How to check if a number is tribonacci number or not?

The Tribonacci sequence is an extension of the Fibonacci sequence where each term is the sum of the previous three terms.

The Tribonacci sequence: 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705,...........

So given a number, can we check if that number is Tribonacci number or not?

I tried but not able to find any formula, I know that Fibonacci number can be checked directly using a formula.

For sufficiently large values (somewhere around $$>200$$ or so), $$n$$ is Tribonacci if and only if $$a^m$$ rounds to $$n$$ for some integer $$m$$, where $$a = \frac 13 \left(1 + \sqrt[3]{19 - 3 \sqrt{3}} + \sqrt[3]{19 + 3 \sqrt{3}}\right),$$ and this can be easily checked by taking logarithms.
So firstly, let us study about the tribonacci constant (analogy of $$\Phi$$ of the Fibonacci series) i am giving here without proof the value-$$k=\frac{1+ \sqrt[3]{19+3\sqrt{33}}+\sqrt[3]{19-3\sqrt{33}}}{3} \approx 1.839$$ which is simply a root of the polynomial $$x^3 -x^2 -x -1$$ you can understand this constant as the converging ratio of any two consecutive numbers of the series.
Now, here is a formula for nth tribonacci numbers which I am giving without proof$$T(n)=\left \lfloor 3b \frac{\left(\frac{1}{3}(a_+ + a_- +1)\right)^n}{b^2 -2b +4} \right \rceil \\ \text{where \lfloor \rceil denotes nearest integer function} \\ a_\pm = \sqrt[3]{19 \pm 3\sqrt{33}} \\ b = \sqrt[3]{586 +102\sqrt{33}}$$ to check if a number is tribonacci or not, just solve for n and conclude