On 2^n – 1

Picking up from my last blog post, where we observed 255 = 15 x 17,

$$2^{n}-1 = (2^{\frac{n}{2}}+1)*(2^{\frac{n}{2}}-1)$$
$$= (2^{\frac{n}{2}}+1)*(2^{\frac{n}{4}}+1)*(2^{\frac{n}{4}}-1)$$
$$= (2^{\frac{n}{2}}+1)*(2^{\frac{n}{4}}+1)*(2^{\frac{n}{8}}+1)*(2^{\frac{n}{8}}-1)$$

If $$n=2^{k}$$, where $$k \in Z$$

$$= (2^{\frac{n}{2}}+1)*(2^{\frac{n}{4}}+1)*(2^{\frac{n}{8}}+1)*(2^{\frac{n}{16}}+1) … (2+1)(2-1)$$
$$= \prod_{i=1}^k (2^{\frac{n}{2^i}}+1)$$

This is incredibly cool!

So,
$$2^4-1$$ = 15 = 5 x 3
$$2^8-1$$ = 255 = 17 x 5 x 3
$$2^{16}-1$$ = 65535 = 257 x 17 x 5 x 3
$$2^{32}-1$$ = 4294967295 = 65537 x 257 x 17 x 5 x 3

It is fascinating to see the above results in binary:

       11
101  x
------
1111  =

       11
101 x
10001 x
--------
11111111 =

       11
101 x
10001 x
100000001 x
-----------
16 1's    =


Primes of the form $$2^{n}-1$$ are called Mersenne primes. Primes of the form $$2^{2^{n}}+1$$ are called Fermat primes. The numbers 3, 5, 17, 257 and 65537 are the only known Fermat primes!

That concludes my random rant. Have fun with $$2^{n}-1$$!