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Using an integer relations algorithm, we get,

$$F_{2n}=-F_{n-1}^2+F_{n+1}^2$$

$$6F_{4n}= -F_{n-2}^4-3F_{n-1}^4+3F_{n+1}^4+F_{n+2}^4$$

The pattern of the subscripts is clear. Expressing the coefficients as a number triangle and including higher powers,

$$F_{2n} = -1,\,1$$

$$6F_{4n} = -1,\,\color{brown}{-3},\,3,\,1$$

$$120F_{6n} = -1,\,\color{brown}{-4},\,20,\,-20,\,4,\,1$$

$$21840F_{8n} = -1,\,\color{brown}{-14},\,91,\,364,\,-364,\,-91,\,14,\,1$$

$$24504480F_{10n} = -1,\,\color{brown}{-33},\, 748,\, 3927,\, -17017,\,17017, \dots\quad\quad\quad\quad$$

Question: Anybody knows the formula for the coefficients?

P.S. One pattern easy to spot is $F_n\,L_{n+1}=0,3,4,14,33,\dots$

See this related post for the slightly different form for odd powers (it contains both the $F_n$ and $F_{pn}$ terms), and Ron Knott's article on fibonomials.

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