Square triangular number

Solution as a Pell equation

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

Define the triangular root of a triangular number N=\tfrac{n(n+1)}{2} to be n. In the form of the quadratic equation, n^2 + n - 2N = 0. From the quadratic formula,

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 kth nontrivial solution to any Pell equation for a particular n, it can be shown by the method of descent that the next solution is x_{k+1} &= 2x_k x_1 - x_{k-1}, \ y_{k+1} &= 2y_k x_1 - y_{k-1}. \end{align}}} 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:

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 , , and respectively.

Explicit formula

In 1778 Leonhard Euler determined the explicit formula

\right)^2.

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

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{align}}}

The corresponding explicit formulas for s_k and t_k are:

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{align}}}

Recurrence relations

The solution to the Pell equation can be expressed as a recurrence relation for the equation's solutions. This can be translated into recurrence equations that directly express the square triangular numbers, as well as the sides of the square and triangle involved. We have

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{align}}}

We have

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{align}}}

Other characterizations

All square triangular numbers have the form b^2c^2, where \tfrac{b}{c} is a convergent to the continued fraction expansion of \sqrt2, the square root of 2.

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

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.

The generating function for the square triangular numbers is: :\frac{1+z}{(1-z)\left(z^2 - 34z + 1\right)} = 1 + 36z + 1225 z^2 + \cdots

Notes

References

1. "Square Triangular Number".
2. Plouffe, Simon. (August 1992). "1031 Generating Functions". University of Quebec, Laboratoire de combinatoire et d'informatique mathématique.