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# Fermat numbers

Fermat numbers are numbers of the form

${\rm {F}}_{n}\equiv 2^{2^{n}}+1,\quad n\geq 0.\,$ Fermat numbers: $2^{2^{n}}+1,\,n\,\geq \,0.\,$ (Cf. A000215)

{3, 5, 17, 257, 65537, 4294967297, 18446744073709551617, 340282366920938463463374607431768211457, 115792089237316195423570985008687907853269984665640564039457584007913129639937, ...}

Fermat numbers $2^{2^{n}}+1,\,n\,\geq \,0,\,$ in base 2 representation. (Cf. A080176 comment)

{11, 101, 10001, 100000001, 10000000000000001, 100000000000000000000000000000001, 10000000000000000000000000000000000000000000000000000000000000001, ...}

## Formulae

${\rm {F}}_{n}=~?.\,$ ## Recurrences

${\rm {F}}_{0}=3,\,{\rm {F}}_{n}=({\rm {F}}_{n-1}-1)^{2}+1,\,n\geq 1.\,$ ${\rm {F}}_{n}=\prod _{k=0}^{n-1}{\rm {F}}_{k}+2,\quad n\geq 0,\,$ where for $n\,=\,0\,$ we have the empty product (giving the multiplicative identity, i.e. 1) + 2, giving $F_{0}\,=\,3\,$ , as expected.

## Properties

The sequence of Fermat numbers is a coprime sequence, since

${\rm {F}}_{n}=\prod _{k=0}^{n-1}{\rm {F}}_{k}+2,\quad n\geq 0,\,$ where for $n\,=\,0\,$ we have the empty product (giving the multiplicative identity, i.e. 1) + 2, giving $F_{0}\,=\,3.\,$ ## Generating function

$G_{\{{\rm {F}}_{n}\}}(x)=~?.\,$ ## Forward differences

${\rm {F}}_{n+1}-{\rm {F}}_{n}=({\rm {F}}_{n})^{2}-3{\rm {F}}_{n}+2=({\rm {F}}_{n}-1)({\rm {F}}_{n}-2),\quad n\geq 0.\,$ ## Partial sums

$\sum _{n=0}^{m}{\rm {F}}_{n}=\sum _{n=0}^{m}(2^{2^{n}}+1)=m+1+\sum _{n=0}^{m}2^{2^{n}}=~?.\,$ ## Partial sums of reciprocals

$\sum _{n=0}^{m}{\frac {1}{{\rm {F}}_{n}}}=\sum _{n=0}^{m}{\frac {1}{2^{2^{n}}+1}}=~?.\,$ ## Sum of reciprocals

$\sum _{n=0}^{\infty }{\frac {1}{{\rm {F}}_{n}}}=\sum _{n=0}^{\infty }{\frac {1}{2^{2^{n}}+1}}=~?.\,$ ## Prime factorization of Fermat numbers

The prime factors of Fermat numbers are of the form (since they are odd)

$k\cdot 2^{m}+1\,$ More interestingly, the prime factors of Fermat numbers are of the form

${\rm {F}}_{n}=2^{2^{n}}+1=(k_{1}\cdot 2n+1)(k_{2}\cdot 2n+1)\cdots \,$ Prime factorization of Fermat numbers
$n\,$ ${\rm {F}}_{n}\,$ Prime factors
0 3 = 1*2^1+1 3 = 1*2^1+1 = (1*(2*1)+1)
1 5 = 1*2^2+1 5 = 1*2^2+1 = (1*(2*2)+1)
2 17 = 1*2^4+1 17 = 1*2^4+1 = (2*(2*4)+1)
3 257 = 1*2^8+1 257 = 1*2^8+1 = (16*(2*8)+1)
4 65537 = 1*2^16+1 65537 = 1*2^16+1 = (2048*(2*16)+1)
5 4294967297 = 1*2^32+1 641 * 6700417 = (5*2^7+1) (52347*2^7+1) = (10*(2*32)+1) (104694*(2*32)+1)
6 18446744073709551617 = 1*2^64+1 274177 * 67280421310721 = (1071*2^8+1) (262814145745*2^8+1) = (2142*(2*64)+1) (525628291490*(2*64)+1)
7 340282366920938463463374607431768211457 = 1*2^128+1 59649589127497217 * 5704689200685129054721 = (116503103764643*2^9+1) (11141971095088142685*2^9+1) = (233006207529286*(2*128)+1) (22283942190176285370*(2*128)+1)

### Fermat primes

It is conjectured that just the first 5 numbers in this sequence are primes (Fermat primes.)

List of Fermat primes: primes of form $2^{2^{k}}+1\,$ , for some $k\,\geq \,0\,$ . (Cf. A019434)

{3, 5, 17, 257, 65537, ?}

#### Products of distinct Fermat primes

Since there are 5 known Fermat primes, {${\rm {F_{0}}}\,$ , ${\rm {F_{1}}}\,$ , ${\rm {F_{2}}}\,$ , ${\rm {F_{3}}}\,$ , ${\rm {F_{4}}}\,$ } = {3, 5, 17, 257, 65537}, then there are

${\binom {5}{0}}+{\binom {5}{1}}+{\binom {5}{2}}+{\binom {5}{3}}+{\binom {5}{4}}+{\binom {5}{5}}=1+5+10+10+5+1=32=2^{5}\,$ products of distinct known Fermat primes. The 31 non-empty products of distinct known Fermat primes give the number of sides of constructible odd-sided polygons (since a polygon has at least 3 sides.)