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# 5

Please do not rely on any information it contains.

5 is a prime number, the only one to be both the largest member of a twin prime pair (with 3) and the smallest member of a twin prime pair (with 7). 5 is a factor of 10, our base of numeration, and was for a time a contender for base.

## Membership in core sequences

 Odd numbers 1, 3, 5, 7, 9, 11, 13, 15, 17, ... A005408 Prime numbers 2, 3, 5, 7, 11, 13, 17, 19, ... A000040 Fibonacci numbers 1, 1, 2, 3, 5, 8, 13, 21, ... A000045 Partition numbers 1, 1, 2, 3, 5, 7, 11, 15, 22, ... A000041

In Pascal's triangle, 5 occurs twice, corresponding to ${\displaystyle {\binom {5}{1}}}$ and ${\displaystyle {\binom {5}{4}}}$. (In Lozanić's triangle, 5 occurs four times).

## Sequences pertaining to 5

 Multiples of 5 0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, ... A008598 Pentagonal numbers 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, 210, ... A000326 Generalized pentagonal numbers 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, 51, 57, 70, 77, ... A001318 Pentagonal pyramidal numbers 1, 6, 18, 40, 75, 126, 196, 288, 405, 550, 726, 936, ... A002411 Centered pentagonal numbers 1, 6, 16, 31, 51, 76, 106, 141, 181, 226, 276, 331, 391, ... A005891 Fermat pseudoprimes to base 5 4, 124, 217, 561, 781, 1541, 1729, 1891, 2821, 4123, ... A005936 ${\displaystyle 3x+1}$ sequence starting at 3 3, 10, 5, 16, 8, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, 2, 1, 4, ... A033478 ${\displaystyle 5x+1}$ sequence starting at 5 5, 26, 13, 66, 33, 166, 83, 416, 208, 104, 52, 26, 13, ... A259207

## Partitions of 5

There are seven partitions of 5, of which three consist of distinct numbers: {1, 4}, {2, 3} and {5}. There is only one nontrivial partition of 5 into primes, and that's {2, 3}.

## Roots and powers of 5

In the table below, irrational numbers are given truncated to eight decimal places.

 ${\displaystyle {\sqrt {5}}}$ 2.23606797 A002163 5 2 25 ${\displaystyle {\sqrt[{3}]{5}}}$ 1.70997594 A005481 5 3 125 ${\displaystyle {\sqrt[{4}]{5}}}$ 1.49534878 A011003 5 4 625 ${\displaystyle {\sqrt[{5}]{5}}}$ 1.37972966 A005534 5 5 3125 ${\displaystyle {\sqrt[{6}]{5}}}$ 1.30766048 A011200 5 6 15625 ${\displaystyle {\sqrt[{7}]{5}}}$ 1.25849895 A011201 5 7 78125 ${\displaystyle {\sqrt[{8}]{5}}}$ 1.22284454 A011202 5 8 390625 ${\displaystyle {\sqrt[{9}]{5}}}$ 1.19581317 A011203 5 9 1953125 ${\displaystyle {\sqrt[{10}]{5}}}$ 1.17461894 A011204 5 10 9765625 ${\displaystyle {\sqrt[{11}]{5}}}$ 1.15755791 A011205 5 11 48828125 ${\displaystyle {\sqrt[{12}]{5}}}$ 1.14352983 A011206 5 12 244140625 A000351

## Logarithms and fifth powers

In the OEIS specifically and mathematics in general, ${\displaystyle \log x}$ refers to the natural logarithm of ${\displaystyle x}$, whereas all other bases are specified with a subscript.

If ${\displaystyle n}$ is not a multiple of 11, then either ${\displaystyle n^{5}-1}$ or ${\displaystyle n^{5}+1}$ is. Hence the formula for the Legendre symbol ${\displaystyle \left({\frac {a}{11}}\right)=a^{5}\mod 11}$.

As above, irrational numbers in the following table are truncated to eight decimal places.

