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# Some series on Fibonacci and Lucas numbers

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In this post, we shall see several curious summation formulas of the Fibonacci numbers $F_n$ and the related Lucas numbers $L_n$ that involves the Bell polynomials $B_n(x)$ and the incomplete gamma function $\Gamma(a,x)$. The Fibonacci numbers are defined as $F_0=0, F_1=1, F_{n+2}=F_{n+1} + F_n, n \ge 0$ and the Lucas numbers are defined as $L_0=2, L_1=1, L_{n+2}=L_{n+1} + L_n, n \ge 0$.

Bell polynomials

The Bell polynomials $B_n(x)$ of polynomials are defined by the exponential generating function

$\displaystyle{\sum_{n=0}^{\infty}\frac{B_n(x)t^n}{n!}=e^{(e^t-1)x}}$.

The first few Bell polynomials are

$B_0(x) = 1$

$B_1(x) = x$

$B_2(x) = x^2 + x$

$B_3(x) = x^3 + 3x^2 + x$

$B_4(x) = x^4 + 6 x^3 + 7x^2 + x$

$B_5(x) = x^5 +10x^4 + 25 x^3 + 15x^2 + x$.

Incomplete gamma function

The incomplete gamma function $\Gamma(a,x)$ is defined as

$\displaystyle{\Gamma(a,x) = \int_{x}^{\infty}t^{a-1}e^{-t}dt}$.

Entry 1-6 involve summation identities involving the  Bell polynomials or the incomplete gamma function.

As usual, $\displaystyle{\phi=\frac{1+\sqrt5}{2}}$ denotes golden ratio. We have the following results:

Entry 1.

$\displaystyle{\sum_{r=1}^{\infty}\frac{F_r r^n}{r!} = \frac{e^{-1/\phi}}{\sqrt5}\{e^{\sqrt5} B_n(\phi)-B_n(-1/\phi)\}}$.

Entry 2.

$\displaystyle{\sum_{r=1}^{\infty}\frac{L_r r^n}{r!} = e^{-1/\phi}\{e^{\sqrt5} B_n(\phi)+B_n(-1/\phi)\}}$.

Entry 3.

$\displaystyle{\sum_{k=0}^{n-1}\frac{F_{k+m}x^k}{k!} = \frac{(-1)^me^{-x/\phi}}{\phi^m\sqrt5(n-1)!}\{(-\phi^2)^m e^{\sqrt5x}\Gamma(n,\phi x) - \Gamma(n,-x/\phi)\}}$

Entry 4.

$\displaystyle{\sum_{k=0}^{n-1}\frac{F_{k+m}(-x)^k}{k!} = \frac{(-1)^m e^{-x\phi}}{\phi^m\sqrt5(n-1)!}\{(-\phi^2)^m \Gamma(n,-\phi x) - e^{\sqrt5x}\Gamma(n,x/\phi)\}}$

Entry 5.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_{r+m} x^r}{r!} = \frac{e^{-x/\phi}}{\sqrt5}\{\phi^m e^{\sqrt5x}-(-\phi)^{-m}\}}$

Entry 6.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_{r+m} (-x)^r}{r!} = \frac{e^{-x\phi}}{\sqrt5}\{\phi^m -e^{\sqrt5x}(-\phi)^{-m}\}}$

Entry 7.

$\displaystyle{\sum_{r=0}^{\infty}\frac{F_r x^r}{r!} = -e^x \sum_{r=1}^{\infty}\frac{F_r (-x)^r}{r!}}$

Entry 8.

$\displaystyle{\sum_{k=0}^{n-1}\frac{L_{k+m}x^k}{k!} = \frac{(-1)^m e^{-x/\phi}}{\phi^m (n-1)!}\{(-\phi^2)^m e^{\sqrt5x}\Gamma(n,\phi x) + \Gamma(n,-x/\phi)\}}$

Entry 9.

$\displaystyle{\sum_{k=0}^{n-1}\frac{L_{k+m}(-x)^k}{k!} = \frac{(-1)^m e^{-x\phi}}{\phi^m (n-1)!}\{(-\phi^2)^m \Gamma(n,-\phi x) + e^{\sqrt5x}\Gamma(n,x\phi)\}}$

Entry 10.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_{r+m} x^r}{r!} = e^{-x/\phi}\{\phi^m e^{\sqrt5x}+(-\phi)^{-m}\}}$

Entry 11.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_{r+m} (-x)^r}{r!} = e^{-x\phi}\{\phi^m + e^{\sqrt5x}(-\phi)^{-m}\}}$

Entry 12.

$\displaystyle{\sum_{r=0}^{\infty}\frac{L_r x^r}{r!} = e^x \sum_{r=1}^{\infty}\frac{L_r (-x)^r}{r!}}$