By shifting property of fibonacci numbers, $$F_{m+n} = F_m · F_{n+1} + F_{m-1} · F_n$$ where $F_k$ denotes the kth Fibonacci number .
I want to extend it to some n numbers .
So , how to find a formula for $F_{k_1+k_2+k_3+...+k_n}$ ?
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What have you tried so far? Have you already considered the obvious $F_{k_1+k_2+k_3} = F_{(k_1+k_2)+k_3}$? – Erick Wong Jun 26 '15 at 15:11
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I haven't really been able to find a closed formula for the question you pose in keeping $k_1,k_2...k_n$ general, but nonetheless have found results which may or may not prove useful to your endeavour
For any integer $n>0$
$F_{2n+2} = (F_{n} + F_{n+2}).F_{n+1}$
$F_{2n+1} = (F_{n})^2 + (F_{n+1})^2 $
For any integer $ n> 2$
$F_{n+2} =F_{n} +F_{n-1}.F_{3}+F_{n-2}.F_{2}$
Malcolm
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While not a closed formula for n in general, I have found results for when n=3,4 which may or may not prove useful to your endeavour.
Using the equation above,
${ F }_{ m+n+1 }={F}_{m+1}{F}_{n+1}+{F}_{m}{F}_{n}$ and ${ F }_{ m+n-1 }={F}_{m}{F}_{n}+{F}_{m-1}{F}_{n-1}$.
The difference between the two equations implies ${ F }_{ m+n }={F}_{m+1}{F}_{n+1}-{F}_{m-1}{F}_{n-1}$.
Therefore,
${ F }_{ a+b+c}={F}_{a+1}{F}_{b+c+1}-{F}_{a-1}{F}_{b+c-1}={F}_{a+1}{F}_{b+1}{F}_{c+1}+{F}_{a+1}{F}_{b}{F}_{c}-{F}_{a-1}{F}_{b}{F}_{c}-{F}_{a-1}{F}_{b-1}{F}_{c-1}={F}_{a+1}{F}_{b+1}{F}_{c+1}+{F}_{a}{F}_{b}{F}_{c}-{F}_{a-1}{F}_{b-1}{F}_{c-1}$
And in a similar way
${F}_{a+b+c+d}={F}_{a+1}{F}_{b+1}{F}_{c+1}{F}_{d+1}+{F}_{a+1}{F}_{b+1}{F}_{c}{F}_{d}+{F}_{a}{F}_{b}{F}_{c+1}{F}_{d+1}-{F}_{a}{F}_{b}{F}_{c-1}{F}_{d-1}-{F}_{a-1}{F}_{b-1}{F}_{c}{F}_{d}-{F}_{a-1}{F}_{b-1}{F}_{c-1}{F}_{d-1}$
Chad Shin
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