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The original post is below. Question 1 was solved in the negative by David Speyer, and the title has now been changed to reflect Question 2, which turned out to be the more difficult one. A bounty of 100 is offered for a complete solution.

Original post. It follows from the prime number theorem and the periodicity properties $f(n+p) \equiv f(n) \mod{p}$ that for each $A < e$ there are only finitely many integer polynomials $f \in \mathbb{Z}[x]$ such that $|f(n)| < A^n$ for all $n \in \mathbb{N}$. On the other hand, for each $k \in \mathbb{N}$ the binomial coefficient $\binom{n}{k}$ is an integer-valued polynomial in $n$ bounded by $2^n$.

Question 1. Are there infinitely many integer polynomials with $|f(n)| < e^n$ for all $n \in \mathbb{N}$?

Question 2. Given $A < 2$, are there only finitely many integer-valued polynomials $f \in \mathbb{Q}[x]$ with $|f(n)| < A^n$ for all $n \in \mathbb{N}$?

Vesselin Dimitrov
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  • For any fixed $k$, $n^k < e^n$ for all $n$ sufficiently large - say, larger than $m$. It follows that $f_k(n) := n^k - m^k$ is smaller than $e^n$ for every $n$ and $k$. Unless I'm making a groggy mistake... – Paul Siegel Aug 11 '13 at 12:53
  • It is not smaller in absolute value, though. Your polynomial has $|f_k(1)| = m^k - 1 > e$. – Vesselin Dimitrov Aug 11 '13 at 13:06
  • Oops! Groggy mistake indeed... – Paul Siegel Aug 11 '13 at 13:10
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    It's probably worth noting here the fact that any integer-valued polynomial is an integral linear combination of the polynomials $\binom{x}{k}$. – Harry Altman Aug 12 '13 at 18:35
  • Indeed, and perhaps the title should now be changed to: "Are there infinitely many integer-valued polynomials dominated by $1.9^n$ on all of $\mathbb{N}$?" – Vesselin Dimitrov Aug 12 '13 at 18:57
  • Regarding question 2: when $k$ is sufficiently large we can't take $f(x)=\binom{x}k$, since $\binom{2k}k$ is asymptotic to $2^{2k}/\sqrt{\pi k}$ as $k\to\infty$. In general if $f(x)$ is integer-valued of degree $k$, with $k$ sufficiently large, then the inequalities $|f(n)|<A^n$ with $n$ near $2k$ will impose huge constraints on $f$. Maybe somehow one can use them to show that there are no $f$'s of suff.large degree? – Michael Zieve Aug 26 '13 at 01:02
  • Also, there are only finitely many integer-valued $f$'s of any fixed degree $k$ which satisfy $|f(n)|<2^n$ for all $n$, since if we write $f(x)=\sum_{i=0}^k a_i \binom{x}i$ with $a_i\in\mathbf{Z}$ then $|f(0)|<2^0$ bounds $|a_0|$, after which $|f(1)|<2^1$ bounds $|a_1|$, and so on. So Question 2 is really asking whether there exist $f$'s of arbitrarily large degree. – Michael Zieve Aug 26 '13 at 01:22
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    Hm, does this mean I should somehow donate half of the 200-point bounty to David Speyer? I suppose I can achieve this purpose by posting a new question "What's the optimal $A$?", putting a 100-point bounty on it, and accepting David's answer which would be basically a link or pointer here... – Noam D. Elkies Aug 26 '13 at 21:59
