Let $\{a_n\}$ be a sequence of complex numbers indexed by the positive integers. Does there always exist an analytic function $f$ such that $f(n) = \{a_n\}$ for $n=1,2,...$? If not, are there any simple necessary or sufficient conditions for the existence of such $f$? This analytic function should be defined on some connected domain in the complex plane containing the positive integers.

To make this concrete, consider Ackermann's function, which is defined recursively: first define the sequence of functions $A_k$, $k=1,2,...$, as

$A_1(n) = 2n$,
$A_k(1) = 2$, $A_k(n) = A_{k-1}(A_k(n-1))$,

and then define Ackermann's function as the diagonal $A(n) = A_n(n)$ for $n \geq 1$. Does there exist an analytic function $f$ such that $f(n) = A(n)$ for $n = 1,2,...$?

Actually, the individual functions $A_k$ are interesting as well. $A_1(n) = 2n$, as given above; $A_2(n) = 2^n$, and $A_3(n) = 2^{2^{\ldots^{2^2}}}$ (with $n$ twos in the expression). Obviously, $A_1$ and $A_2$ have analytic extensions. According to Wikipedia (which uses a slightly different definition and notation), analytic extensions of $A_3$ or $A_k$ for any other $k$ aren't known, but from the language, it isn't clear whether the existence of an extension is itself in question, or whether one simply hasn't been found yet. Also, it doesn't say anything about the diagonal $A(n)$ (unless I missed it).

There are many other obvious sequences that don't seem to have obvious analytic extensions, like the prime-counting function (just to name one!). As far as my knowledge is concerned, this seems more like the rule than the exception. My knowledge here is admittedly very limited, though, so anything at all that you can share will probably teach me something.

  • $\begingroup$ There is nice trick to see that for any function $f(n)$ on the nonnegative integers we can find an entire function $F(x)$ which is real and increasing on the positive reals and satisfies $F(n)\geq f(n)$ for all nonnegative integers $n$, namely, if $f(n)\geq n$ (say) then $F(x)=\sum_{n\geq 0} f(n)(x/n)^{f(n)}$. $\endgroup$ Aug 17, 2017 at 2:00

4 Answers 4


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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    $\begingroup$ Ah, cool. Ironically, this also turns out to be an exercise in my old complex analysis textbook! Its construction, which they name the "Pringsheim interpolation formula," is very similar to yours. $\endgroup$ Oct 28, 2009 at 1:21

Actually, there is an even stronger result, often called the interpolation theorem, which follows from a well-known theorem of Mittag-Leffler:

Let $(z_n)$ be a sequence of complex numbers with no limit point. For each $n$, let $l(n)$ be any integer greater or equal to $1$ and for $0 \leq k \leq l(n)$, let $(a_{n, k})$ be complex numbers. Then there exists an entire function $g$ such that

$$g^{(k)}(z_n) = a_{n,k}$$

for every $n \geq 1$ and every $0 \leq k \leq l(n)$.

That is, you can fix values for the derivative at the $z_j$'s.

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    $\begingroup$ Of course, one needs that the $z_n$ are pairwise distinct. $\endgroup$ Jan 10, 2020 at 11:34

If I remember correctly you can get even a stronger result. Let $a_n$, $b_n$ and $c_n$ be sequences of complex numbers and $i_n$ be a sequence of natural numbers. Then I believe there exists a meromorphic function that takes the value $b_n$ on $a_n$ for all $n$, and has a pole of order $i_n$ at $c_n$ for all $n$. Assuming that the $a_n$ and $c_n$ are disjoint and have no accumulation point. It's been a while since I thought about complex analysis but I seem to remember learning this.


Look for Pringsheim´s Interpolation Formula at Stein´s Complex Analysis book in page 156 exercice 17, that would give you the existence, and the conditions under it holds.


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