 ${\displaystyle \log _{5}2}$ 0.43067655 A152675 ${\displaystyle \log _{2}5}$ 2.32192809 A020858 2 5 32 ${\displaystyle \log _{5}e}$ 0.62133493 ${\displaystyle \log 5}$ 1.60943791 A016628 ${\displaystyle e^{5}}$ 148.413 A092511 ${\displaystyle \log _{5}3}$ 0.68260619 A152914 ${\displaystyle \log _{3}5}$ 1.46497352 A113209 3 5 243 ${\displaystyle \log _{5}\pi }$ 0.71126066 ${\displaystyle \log _{\pi }5}$ 1.40595430 ${\displaystyle \pi ^{3}}$ 306.02 A092731 ${\displaystyle \log _{5}4}$ 0.86135311 A153101 ${\displaystyle \log _{4}5}$ 1.16096404 A153201 4 5 1024 ${\displaystyle \log _{5}5}$ 1.00000000 5 5 3125 ${\displaystyle \log _{5}6}$ 1.11328275 A153461 ${\displaystyle \log _{6}5}$ 0.89824440 A153202 6 5 7776 ${\displaystyle \log _{5}7}$ 1.20906195 A153616 ${\displaystyle \log _{7}5}$ 0.82708747 A153203 7 5 16807 ${\displaystyle \log _{5}8}$ 1.29202967 A153739 ${\displaystyle \log _{8}5}$ 0.77397603 A153204 8 5 32768 ${\displaystyle \log _{5}9}$ 1.36521238 A154008 ${\displaystyle \log _{9}5}$ 0.73248676 A153205 9 5 59049 ${\displaystyle \log _{5}10}$ 1.43067655 A154156 ${\displaystyle \log _{10}5}$ 0.69897000 A153268 10 5 100000

(See A000584 for the fifth powers of integers).

## Values for number theoretic functions with 5 as an argument

 ${\displaystyle \mu (5)}$ –1 ${\displaystyle M(5)}$ –2 ${\displaystyle \pi (5)}$ 3 ${\displaystyle \sigma _{1}(5)}$ 6 ${\displaystyle \sigma _{0}(5)}$ 2 ${\displaystyle \phi (5)}$ 4 ${\displaystyle \Omega (5)}$ 1 ${\displaystyle \omega (5)}$ 1 ${\displaystyle \lambda (5)}$ 4 This is the Carmichael lambda function. ${\displaystyle \lambda (5)}$ –1 This is the Liouville lambda function. ${\displaystyle \zeta (5)}$ 1.036927755143369926331... (see A013663). 5! 120 ${\displaystyle \Gamma (5)}$ 24

## Factorization of some small integers in a quadratic integer ring adjoining the square roots of −5, 5

${\displaystyle \mathbb {Z} [{\sqrt {5}}]}$ has quite a few things in common with ${\displaystyle \mathbb {Z} [{\sqrt {-3}}]}$. Both are actually subdomains because neither is integrally closed. Since ${\displaystyle 5\equiv 1\mod 4}$ (and likewise for –3), it is not enough to consider algebraic integers of the form ${\displaystyle a+b{\sqrt {5}}}$ (with ${\displaystyle a,b\in \mathbb {Z} }$); we must also consider algebraic integers of the form ${\displaystyle {\frac {a}{2}}+{\frac {b{\sqrt {5}}}{2}}}$ (with ${\displaystyle ab\equiv 1\mod 2}$).

As it turns out, ${\displaystyle {\frac {1}{2}}+{\frac {\sqrt {5}}{2}}}$ is an algebraic integer in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {5}})}}$. If this number looks familiar, that's probably because it's the golden ratio, ${\displaystyle \phi }$. This means that ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {5}})}=\mathbb {Z} [\phi ]}$, just as ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-3}})}=\mathbb {Z} [\omega ]}$. And furthermore, just as ${\displaystyle \omega }$ is a unit in ${\displaystyle \mathbb {Z} [\omega ]}$, so is ${\displaystyle \phi }$ a unit in ${\displaystyle \mathbb {Z} [\phi ]}$, and both are unique factorization domains.