  • Well, actually, he would have to donate it to you - for I already awarded you the old 100 bounty. Then I intended to set a new 100 bounty for him, but it turned out this wasn't an option, so I went for 200. (The latter hasn't yet been awarded, so if I award it to you instead, what you say could be an option if you set a 150-point bounty, if that's possible, for his answer :). – Vesselin Dimitrov Aug 26 '13 at 22:10
  • Ah, I misunderstood. I'm entirely fine with David getting double my bounty, since his answer is definitive (besides being a double answer - both an upper and a lower bound which "happen" to coincide!). – Noam D. Elkies Aug 26 '13 at 23:20
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    Looking beyond Mathoverflow, is it worth writing this up for publication as a joint paper? – Noam D. Elkies Aug 26 '13 at 23:20
  • @ Noam Elkies: I think so; the relation to Fekete's theorem which you and David discovered seems particularly interesting to me. My contribution here, though, is minimal and reduces to having merely asked the question. – Vesselin Dimitrov Aug 27 '13 at 09:35
  • Here is a generalization of Ruzsa's conjecture which, though not involving polynomial coefficients anymore, is nonetheless in the spirit of the two questions. It is suggested to me by the connection with Fekete's theorem (and by that of Polya-Carlson-Bertrandias). Let $G \ni 0$ be a pointed (say) simply connected domain, and $\rho$ its conformal mapping radius. Let $h : P \to \mathbb{N}_0 \cup {\infty}$ be a function on the set $P$ of primes, and consider the set $S(G,h)$ of those $f \in \mathbb{Z}[[x]]$ which are meromorphic on $G$ and whose reduction $f \mod{p}$ at each prime $p$ (cont.) – Vesselin Dimitrov Aug 27 '13 at 09:55
  • (cont.) is a rational fraction of degree $\leq h_p$. Then: 1) If $\log{\rho} + \liminf_n \frac{1}{n} \sum_{p : h_p < n} \log{p} > 0$, the set $S(G,h)$ should only contain rational functions in $\mathbb{Z}[[x]]$. 2) If however the inequality in 1) is not satisfied, the set $S(G,h)$ should be uncountable. – Vesselin Dimitrov Aug 27 '13 at 10:12
  • (cont.) This generalizes in an obvious way to global fields and to algebraic curves of higher genus. It would thus be a sharp algebraicity criterion, of which I can only prove a weaker version involving an inequality stronger than 1). It was this type of problems which motivated me to ask question 1. – Vesselin Dimitrov Aug 27 '13 at 10:16
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    @Vesselin Dimitrov: Your contribution is not minimal; not only is asking a good question valuable in itself, but you also gave the example showing $A \leq 2$ which suggested that this "Question 2" has quite a different flavor from "Question 1" on ${\bf Z}[x]$. – Noam D. Elkies Aug 27 '13 at 14:44
  • @ Noam Elkies: Thank you! It was of indeed the example of $\binom{n}{k}$ which led me to ask question 2. (Otherwise I was considering rather different kinds of stuff, more in line with the first question.) – Vesselin Dimitrov Aug 27 '13 at 15:02

