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It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. There is a function f vanishing to order 1 at each z_n (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)

EDIT: It's easy to show that such c_n exist. If you write b_n=a_n/(z_n f'(z_n)), then for any fixed z, the terms of the sum will be approximately b_n exp(c_n z_n) for n large. You can obviously pick c_n so that this converges.

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It's a standard theorem in complex analysis that if z_n is a sequence that goes to infinity, there is an entire function taking any prescribed values at the z_n. There is a function f vanishing to order 1 at each z_n (for z_n=n, you could take f(z)=\sin \pi z), and then consider \sum_n a_nf(z)/(f'(z_n)(z-z_n)). This may not converge, but you can tweak it by multiplying each term by something that is 1 at z_n (eg, exp(c_n(z-z_n)) for c_n chosen appropriately) to make it converge.

(I don't know off the top of my head how to choose the c_n; this is copied from Exercise 1 on page 197 of Ahlfors's Complex Analysis.)