# a(27) in A059097: A new lower higher bound

Alec Mihailovs alec at mihailovs.com
Tue Nov 22 06:12:36 CET 2005

```From: "David Wilson" <davidwwilson at comcast.net>

> Indeed, I contend that all but
> a finite number of c(2n, n) are divisible by 9 or 25.

From a probabilistic point of view, that can't be true.

C(2n,n) is not divisible by 3 iff all the digits of n written in base 3, are
either 0, or 1. The probability that a k-digit number in base 3 has all
digits 0 or 1, is (2/3)^k (allowing zeros at the beginning).

Similarly, C(2n,n) is not divisible by 5 iff all the digits of n written in
base 5, are 0, 1, or 2. The probability that a k-digit number in base 5 hase
all digits 0, 1, or 2, is (3/5)^k (again allowing zeros at the beginning).

Thus, the probability that C(2n,n) is not divisible by 3 and by 5 for n <N
is greater than (2/3)^(log3(N)+1) * (3/5)^(log5(N)+1) = 0.4 * (2/3)^log3(N)
* (3/5)^log5(N). In this case, the expected number of such n < N is greater
than 0.4 * (2/3)^log3(N) * (3/5)^log5(N) * N = 0.4*N^0.313536.

Since the expected number grows with N, there should be infinite number of
such n that C(2n,n) is not divisible by 3 and by 5.

Alec Mihailovs
http://math.tntech.edu/alec/

```