But then we come to a contrast: ${\displaystyle \mathbb {Z} [\omega ]}$ has just six units, but all the infinitely many powers of ${\displaystyle \phi }$ with integer exponents are units in ${\displaystyle \mathbb {Z} [\phi ]}$ (and that includes ${\displaystyle \phi ^{2}=\phi +1}$). So, where a given number may seem to have two distinct factorizations in ${\displaystyle \mathbb {Z} [\phi ]}$, these can be shown to be the same by multiplying or dividing by the appropriate power of ${\displaystyle \phi }$.

${\displaystyle \mathbb {Z} [\phi ]}$ also contrasts with ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$, which is not a UFD but has only two units: 1 and –1. This allows us to be very certain when asserting that a given number has two distinct factorizations in ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$.

In the table below, some factorizations in ${\displaystyle \mathbb {Z} [\phi ]}$ will also be expressed using ${\displaystyle {\sqrt {5}}}$; this does not constitute a distinct factorization since ${\displaystyle \mathbb {Z} [\phi ]}$ has unique factorization and ${\displaystyle {\sqrt {5}}=-1+2\phi }$. This is done for the sake of clarification, to facilitate comparison to other domains not usually expressed in terms of multiples of a unit that has both a rational and an irrational part.

 ${\displaystyle n}$ ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ ${\displaystyle \mathbb {Z} [\phi ]}$ 1 Unit 2 Irreducible Prime 3 4 2 2 5 ${\displaystyle (-1)({\sqrt {-5}})^{2}}$ ${\displaystyle (-1+2\phi )^{2}}$ 6 2 × 3 OR ${\displaystyle (1-{\sqrt {-5}})(1+{\sqrt {-5}})}$ 2 × 3 7 Irreducible Prime 8 2 3 9 3 2 OR ${\displaystyle (2-{\sqrt {-5}})(2+{\sqrt {-5}})}$[1] 3 2 10 ${\displaystyle (-1)2({\sqrt {-5}})^{2}}$ ${\displaystyle 2(-1+2\phi )^{2}}$ 11 Prime ${\displaystyle (4-\phi )(3+\phi )}$ 12 2 2 × 3 13 Prime 14 2 × 7 OR ${\displaystyle (3-{\sqrt {-5}})(3+{\sqrt {-5}})}$ 2 × 7 15 ${\displaystyle (-1)3({\sqrt {-5}})^{2}}$ ${\displaystyle 3(-1+2\phi )^{2}}$ 16 2 4 17 Prime 18 2 × 3 2 19 Prime ${\displaystyle (5-\phi )(4+\phi )}$ 20 ${\displaystyle (-1)2^{2}({\sqrt {-5}})^{2}}$ ${\displaystyle 2^{2}(-1+2\phi )^{2}}$

Of course nothing actually forbids us from regarding ${\displaystyle \mathbb {Z} [{\sqrt {5}}]}$ as a domain in its own right, even though it's not integrally closed. As it turns out, it does not have unique factorization, since, for example, ${\displaystyle 4=2^{2}=(-1)(1-{\sqrt {5}})(1+{\sqrt {5}})}$. This fact may seem incompatible with ${\displaystyle \mathbb {Z} [\phi ]}$ being a UFD, but there is no conflict: the latter factorization is equivalent to ${\displaystyle (-1)(2-2\phi )(2\phi )}$ which can be rewritten as ${\displaystyle (-1)(2^{2}\phi -2^{2}\phi ^{2})}$ which leads to ${\displaystyle (-1)2^{2}\phi }$ before finally boiling down to 2 2 after all the units are stripped away.

