Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers
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Theorem
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Then:
- $\ds \sum_{k \mathop \in \Z} \dbinom m k_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 1} = 0$
where:
- $\dbinom m k_\FF$ denotes a Fibonomial coefficient
- $F_{n + k}$ denotes the $n + k$th Fibonacci number
- $\ceiling {\, \cdot \,}$ denotes the ceiling function
Proof
Lemma 1
- $\ds \sum_{k \mathop \in \Z} \dbinom m k_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 2} F_k = F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} {k - 1}_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 2} = 0$
Lemma 2
- $\ds \sum_{k \mathop \in \Z} \dbinom m k_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 2} \paren {-1}^k F_{m - k} = \paren {-1}^m F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} k_\FF \paren {-1}^{\ceiling {\paren {m - 1 - k} / 2} } {F_{n + k} }^{m - 2} = 0$
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Examples
Example: $m = 3$
- ${F_n}^n - 2 {F_{n + 1} }^2 - 2 {F_{n + 2} }^2 + {F_{n + 3} }^2 = 0$
Historical Note
Donald E. Knuth, in his The Art of Computer Programming: Volume 1: Fundamental Algorithms, 3rd ed. of $1997$, attributes this result to Dov Jarden and Theodore Samuel Motzkin.
However, he provides a citation for Jarden's work only, and it is unclear from where Motzkin's contribution is sourced.
Sources
- 1962: John Riordan: Generating functions for powers of Fibonacci numbers (Duke Math. J. Vol. 29, no. 1: pp. 5 – 12)
- 1966: Dov Jarden: Recurring Sequences (2nd ed.)
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.8$: Fibonacci Numbers: Exercise $30$