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Can someone give me a proof or a hint on why the recurrence equation: $$g(k+2)=k*g(k+1)-g(k)$$ has the solution: $$g(k)=c_1 {_0\tilde F_1}(;k;-1)+c_2 Y_{k-1}(2)$$ where ${_0\tilde F_1}(;a;x)$ is the regularized hypergeometric function and $Y_{k-1}(2)$ is the Bessel function of the second kind? Thanks in advance.

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up vote 2 down vote accepted

Your recurrence equation is an instance of the recurrence equation, satisfied by Bessel functions: $$ x J_{\nu-1}(x) + x J_{\nu+1}(x) = 2\nu J_{\nu}(x), \quad x Y_{\nu-1}(x) + x Y_{\nu+1}(x) = 2\nu Y_{\nu}(x) $$ for $x=2$. Moreover, Casoratian of $J_{k-1}(2)$ and $Y_{k-1}(2)$ is non-zero for all $k$ $$ C(J_{k-1}(2), Y_{k-1}(2)) = J_{k-1}(2) Y_k(2) - Y_{k-1}(2) J_k(2) \stackrel{\text{Wronskian}}{=} \left.-\frac{2}{\pi x}\right|_{x=2} = -\frac{1}{\pi} \not= 0 $$ therefore $J_{k-1}(2)$ and $Y_{k-1}(2)$ are linearly independent solutions of the recurrence equation, and hence $g(k) = c_1 J_{k-1}(2) + c_2 Y_{k-1}(2)$ is the fundamental system.

Incidentally: $$ {}_0 \tilde{F}_1\left(;k;-x^2\right) = x^{k-1} J_{k-1}(2x) $$

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