Square triangular number

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For squares of triangular numbers, see squared triangular number

A square triangular number (or triangular square number) is a number which is both a triangular number and a perfect square. There is 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

N k = 34 N k - 1 - N k - 2 + 2

with

N 0 = 0

and

N 1 = 1

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 = m 2 .

With a bit of algebra this becomes

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

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

k 2 = 2 h 2 + 1

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

The k^th triangular square N_k is equal to the s^th perfect square and the t^th triangular number, such that

$s(N) = \sqrt{N},$

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

t is given by the formula

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

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))

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