Examples of patterns that eventually fail
I'll translate an entry in the blog Gaussianos ("Gaussians") about Polya's conjecture, titled:
A BELIEF IS NOT A PROOF.
We'll say a number is of even kind if in its prime factorization, an even number of primes appear. For example $6 = 2\cdot 3$ is a number of even kind. And we'll say a number is of odd kind if the number of primes in its factorization is odd. For example, $18 = 2·3·3$ is of odd kind. ($1$ is considered of even kind).
Let $n$ be any natural number. We'll consider the following numbers:
- $E(n) =$ number of positive integers less or equal to $n$ that are of even kind.
- $O(n) =$ number of positive integers less or equal to $n$ that are of odd kind.
Let's consider $n=7$. In this case $O(7) = 4$ (number 2, 3, 5 and 7 itself) and $E(7) = 3$ ( 1, 4 and 6). So $O(7) >E(7)$.
For $n = 6$: $O(6) = 3$ and $E(6) = 3$. Thus $O(6) = E(6)$.
In 1919 George Polya proposed the following result, know as Polya's Conjecture:
For all $n > 2$, $O(n)$ is greater than or equal to $E(n)$.
Polya had checked this for $n < 1500$. In the following years this was tested up to $n=1000000$, which is a reason why the conjecture might be thought to be true. But that is wrong.
In 1962, Lehman found an explicit counterexample: for $n = 906180359$, we have $O(n) = E(n) – 1$, so:
$$O(906180359) < E(906180359).$$
By an exhaustive search, the smallest counterexample is $n = 906150257$, found by Tanaka in 1980.
Thus Polya's Conjecture is false.
What do we learn from this? Well, it is simple: unfortunately in mathematics we cannot trust intuition or what happens for a finite number of cases, no matter how large the number is. Until the result is proved for the general case, we have no certainty that it is true.
From "Experimentation in Mathematics" Borwein, Bailey and Girgensohn 2004 : $$\sum_{n=1}^{\infty} \lfloor n\cdot e^{\frac{\pi}3\sqrt{163}}\rfloor 2^{-n}=1280640\ \ \text{(correct to at least half a billion digits!)}$$ Using the $\mathrm{sinc}$ function ($\mathrm{sinc}(x)=\frac{\sin(x)}x$ and this paper) : $$\int_0^{\infty} \mathrm{sinc}\left(\frac x1\right)\,\mathrm dx=\frac{\pi}2$$ $$\int_0^{\infty} \mathrm{sinc}\left(\frac x1\right)\cdot \mathrm{sinc}\left(\frac x3\right)\,\mathrm dx=\frac{\pi}2$$ $$\int_0^{\infty} \mathrm{sinc}\left(\frac x1\right)\cdot \mathrm{sinc}\left(\frac x3\right)\cdot \mathrm{sinc}\left(\frac x5\right)\,\mathrm dx=\frac{\pi}2$$ $$\cdots$$ $$\int_0^{\infty} \mathrm{sinc}\left(\frac x1\right)\cdot \mathrm{sinc}\left(\frac x3\right)\cdot \mathrm{sinc}\left(\frac x5\right)\cdots \mathrm{sinc}\left(\frac x{13}\right)\,\mathrm dx=\frac{\pi}2$$ $$\int_0^{\infty} \mathrm{sinc}\left(\frac x1\right)\cdot \mathrm{sinc}\left(\frac x3\right)\cdots \mathrm{sinc}\left(\frac x{15}\right)\,\mathrm dx=\frac{467807924713440738696537864469}{ 935615849440640907310521750000}\pi$$
In fact the story doesn't end here! It was found (see Baillie and Borweins' "Surprising Sinc Sums and Integrals") that you could replace the integrals by the corresponding $\frac 12 + \sum_1^{\infty}$ series : $$\frac 12 + \sum_{m=1}^{\infty} \prod_{k=0}^N \mathrm{sinc}\left(\frac m{2k+1}\right)=\int_0^{\infty} \prod_{k=0}^{N} \mathrm{sinc}\left(\frac x{2k+1}\right)\,\mathrm dx.$$
for the previous values of ($N=0,1,2,3\cdots 7$) but also for larger values of $N$ up to $40248$. For $N\gt 40248$ the left part is always larger than the integral at the right!
At this point the reciprocals of the odd integers could be replaced by other values (see the paper for the conditions required for the equality to hold) for example by the reciprocals of the prime numbers. Now, because of the slow divergence in this case, the equality breaks down only for $N \approx 10^{176}$ (when the sum of values slowly crosses the $2\pi$ barrier) and with an error smaller than $\displaystyle 10^{-10^{86}}$.