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T. Amdeberhan
T. Amdeberhan
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Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$$$F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

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T. Amdeberhan
T. Amdeberhan
  • 43.1k
  • 5
  • 57
  • 217

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdots F_{n-k}!}$$\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then,, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdots F_{n-k}!}$. Then,, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdot F_{n-k}!}$. Then, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$

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T. Amdeberhan
T. Amdeberhan
  • 43.1k
  • 5
  • 57
  • 217

Products and sum of cubes in Fibonacci

Consider the familiar sequence of Fibonacci numbers: $F_0=0, F_1=1, F_n=F_{n-1}+F_{n-2}$.

Although it is rather easy to furnish an algebraic verification of the below identity, I wish to see a different approach. Hence,

QUESTION. Is there a combinatorial or more conceptual reason for this "pretty" identity? $$F_nF_{n-1}F_{n-2}=\frac{F_{n+2}^3-F_{n+1}^3-F_n^3}3.$$

Caveat. I'm open to as many alternative replies, of course.

Remark. The motivation comes as follows. Define $F_n!=F_1\cdots F_n$ and $F_0!=1$. Further, $\binom{n}k_F:=\frac{F_n!}{F_k!\cdots F_{n-k}!}$. Then,, I was studying these coefficients and was lead to $$\binom{n}3_F=\frac{F_n^3-F_{n-1}^3-F_{n-2}^3}{3!}.$$