3 Answers3

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$\def\ZZ\mathbb{Z}$Question 1: No. Let $C>1$. I will show that there are only finitely many $f(x)$ in $\ZZ[x]$ so that $|f(n)| \leq C^n$ for all $n \in \ZZ_{\geq 0}$.

Choose $d$ large enough that, for all $k>d$, we have $k! > 2 C^k$. I claim that a polynomial with $|f(n)|<C^n$ is determined by its values on $0$, $1$, ..., $d$. Suppose, to the contrary, that $f(n) \neq g(n)$ but that they agree for $0 \leq n \leq d$. Let $k$ be the first integer where $f$ and $g$ disagree.

So $f(x)-g(x)$ is divisible by $x(x-1)(x-2) \cdots (x-k+1)$, so $f(x) - g(x) = x(x-1) \cdots (x-k+1) h(x)$ for some $h$ with integer coefficients. So $f(k) - g(k) = k! h(k) \equiv 0 \bmod k!$.

But, by assumption, $|f(k)|$ and $|g(k)| < C^k < k!/2$. So it is impossible that $f(k) \neq g(k)$ and $f(k) \equiv g(k) \bmod k!$. This contradiction concludes the proof.

Question 2 is still stumping me.

David E Speyer
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  • Very neat. I had failed to fully exploit the polynomial assumption. My argument for $C < e$ was otherwise similar; it amounted to noting that $f(k)-g(k)$ was divisible by the product of the primes less than $k$. The latter argument works for arbitrary mappings $f : \mathbb{N} \to \mathbb{Z}$ with the property $f(n+p) \equiv f(n) \mod{p}$ for all $n \in \mathbb{N}$ and all primes $p$: for $C < e$ there are only finitely many such mappings (conjectured to be polynomials anyway), while for $C = e$ there are uncountably many. – Vesselin Dimitrov Aug 12 '13 at 18:19
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    Thus, you have shown that for each $n_0 < \infty$, there are only finitely many $f \in \mathbb{Z}[x]$ with $|f(n)| < n!/2$ for all $n \geq n_0$. This is close to optimal upon considering $k!\binom{n}{k}$, although (especially in view of the proof) it could still be an interesting problem to see whether the coefficient $1/2$ of $n!$ is optimal, or if it could be improved to any constant smaller than $1$. – Vesselin Dimitrov Aug 12 '13 at 18:23
  • (To correct my first comment above: I meant to say that for each $C > e$ there are uncountably many such mappings $f : \mathbb{N} \to \mathbb{Z}$ --- not for $C = e$.) – Vesselin Dimitrov Aug 12 '13 at 18:35
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Question 2: The constant $A$ can be brought down to $\sqrt 3$, and probably a bit below that but not all the way down to $1+\epsilon$.

Instead of the polynomial $f(n) = {n \choose m}$, use a finite difference of such polynomials, $$ f(n) = \sum_{i=0}^m (-1)^i {m \choose i} {n \choose m+i}. $$ This is the $x^n$ coefficient of $$ \frac1{1-x} \left( \frac{x(1-2x)}{(1-x)^2} \right)^m $$ and can be estimated by contour integration on $|x| = 3^{-1/2}$; the maximum of $|f(n)|^{1/n}$ occurs near $n=3m$, for which the critical points are at $x = (3 \pm \sqrt{-3}) / 6$.

Note that for $f(n) = {n \choose m}$ the generating function was $\frac1{1-x} (x/(1-x))^m$, and the maximum occurred near $n=2m$, for which the critical point was at $x = 1/2$. The factor $1-2x$ kills that maximum, and it seems that using $x/(1-x)$ and $(1-2x)/(1-x)$ to the same power is optimal. To reduce $A$ further, try to include also some power of $(3x^2-3x+1)/(1-x)^2$ to kill the new critical point.