It is worth noting that ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ has class number 2. What this means is that a number may have more than one distinct factorization into irreducibles, but in every case each of those factorizations will have the same amount of irreducibles. For example, ${\displaystyle 21=3\times 7=(4-{\sqrt {-5}})(4+{\sqrt {-5}})}$, wherein we see that both factorizations consist of two irreducibles each.

Now, the number 30 may seem like a counterexample, given that ${\displaystyle (-1)\times 2\times 3\times ({\sqrt {-5}})^{2}=(5-{\sqrt {-5}})(5+{\sqrt {-5}})=30}$. But that's kind of like saying that 5 × 6 is a distinct factorization of 30, because neither 5 nor 6 is irreducible in ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$. So, in order to prove that ${\displaystyle (5-{\sqrt {-5}})(5+{\sqrt {-5}})}$ is not a distinct factorization of 30 consisting of only two irreducibles rather than four, we need to find two numbers with a norm of 5 and another two with a norm of 6. Naturally this leads us to find ${\displaystyle {\sqrt {-5}}}$ and ${\displaystyle (1-{\sqrt {-5}})}$, and we verify that ${\displaystyle ({\sqrt {-5}})(1-{\sqrt {-5}})=(5+{\sqrt {-5}})}$.

Therefore, the two distinct factorizations of 30 are ${\displaystyle (-1)\times 2\times 3\times ({\sqrt {-5}})^{2}}$ and ${\displaystyle ({\sqrt {-5}})^{2}(1-{\sqrt {-5}})(1+{\sqrt {-5}})}$. In one of the factorizations we have felt it necessary to include the unit ${\displaystyle -1}$, though we could certainly avoid this, if we so wished, by "shopping" the associates. And if in the other factorization we were so inclined, we could prefix "${\displaystyle 1\times }$". But aside from this fiddling with the units, we essentially have two distinct factorizations each nevertheless consisting of four irreducibles.

It should be quite clear at this point that ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ is not a Euclidean domain, since it's not a unique factorization domain. Of course that doesn't mean that the Euclidean algorithm will always fail to find the greatest common divisor of a given pair of numbers. Presumably if both ${\displaystyle a}$ and ${\displaystyle b}$ are purely real integers the Euclidean algorithm will still find ${\displaystyle \gcd(a,b)}$.

What if ${\displaystyle a}$ is a purely real integer but ${\displaystyle b}$ is complex? Let's try ${\displaystyle \gcd(2,1+{\sqrt {-5}})}$. Since ${\displaystyle N(2)=4}$ and ${\displaystyle N(1+{\sqrt {-5}})=6}$, perhaps we can find numbers ${\displaystyle q}$ and ${\displaystyle r}$ in this domain such that ${\displaystyle 1+{\sqrt {-5}}=2q+r}$, with ${\displaystyle q\neq 0}$ and ${\displaystyle N(r)<4}$.

However, the four multiples of 2 nearest to ${\displaystyle 1+{\sqrt {-5}}}$ all give ${\displaystyle r}$ such that ${\displaystyle N(r)=6}$, as the following table shows:

 ${\displaystyle 2q}$ ${\displaystyle r}$ ${\displaystyle N(r)}$ 2 ${\displaystyle -1+{\sqrt {-5}}}$ 6 ${\displaystyle 2+2{\sqrt {-5}}}$ ${\displaystyle -1-{\sqrt {-5}}}$ 6 ${\displaystyle 2{\sqrt {-5}}}$ ${\displaystyle 1-{\sqrt {-5}}}$ 6 0 ${\displaystyle 1+{\sqrt {-5}}}$ 6

The intuition here is that as we pick multiples of 2 that are further away from ${\displaystyle 1+{\sqrt {-5}}}$, the norm of the remainder will be increasingly larger.

But by using the basic property of parity, we need not rely on an intuition. If ${\displaystyle 2q=m+n{\sqrt {-5}}}$, with ${\displaystyle m,n\in \mathbb {Z} }$, it is obvious that both ${\displaystyle m}$ and ${\displaystyle n}$ are even. Then we have ${\displaystyle r=(1+{\sqrt {-5}})-2q=(-m+1)+(-n+1){\sqrt {-5}}}$.

