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Which short closed-form formulas for the Fibonacci numbers of higher order $F(m;n)$ (Wikipedia: Generalizations of Fibonacci numbers), or of its shifted form $F(m;n+m-1)$, are there?

I already found two formulas. I found them with help of Math.StackExchange (see Math.StackExchange: Simplifying my sum which contains binomials).

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IV_
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A linear recurrence relation with $n$ terms with constant factors has a closd form solution in the form of $n$ exponential functions (similar to what happens with differential equation). So, write it down as

$$a_{n}-a_{n-1}\ldots-a_{n-k}=0$$ write $a_{n}=A q^n$, and solve the characteristic equation for all possible $q$ (the solution is then a superposition of these that matches the initial terms).

orion
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  • A closed-form expression for the number $F(m;n)$ or $F(m;n+m-1)$ in dependence of $m$ and $n$ is meant, not a generating function for all this numbers. – IV_ Jan 30 '16 at 19:20
  • $a_{n}=Aq^n$ means $A$ will be eliminated. Therefore $A$ is not needed. Have I understood your comment correctly? – IV_ Jan 31 '16 at 14:07
  • True, not needed to get $q$ - like for all homogeneous linear equations, scaling a solution gives another solution. – orion Jan 31 '16 at 21:25