The following Tn formulas calculate the n-th term of the Tribonacci sequence. The (2) is unknown and uses only one cubic root, since we can write [tex]b = (p + (4/p) + 1)/3[/tex], but how it is proved that (1) = (2)?
[tex]p = \sqrt [3] {19 + 3 \sqrt {33}}, p' = \sqrt [3] {19 - 3 \sqrt {33}}[/tex]
[tex]q = \sqrt [3] {586 + 102 \sqrt {33}}, q' = \sqrt [3] {586 - 102 \sqrt {33}}[/tex]
[tex]pp' = qq' = 4[/tex]
[tex]b = (p + p' + 1)/ 3, d = (q + q' + 1)/ 3[/tex]
[tex]T_n = [ b^{n - 1}/ (d - 1)] (1)[/tex]
[tex]T_n = [(b - 1)b^n/(4b - 6)] (2)[/tex]

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