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A divisor $d$ of $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}}$ is unitary if and only if $d = p_{1}^{\varepsilon_{1}} \cdots p_{n}^{\varepsilon_{n}}$, where each exponent $\varepsilon_{i}$ is either $0$ or $r_{i}$. Let $D_{k} = ${ $d$ } be the subset of unitary divisors of an integer $k > 1$ satisfying $\omega(d) = \omega(k) - 1$.

Definition. A positive integer $k = p_{1}^{r_{1}} \cdots p_{n}^{r_{n}} > 1$ is hyperbolic if and only if $\sum_{i = 1}^{n} p_{i}^{-r_{i}} < 1$ or, equivalently, $\sum_{d \in D_{k}} d < k$.

See my OEIS entry.

For example, $3$, $10$ and $20$ are hyperbolic, but $30$ and $510510 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$ are not.

Assuming my calculations are correct, I'll state the following with some confidence:

Indeed, many positive integers are hyperbolic. Of the first $10^{8}$ integers, $70334760$ are hyperbolic. Non-trivial prime powers, squares or higher, are also hyperbolic. If $k$ is a hyperbolic integer, then so are its proper, non-trivial unitary divisors; however, the same cannot be inferred for all divisors (consider the hyperbolic integer $900$ and its non-hyperbolic divisor $30$). In fact, an arbitrary product of any number of non-trivial unitary divisors of a hyperbolic integer is again hyperbolic.

The set of hyperbolic integers is closed under exponentiation, but not under addition (e.g., $10 + 20 = 30$) or under multiplication (e.g., $3 \times 10 = 30$).

Define the Prime zeta function, $\zeta_{P}(s) = \sum_{p \text{ prime}} p^{-s}$, which converges absolutely for $\mathsf{Re}(s) > 1$. Recall that the multiplicity of a prime divisor $p$ of $k$ is the largest exponent $r$ such that $p^{r}$ divides $k$ but $p^{r+1}$ does not. If the minimum multiplicity of an integer $k$ is $2$ or greater, then $k$ is hyperbolic as can be seen by the elementary bound \begin{eqnarray} \sum_{i = 1}^{n} p_{i}^{-r_{i}} < \zeta_{P}(2) \approx 0.452247 .... \end{eqnarray} Thus, the question of hyperbolicity is non-trivial only for integers with minimum multiplicity $1$.

Numerical evidence suggests that the natural density of the hyperbolic integers is greater than $0.988284 \dots$, and I conjecture that almost all integers are indeed hyperbolic (i.e., the natural density is 1).

Question: Is anything presently known about such integers? (References welcome!)

Question: Is there a simple proof showing (or refuting) that almost all integers are hyperbolic?


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Say $k$ has the prime factorization $p_1^{r_1} p_2^{r_2} \cdots p_n^{r_n}$. Then let $H(k)$ be the "hyperbolicity" of this integer, $H(k) = \sum_{i=1}^n p_i^{-r_i}$. Hyperbolic integers are those integers $k$ that have $H(k) < 1$.

Now, $H(k)$ for some "random" integer $k$ looks like a sum of independent random variables $X_p$, one for each prime $p$, where $P(X_p = p^{-k})$ is the probability that a random integer is divisible by $p^k$ but not $p^{k+1}$. Thus $P(X_p = p^{-k}) = (p-1)/p^{k+1}$.

In particular $$ E(X_p) = \sum_{k \ge 1} p^{-k} P(X_p = p^{-k}) = \sum_{k \ge 1} {p-1 \over p^{2k+1}} = {1 \over p^2 + p} $$

Things get a bit tricky in finding the distribution of $\sum_p X_p$; in particular a central limit theorem won't work, because the contribution of each of the small primes is too large. But there is some limiting distribution, and I suspect it at least bears some resemblance to the limiting distribution of the actual hyperbolicities of integers.

Incidentally, my tests seem to point to a natural density of hyperbolic integers somewhere in the neighborhood of 99 percent. I'd conjecture that integers with hyperbolicity less than $k$ have a natural density. There are similar results on the distribution of $\sigma(n)/n$ where $\sigma(n)$ is the sum of divisors of $n$; see for example Marc Deleglise, Bounds for the density of abundant integers.

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Thank you for reference! It's been quite helpful. – user02138 Nov 4 '10 at 4:07

[[EDIT: Let $ k = p_1^{r_1} \cdots p_n^{r_n} $ as in your post. Then define the hyperbolicity $ H(k) = \sum_{i=1}^n p_i^{-r_i} $.]]

