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 is an infinity of triangular squares, given by the formula

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

or by the linear recursion

<math>N_k = 34N_{k-1} - N_{k-2} + 2</math> with <math>N_0 = 0</math> and <math>N_1 = 1</math>

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

<math>n(n+1)/2 = m^2.</math>

With a bit of algebra this becomes

<math>(2n+1)^2=8m^2+1,</math>

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

<math>k^2=2h^2+1</math>

which is an instance of Pell's equation.

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

<math> s(N) = \sqrt{N}, </math>

<math> t(N) = \lfloor \sqrt{2 N} \rfloor. </math>

t is given by the formula

<math> 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]. </math>

As k becomes larger, the ratio t/s approaches the square root of two:

<math> \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} </math>

External references

• Triangular numbers that are also square at cut-the-knot
• Sequence A001110 from the On-Line Encyclopedia of Integer Sequences.fr:Nombre carré triangulaire

it:Numero quadrato triangolare he:מספר משולשי ריבועי sl:Trikotniško kvadratno število zh:三角平方數