A reverse birthday problem: no pair out of 1 million aliens shares a birthday; what is their year length? Assume a planet with a very very long year of $N$ days.  There are 1 million aliens at a party in a room, and no one at all shares a birthday.  What can be inferred about the size of $N$?
(This more compact question supersedes this poorly phrased one.)
 A: Assuming all birthdays are equally likely and the birthdays are independent, the chance that $k+1$ aliens do not share a birthday is
$$p(k;N) = 1\left(1-\frac{1}{N}\right)\left(1-\frac{2}{N}\right)\cdots\left(1-\frac{k}{N}\right).$$
Its logarithm can be summed asymptotically provided $k$ is much smaller than $N$:
$$\log(p(k;N)) = -\frac{k(k+1)}{2N} - \frac{k + 3k^2 + 2k^3}{12N^2} - O(k^4 N^{-3}).\tag{1}$$
To be $100-100\alpha\%$ confident that $N$ is no less than some value $N^{*}$, we need $(1)$ to be greater than $\log(1-\alpha)$.  Small $\alpha$ ensure $N$ is much larger than $k$, whence we may approximate $(1)$ accurately as $-k^2/(2N)$.  This yields
$$-\frac{k^2}{2N} \gt \log(1-\alpha),$$
implying
$$N \gt\frac{-k^2}{2\log(1-\alpha)} \approx \frac{k^2}{2\alpha}=N^{*}\tag{2}$$
for small $\alpha$.
For instance, with $k=10^6-1$ as in the question and $\alpha=0.05$ (a conventional value corresponding to $95\%$ confidence), $(2)$ gives $N \gt 10^{13}$.  
Here's a more expansive interpretation of this result. Without approximating in formula $(2)$, we obtain $N=9.74786\times 10^{12}$. For this $N$ the chance of no collision in a million birthdays is $p(10^6-1, 9.74786\times 10^{12})=95.0000\ldots\%$ (computed without approximation), essentially equal to our threshold of $95\%$.  Thus for any $N$ this large or larger it's $95\%$ or more likely there will be no collisions, which is consistent with what we know, but for any smaller $N$ the chance of a collision gets above $100 - 95= 5\%$, which starts to make us fear we might have underestimated $N$.
As another example, in the traditional Birthday problem there is a $4\%$ chance of no collision in $k=6$ people and $5.6\%$ chance of no collision in $k=7$ people.  These numbers suggest $N$ ought to exceed $360$ and $490$, respectively, right in the range of the correct value of $366$.  This shows how accurate these approximate, asymptotic results can be even for very small $k$ (provided we stick to small $\alpha$).
