Encontrar $\mathrm{lcm}(F_{m},F_{n},F_{k})$.

Question

Por lo que considerar tres números de Fibonacci.

Mi pregunta es:

Encontrar $\mathrm{lcm}(F_{m},F_{n},F_{k})$, donde lcm es el mínimo común múltiplo.

Answer 1

Sugerencia: Utilice el siguiente hecho: $$\mathrm{lcm}(a,b,c) = \frac{abc \cdot \gcd(a,b,c)}{\gcd(a,b)\gcd(a,c)\gcd(b,c)}$$

a lo largo de con $\gcd(F_m,F_n)=F_{\gcd(m,n)}$$\gcd(a,b,c)=\gcd(a,\gcd(b,c))$.

Answer 2

Lema. $\gcd(a,b,c)=\gcd(\gcd(a,b),\gcd(b,c))$.

Prueba. Vamos, $$a=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_n^{\alpha_n}$$$$b=p_1^{\beta_1}p_2^{\beta_2}\ldots p_n^{\beta_n}$$$$c=p_1^{\gamma_1}p_2^{\gamma_2}\ldots p_n^{\gamma_n}$$ where $p_i$'s are all distinct primes dividing at least one of $a,b$ or $c$. Then, from the definition of $\gcd$ el lado derecho es equivalente a, \begin{align}\gcd(\gcd(a,b),\gcd(b,c))&=\gcd(p_1^{\min(\alpha_1,\beta_1)}p_2^{\min(\alpha_2,\beta_2)}\ldots p_n^{\min(\alpha_n,\beta_n)},p_1^{\min(\beta_1,\gamma_1)}p_2^{\min(\beta_2,\gamma_2)}\ldots p_n^{\min(\beta_n,\gamma_n)})\\&=p_1^{\min(\alpha_1,\beta_1,\gamma_1)}p_2^{\min(\alpha_2,\beta_2,\gamma_2)}\ldots p_n^{\min(\alpha_n,\beta_n,\gamma_n)}\\&=\gcd(a,b,c)\end{align}

En consecuencia, \begin{align}\text{lcm}(F_m,F_n,F_k)&=\dfrac{F_mF_nF_k\gcd(F_m,F_n,F_k)}{\gcd(F_m,F_n)\gcd(F_n,F_k)\gcd(F_k,F_m)}\\&=\dfrac{F_mF_nF_k\gcd(\gcd(F_m,F_n),\gcd(F_n,F_k))}{\gcd(F_m,F_n)\gcd(F_n,F_k)\gcd(F_k,F_m)}\\&=\dfrac{F_mF_nF_kF_{\gcd(m,n,k)}}{F_{\gcd(m,n)}F_{\gcd(n,k)}F_{\gcd(k,m)}}\end{align}