Noam D. Elkies
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    So I was wrong on both questions :-). Since you believe the limit infimum [of values of $A$] to be nonetheless strictly $> 1$, interesting what the actual value would be. Regarding integer polynomials the right question would have been: what is the limit infimum $\delta$ of the values of $c > 0$ such that there are infinitely many $f \in \mathbb{Z}[t]$ with $|f(n)|/n! < c$ for all $n \in \mathbb{N}$? We know from David Speyer's solution that $\delta \in [1/2,1]$. – Vesselin Dimitrov Aug 26 '13 at 04:29
  • Thanks also to David Speyer for his solution to Question 1. But the bounty was offered for Question 2, so I'm marking this as the accepted answer. – Vesselin Dimitrov Aug 26 '13 at 04:30
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    Thanks. Since $25 \geq 2 \cdot 11$ (and $11 \geq 10$) this also gives David Speyer the rare "Populist" badge, which is his third gold and which David may actually prefer to the bounty points :-) – Noam D. Elkies Aug 26 '13 at 04:47
  • Was there a motivating application for either question? (It's a natural problem in any case but it would be nice if the answer has some further use.) – Noam D. Elkies Aug 26 '13 at 04:51
  • Not an application, but it came up in relation to an old conjecture of Ruzsa characterizing the subring $\mathbb{Z}[1/(1-x)] \subset \mathbb{Q}[[x]]$ by the two properties: 1) radius of convergence (and I would add: meromorphy) is strictly $> 1/e$; and 2) for a set of primes $p$ of full density, the mod $p$ reduction is $A_p(x)/(1-x)^p$, with $A_p$ a polynomial of degree $< p$. Perelli and Zannier have shown that such an $f \in \mathbb{Q}[[x]]$ is at least $D$-finite. (continued.) – Vesselin Dimitrov Aug 26 '13 at 05:36
  • (continued.) And a closer look at their argument shows that not only the order of the differential equation can be taken to depend only on the radius of convergence, but that if the coefficients satisfy $|a(n)| < A^n$ for all $n \geq n_0$ and $A < e$, then in fact $f$ satisfies one of a finite set (depending on $A$ and $n_0$) of linear homogeneous differential equations. This implies that the degree of a polynomial in Question 1 is bounded for $A < e$. – Vesselin Dimitrov Aug 26 '13 at 05:40
  • (And I meant, after all, to write $\mathbb{Z}[1/(1-x)] \subset \mathbb{Z}[[x]]$ and $f \in \mathbb{Z}[[x]]$ in the above comment. Sorry about this, as well as for the switch of notation, $f$ now denoting the generating function.) – Vesselin Dimitrov Aug 26 '13 at 06:23
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    Let me spell out what I think Noam is implying. Let $p(x)$ be a polynomial with integer coefficients and degree $d$. Let $f_m(n)$ be the coefficient of $x^n$ in $p(x)^m/(1-x)^{dm+1}$. Then $f_m$ is an integer valued polynomial. If $|p(x)/(1-x)^d| \leq 1$ on the circle $|x| = r$, then integrating on this contour gives a bound $|f_m(n)| \leq C r^{-n}$. So our goal is to find polynomials $p$ which can push the radius $r$ as far out as possible. – David E Speyer Aug 26 '13 at 14:46
  • This is the opposite of http://mathoverflow.net/questions/32961, which aims to find monic polynomials $f$ and $g$ so that ${ |f(z)|=|g(z)| }$ stays as close to $0$ as possible. – David E Speyer Aug 26 '13 at 15:22
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    The integers $f(n)$, when $n = 3m$ seem to be divisible by all primes between $2m$ and $4m$, which leads to a Chebyshev estimate slightly better than the one from the middle binomial coefficient. I haven't looked at David's new example but this suggests that getting $A$ down to $1$ might involve the PNT. – Felipe Voloch Aug 26 '13 at 18:12
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    @FelipeVoloch It looks like $|f(2n)|$ is $(6n)! (2n)!/((4 n)! (3n)! n!)$, see https://oeis.org/A211419 . – David E Speyer Aug 26 '13 at 18:41
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    Question: This constructs a constant $C$ and infinitely many integer values polynomials with $|f_m(n)|<C r^{-n}$. Clearly, at the cost of increasing $r$ infinitesimally, we can find an $N$ so that we get $|f_m(n)| < r^{-n}$ for $n>N$. But how do we guarantee ourselves $|f_m(n)|<r^{-n}$ for all $n$? – David E Speyer Aug 26 '13 at 19:44
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    For $m$ large, the average of $|f_m(z)|$ over $|z|=r$ is $o(1)$ because $|f_m|$ behaves like the $m$-th power of a function that's at most $1$ but attains that value at only finitely many points. (Most likely $O(m^{-1/2})$ unless for some reason $1 - \lim_{m\rightarrow\infty} |f_m|^{1/m}$ has a quadruple or higher-order zero at some point of $|z|=r$.) – Noam D. Elkies Aug 26 '13 at 19:55
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The optimal growth rate is $\tau:= (1+\sqrt{5})/2$. Specifically, for any $\epsilon>0$, there are infinitely many integer valued polynomials bounded by $(\tau+\epsilon)^n$, but only finitely many below $(\tau-\epsilon)^n$. The first part of this answer (written first) proves the finiteness; the second uses Noam Elkies' idea combined with a theorem of Fekete to prove the infinitude.