We can refute the idea in your second question in a much more general manner than you might expect. Note that since $ \zeta_P(1) $ diverges, numbers can have arbitrarily large hyperbolicity; similarly, since $ H(2^n) = 2^{-n} $, it follows that numbers can have arbitrarily small hyperbolicity. Define $ d(x) $ to be the asymptotic density of natural numbers satisfying $ H(n) > x $. We will demonstrate the strict inequality $ d(x) > 0 $ holds for all $ x > 0 $, and even deduce a satisfying lower bound for sufficiently, erm... not-small $x $.

Fix $ x \in (0,\infty) $. Let $ k $ be the smallest natural number such that $ 1/2 + 1/3 + 1/5 + \dots + 1/p_k > x $. Then for $ n = 2(3)(5)\cdots (p_k) $, $ H(n) > x $. Note that hyperbolicity, taken as an arithmetic function, is multiplicative, though not completely. In fact, by categorizing which prime factors belong where and when they're duplicated, one can see the following formula:

$$ H(ab) = H(a) + H(b) - H(\gcd(a,b)^2) + H(\gcd(a,b)) $$

This isn't necessary for our argument, I'm just putting that out there for posterity.

At any rate, we may conclude that for any number $ m $ which is coprime with $ n $, we can obtain another sufficently hyperbolic number via multiplication: $ H(nm) > H(n) > x $. Note that the density of numbers coprime with $ n $ can be expressed as

$$ y = (1-1/2)(1-1/3)(1-1/5)\cdots (1-1/p_k). $$

This formula echoes the Sieve of Eratosthenes: it follows from the fact that the probability a number is divisible by a prime is independent of the probability it is divisible by a different prime.

Finally, if you take the set of all numbers $ m $ coprime with $n $, and multiply them all by $ n $, they'll become $ n $ times more sparse, hence the density of all numbers of the form $ n m $ is $ y / n $. This proves $ d(x) \ge y/n $, which is a pretty good lower bound so long as $ x $ isn't too small (this is my subjective impression). For $ x = 1 $, as in your original question, we have the close lower bound $ (1/30)(1/2)(2/3)(4/5) $ or about 0.89%.

So far I haven't been able to determine whether or not there exists any $ \epsilon > 0 $ such that $ d(\epsilon) = 1 $. If there is one, then there would exist a maximum $ u $ such that $ d(\epsilon) = 1 $ for any $ \epsilon \in (0,u) $, which would make an interesting new constant. Note that $ d(x)$ is decreasing, though I'm not sure if it's continuous. What would really be interesting is if $ d(x) $ were differentiable on some interval.

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I'm interested in more on this concept. I'll add some more trivia I've gathered since posting into the above answer (I figured out how to prove $ d(x) $ never equals 1). One thing I don't get, though, is why you're focused in on $ x = 1 $; seems sort of arbitrary to me. Then again maybe you have some special facts about that case. – anon Jul 9 '11 at 22:25
If you consider the sets {2,3,5}, {2,3,7,11}, and {2,3,7,13} you get up to 0.973%. It should not be hard to get this over 1%, but the inclusion-exclusion gets tricky. – deinst Jul 10 '11 at 1:49
Scratch that, I haven't proved $d(x) < 1$, I made an error. Note that almost-but-not-quite-1 asymptotic density is true for almost all $x$ here, not just $x=1$. Anyway I found a few other bits. Let $n=\prod p^r$ and $q=\prod p$. Then $$\sum_{n=1}^{\infty}H(n)n^{-s}=\zeta'(s)-\zeta(s)\sum_{p}\frac{1}{p^s(p^{s+1}-1‌​)}$$ and $$\exp\left(-\sum_{k=1}^\infty \frac{H(n^k)}{k}\right) = \frac{\phi(q)\sigma_1(n/q)}{n}$$ – anon Jul 10 '11 at 2:14
Also, based on my intuition I have a conjecture as to the moment generating function of $H$'s distribution. I essentially factor it into an Euler product: $$\mathbb{E}(e^{tH}) = \prod_{p}\left(1-\frac{1}{p}\right)\left(1+\sum_{r=1}^\infty p^{-r} \exp(p^{-r} t) \right). $$ This is remarkably eerie, is it not? – anon Jul 10 '11 at 2:43
@anon: This problem is starting to turn into beautiful mathematics. – user02138 Jul 10 '11 at 3:18

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