^2,\text{ with }N_0 = 1\text{ and }N_1 = 36.$

We have

s_k = 6s_{k − 1} − s_{k − 2}, with s₀ = 0 and s₁ = 1;

t_k = 6t_{k − 1} − t_{k − 2} + 2, with t₀ = 0 and t₁ = 1.

Other characterizations

All square triangular numbers have the form b²c², where b / c is a convergent to the continued fraction for the square root of 2.

A. V. Sylwester gave a short proof that there are an infinity of square triangular numbers, to wit: If the triangular number n(n+1)/2 is square, then so is the larger triangular number

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

The generating function for the square triangular numbers is:

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

Numerical data

As k becomes larger, the ratio t_k / s_k approaches $\sqrt{2} \approx 1.41421$ and the ratio of successive square triangular numbers approaches $17+12\sqrt{2} \approx 33.97056$.

$\begin{array}{rrrrll} k & N_k & s_k & t_k & t_k/s_k & N_k/N_{k-1} \\ 0 & 0 & 0 & 0 & & \\ 1 & 1 & 1 & 1 & 1 & \\ 2 & 36 & 6 & 8 & 1.33333 & 36\\ 3 & 1\,225 & 35 & 49 & 1.4 & 34.02778\\ 4 & 41\,616 & 204 & 288 & 1.41176 & 33.97224\\ 5 & 1\,413\,721 & 1\,189 & 1\,681 & 1.41379 & 33.97061\\ 6 & 48\,024\,900 & 6\,930 & 9\,800 & 1.41414 & 33.97056\\ 7 & 1\,631\,432\,881 & 40\,391 & 57\,121 & 1.41420 & 33.97056\\ \end{array}$

External links

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