Fix $\epsilon>0$. I will show that there are only finitely many integer values polynomial $f(z)$ with $|f(n)| < (\tau-\epsilon)^n$.

Let $f$ be such a polynomial of degree $d$. Set $$\frac{p(z)}{(1-z)^{d+1}} = \phi(z) = \sum_{n=0}^{\infty} f(n) z^n$$ Then $p(z)$ has integer coefficients, $p(1) \neq 0$, and we can uniquely recover $f$ from $p$. Moreover, there is some $M$ and some $\delta_1>0$ (dependent on $\epsilon$) so that $|\phi(z)| < M$ on $|z|=\tau^{-1}+\delta_1$.

We make the change of variables $u = 1/(1-z)$, so $z=1-1/u$. We have $\phi(1-1/u) = p(1-1/u) u^{d+1}$. Set $q(u) = p(1-1/u) u^{d+1}$. From the properties of $p$ above, $q$ is a polynomial with integer coefficients of degree $d+1$, and $|q(1/(1-z))| < M$ when $|z|=\tau^{-1}+\delta_1$. The map $z \mapsto 1/(1-z)$ sends $|z|=\tau^{-1}+\delta_1$ to a circle which contains the circle of radius $1+\delta_2$ around $\tau$ (for some $\delta_2>0$). So, using the maximum modulus principle, $|q(u)|<M$ on the circle of radius $1+\delta_2$ around $\tau$.

We therefore make one more change of coordinates, $v=u-\tau$ and $s(v) = q(v+\tau)$, to get a polynomial $s$ with $|s(v)|<M$ on the circle of radius $1+\delta_2$ around $0$. Although $s$ does not have integer coefficients, its leading term $v^{d+1}$ is a nonzero integer.

Choose $D_1$ sufficiently large enough that $2 \pi M (1+\delta_2)^{-D_1-1} <1$. Then, for $D_2 \geq D_1$, taking a contour integral around $|v|=1+\delta_2$ shows that the coefficient of $v^{D_2}$ in $s(v)$ has absolute value $<1$. Since the coefficient of $v^{d+1}$ is a nonzero integer, this establishes that $d<D_1$. So we have bounded the degree of $f$. Thus, $f$ is determined by its values at $D_1$ integers, and there are only finitely many possible polynomials $f$.

Now for the reverse bound. This argument is closely based on the proof of Fekete's Theorem here. (The original paper is here, but I don't speak German so I haven't checked whether they are the same argument.)

Our first goal is to establish the following: Let $r < 1$. There exists a nonzero polynomial $q(u)$ with integer coefficients so that $|q(u)|<1$ on the circle $|u-\tau|<r$.

Choose an integer $T$ large enough that, for any $N > T$, we have $$r^N + (1/2) r^{N-1} + (1/2) r^{N-2} + \cdots + (1/2) r^{T+1} + (1/2) r^T < 1/3.$$

Take $N$ larger than $T$. Define $q^N_N(u) = (u-\tau)^N$. Define $q^N_i(u)$ to be the unique polynomial of the form $$q^N_i(u) = q^N_{i+1}(u) + \theta_i \cdot (u-\tau)^{i}$$ so that $|\theta_i| \leq 1/2$ and the coefficient of $u^{i}$ in $q^N_i$ is an integer. Set $q^N(u) = q^N_T(u)$. So the coefficient of $u^k$ in $q^N(u)$ is an integer for $T \leq k \leq N$. For $u$ on the circle $|u-\tau|=r$, we get $$|q^N_T(u)| \leq r^N + (1/2) r^{N-1} + \cdots + (1/2) r^T < 1/3.$$

Let $(c^N_{T-1}, C^N_{T-2}, \ldots, c^T_0)$ be the last $T$, noninteger, coefficients of $q^N$. By the Pigeonhole principle, we can find $q^M$ and $q^N$ so that $$\sum_i |\{ c^N_i - c^M_i \}| r^i < 1/3$$ where $\{ x \}$ is the distance from $x$ to the nearest integer. We define $q(u)$ to be the result of taking $q^N(u) - q^M(u)$ and rounding the last $T$ coefficients to the nearest integer. We have now constructed $q$.

