Square triangular number

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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 triangular squares, given by the formula

 N_k = {1 \over 32} \left( \left( 1 + \sqrt{2} \right)^{2k} - \left( 1 - \sqrt{2} \right)^{2k} \right)^2 .

or by the linear recursion

Nk = 34Nk − 1Nk − 2 + 2 with N0 = 0 and N1 = 1

The first few square triangular numbers are 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, 55420693056, 1882672131025, ... (sequence A001110 in OEIS)

The problem of finding square triangular numbers reduces to Pell's equation in the following way. Every triangular number is of the form n(n + 1)/2. Therefore we seek integers n, m such that

n(n + 1) / 2 = m2.

With a bit of algebra this becomes

(2n + 1)2 = 8m2 + 1,

and then letting k = 2n + 1 and h = 2m, we get the Diophantine equation

k2 = 2h2 + 1

which is an instance of Pell's equation and is solved by the Pell numbers.

We get the recursion

mk = 6mk − 1mk − 2.

Also, note that

m^2_{k}-1=m_{k+1}m_{k-1}

since m0 = 1 and m1 = 6.

The kth triangular square Nk is equal to the sth perfect square and the tth triangular number, such that

 s(N) = \sqrt{N},
 t(N) = \lfloor \sqrt{2 N} \rfloor.

t is given by the formula

 t(N_k) = {1 \over 4} \left[ \left( \left( 1 + \sqrt{2} \right)^k + \left( 1 - \sqrt{2} \right)^k \right)^2 - \left( 1 + (-1)^k \right)^2 \right].

or by the recursion

 t_k = 2\sqrt{2t_{k-1}(t_{k-1}+1)} + 3t_{k-1} + 1

As k becomes larger, the ratio t/s approaches the square root of two: Also ratio of successive square triangulars converges to 17+12(sqrt(2))

 \begin{matrix} N=1 & s=1 & t=1 & t/s=1
\\ N=36 & s=6 & t=8 & t/s = 1.3333333
\\ N=1225 & s=35 & t=49 & t/s = 1.4
\\ N=41616 & s=204 & t=288 & t/s = 1.4117647
\\ N=1,413,721 & s=1189 & t=1681 & t/s = 1.4137931
\\ N=48,024,900 & s=6930 & t=9800 & t/s = 1.4141414
\\ N=1,631,432,881 & s=40391 & t=57121 & t/s = 1.4142011
\end{matrix}

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