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Fermat numbers

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Fermat numbers are numbers of the form

{\rm F}_n \equiv 2^{2^n} + 1,\quad n \ge 0. \,

Fermat numbers: \scriptstyle 2^{2^n} + 1,\, n \,\ge\, 0. \, (Cf. A000215)

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

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

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

Contents

Formulae

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

Recurrences

{\rm F}_{0} = 3,\, {\rm F}_{n} = ({\rm F}_{n-1} - 1)^2 + 1,\, n \ge 1. \,
{\rm F}_{n} = \prod_{k=0}^{n-1} {\rm F}_{k} + 2,\quad n \ge 0, \,

where for \scriptstyle n \,=\, 0 \, we have the empty product (giving the multiplicative identity, i.e. 1) + 2, giving \scriptstyle 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 \ge 0, \,

where for \scriptstyle n \,=\, 0 \, we have the empty product (giving the multiplicative identity, i.e. 1) + 2, giving \scriptstyle 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 \ge 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)[1]

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 \scriptstyle 2^{2^k} + 1 \,, for some \scriptstyle k \,\ge\, 0 \,. (Cf. A019434)

{3, 5, 17, 257, 65537, ?}

Products of distinct Fermat primes

Since there are 5 known Fermat primes, {\scriptstyle {\rm F_0} \,, \scriptstyle {\rm F_1} \,, \scriptstyle {\rm F_2} \,, \scriptstyle {\rm F_3} \,, \scriptstyle {\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.)

See also

Notes

  1. Wilfrid Keller, Prime factors k · 2^n + 1 of Fermat numbers F_m and complete factoring status.
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