We now undo the above argument. Since $|q(u)|<1$ for $|u-\tau|<r$, we have $|q(1/(1-z))|<1$ on the disc with diameter $(1-(\tau+r)^{-1}, 1-(\tau-r)^{-1})$. This contains the circle of radius $\tau^{-1} - \delta_1$ about $0$, where $\delta_1 \to 0$ as $r \to 1$. So Noam's argument constructs infinitely many polynomials bounded by $(\tau+\delta_2)^n$.

Just for the fun of it, I used the above construction to find a polynomial $\sum_{i=1}^{20} \theta_i (u-\tau)^i$ with $|\theta_i| < 1/2$ and all coefficients other than the constant term integral. The constant term turned out to be $-3878005 + 1739105 \sqrt{5} \approx 10752.00000977$. If I round that off to $10752$, the resulting polynomial factors as $(2 - u)^9 (1 - u)^5 (3 - 3 u + u^2) (7 - 15 u + 14 u^2 - 6 u^3 + u^4)$. Making the variable substitution suggests that our next family of polynomials should be the coefficients of $$\frac{1}{1-z} \left( \frac{z^5 (1 - 2 z)^9 (1 - 3 z + 3 z^2) (1 - 5 z + 11 z^2 - 13 z^3 + 7 z^4)}{(1 - z)^{20}} \right)^m.$$ Of the four roots of $7 - 15 u + 14 u^2 - 6 u^3 + u^4$, two are at distance $0.883514$ from $\tau$ and two are at distance $1.02472$. Much past $N=20$, my naive implementation times out.

Community
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David E Speyer
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    The best exponent $k$ in $z^k(1-2z)^k (1-3z+z^2) / (1-z)^{2+2k}$ is not $k=2$ but $k \approx 3.7559$ [it doesn't have to be rational; just use $z^A (1-2z)^A (1-3z+z^2)^B / (1-z)^{2A+2B}$ with $A/B \rightarrow k$]. This gives a bound about $1.68053$, which is the largest root of $x^8 - 7x^6 + 23x^4 - 49x^2 + 49$. One can do better yet by using different multiplicities for the factors $z$ and $1-2z$, which brings the bound down to about $1.656246$. – Noam D. Elkies Aug 26 '13 at 18:42
  • @NoamD.Elkies Thanks! In the comment above, should $z^2-3z+1$ read $3 z^2 - 3z +1$? – David E Speyer Aug 26 '13 at 18:55
  • You're welcome, and yes I meant $1-3z+3z^2$, not $1-3z+z^2$.
    (And I was wondering if it might be converging to $\tau$...)     – Noam D. Elkies Aug 26 '13 at 19:45
  • So this is a special case of the general problem: Find an integer polynomial which takes values of norm $\leq 1$ on a circle of radius $r<1$ around a real point $x$ in the complex plane. (With $r=A/(A^2-1)$ and $x=A^2/(A^2-1)$) Since it doesn't seem like we are dealing with especially easy $x$, I don't think this will converge to $\tau$ unless the problem is solvable for all $r$ and $x$. But if it were, this would probably be in the literature David Speyer surveyed... – Will Sawin Aug 26 '13 at 20:41
  • @WillSawin Check the update, this problem is solvable for all $r$ and $x$. Fekete's result is that, if $h(x)$ is a monic polynomial with real coefficients, then we can find a a monic polynomial with integer coefficients which is $<1$ on ${ z: |h(z)|<r }$. Or, at least, I think that's what the result unpacks to, and I wrote out the argument above for the circle to see if I got it right. – David E Speyer Aug 26 '13 at 20:50
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    A generalization of Fekete's theorem inspired by arithmetic geometry (and, I think, in a particularly illuminating way) appears in Chinburg's paper "Capacity theory on varieties": http://archive.numdam.org/ARCHIVE/CM/CM_1991__80_1/CM_1991__80_1_75_0/CM_1991__80_1_75_0.pdf See Theorem 1.2 and the Minkowski application in the paragraph following it. – Vesselin Dimitrov Aug 26 '13 at 21:10
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    In any case, this is amazing. Now I have to award the bounty. Since the solution is a joint one, here is what I will do. I will award the current bounty to Noam Elkies' answer, and start a new bounty of the same worth to reward this complete answer. – Vesselin Dimitrov Aug 26 '13 at 21:39
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    A bounty of 200 will be awarded to this answer in 24 hours. (It turns out that 1) one may not set two bounties of the same worth on the same question, and 2) one has to wait for 24 hours before the bounty can be awarded.) – Vesselin Dimitrov Aug 26 '13 at 21:53
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    Do you know if there are finitely many integer-valued $f$ such that $f(n) \leq \tau^n$? – Will Sawin Aug 27 '13 at 15:24
  • I don't know. My vague guess would be no, because the recipe for constructing polynomials below $(\tau+\epsilon)^n$ has so many "large but finite" error terms that I feel like you shouldn't be able to squeeze all the way down to $\tau^n$. But it would certainly be interesting to know! – David E Speyer Aug 27 '13 at 15:52
  • Neat computation. The factor $P_4=1-5z+11z^2-13z^3+7z^4$ is the minimal polynomial of the critical points I found for the optimal $z^k(1-2z)^k (1-3z+3z^2) / (1-z)^{2+2k}$. The exponents $5,9,1,1$ for $z$, $1-2z$, $1-3z+3z^2$, and $P_4$ must be just an approximation to the optimal proportion: even without $P_4$ the optimal ratio is not rational. There's no need to stop at $N=20$: tell gp tau=(sqrt(5)+1)/2; zdif(x)=x-round(x); and then speyer(N, q) = q=(u-tau)^N; forstep(i=N-1,0,-1, q -= zdif(polcoeff(q,i))*(u-tau)^i); round(q) will compute well beyond speyer(20) ... [cont.] – Noam D. Elkies Aug 28 '13 at 00:04
  • [cont.'d] ... before complaining about loss of precision. For $N \leq 37$ we see only the same four factors, except for a degree-$11$ irreducible in the $N=30$ polynomial; then there are some new factors of degree $\geq 5$ for $38 \leq N \leq 50$, and for $N>50$ it more-or-less breaks down, usually giving just a few factors of $z$, $1-2z$, $1-3z+3z^2$, and sometimes $P_4$, and then a huge residual factor. The neat exception is $N=64$ where we get a pair of degree-$13$ irreducibles. But there might be a tweak that continues giving good factors past $N=50$ and $P_4$. – Noam D. Elkies Aug 28 '13 at 03:23
  • @NoamD.Elkies So, how do you find the optimal ratio of exponents? – David E Speyer Aug 28 '13 at 13:16
  • @Will Sawin: For each $m\leq49$ there is an integer valued polynomial of degree $m$ and $\lvert f(n)\rvert\le\tau^n$ for all non-negative integers $n$. However, this family doesn't work for larger degrees anymore. – Peter Mueller Sep 02 '13 at 13:54
  • @PeterMueller: Interesting! Which family? – Will Sawin Sep 02 '13 at 20:11
  • @Will Sawin: Those polynomials which have generating function $\sum f(n)x^n=\frac{x^{m-k}(2x-1)^{k-2r}(3x^2-3x+1)^r}{(1-x)^{m+1}}$, where $0\le 2r\le k\le m$. For $m\le49$, there are $k$ (around $0.7m$) and a small $r$ such $\lvert f(n)\rvert\le\tau^n$. – Peter Mueller Sep 02 '13 at 20:37