Since ${\displaystyle (-m+1)\neq 0}$ and ${\displaystyle (-n+1)\neq 0}$ either, it follows that 6 is the minimum norm of ${\displaystyle r}$. Hence the Euclidean algorithm can't proceed because it can't even start.

## Factorization of 5 in some quadratic integer rings

In ${\displaystyle \mathbb {Z} }$, 5 is a prime number. But it has different factorizations in some quadratic integer rings.

 ${\displaystyle \mathbb {Z} [i]}$ ${\displaystyle (2-i)(2+i)}$ ${\displaystyle \mathbb {Z} [{\sqrt {-2}}]}$ Prime ${\displaystyle \mathbb {Z} [{\sqrt {2}}]}$ Prime ${\displaystyle \mathbb {Z} [\omega ]}$ ${\displaystyle \mathbb {Z} [{\sqrt {3}}]}$ ${\displaystyle \mathbb {Z} [{\sqrt {-5}}]}$ ${\displaystyle (-1)({\sqrt {-5}})^{2}}$ ${\displaystyle \mathbb {Z} [\phi ]}$ ${\displaystyle (-1+2\phi )^{2}}$ ${\displaystyle \mathbb {Z} [{\sqrt {-6}}]}$ Irreducible ${\displaystyle \mathbb {Z} [{\sqrt {6}}]}$ ${\displaystyle (-1)(1-{\sqrt {6}})(1+{\sqrt {6}})}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-7}})}}$ Prime ${\displaystyle \mathbb {Z} [{\sqrt {7}}]}$ Prime ${\displaystyle \mathbb {Z} [{\sqrt {-10}}]}$ Irreducible ${\displaystyle \mathbb {Z} [{\sqrt {10}}]}$ Irreducible ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-11}})}}$ ${\displaystyle \left({\frac {3}{2}}-{\frac {\sqrt {-11}}{2}}\right)\left({\frac {3}{2}}+{\frac {\sqrt {-11}}{2}}\right)}$ ${\displaystyle \mathbb {Z} [{\sqrt {11}}]}$ ${\displaystyle (4-{\sqrt {11}})(4+{\sqrt {11}})}$ ${\displaystyle \mathbb {Z} [{\sqrt {-13}}]}$ Irreducible ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {13}})}}$ Prime ${\displaystyle \mathbb {Z} [{\sqrt {-14}}]}$ ${\displaystyle \mathbb {Z} [{\sqrt {14}}]}$ ${\displaystyle (-1)(3-{\sqrt {14}})(3+{\sqrt {14}})}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}}$ ${\displaystyle \mathbb {Z} [{\sqrt {15}}]}$ Irreducible ${\displaystyle \mathbb {Z} [{\sqrt {-17}}]}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {17}})}}$ Prime ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-19}})}}$ ${\displaystyle \left({\frac {1}{2}}-{\frac {\sqrt {-19}}{2}}\right)\left({\frac {1}{2}}+{\frac {\sqrt {-19}}{2}}\right)}$ ${\displaystyle \mathbb {Z} [{\sqrt {19}}]}$ ${\displaystyle (9-2{\sqrt {19}})(9+2{\sqrt {19}})}$.

We can say with certainty that 5 is prime in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-43}})}}$, ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-67}})}}$ and ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {-163}})}}$. For any other imaginary quadratic rings, we can only say it is irreducible, at least not without some examination of ideals in a specific ring.

## Representation of 5 in various bases

 Base 2 3 4 5 6 through 36 Representation 101 12 11 10 5

In the balanced ternary numeral system, 5 is {1, –1, –1}, meaning ${\displaystyle 3^{2}-3^{1}-3^{0}}$. In negabinary, 5 has the same representation as in binary. In quater-imaginary base, 10 is 10301, since ${\displaystyle (2i)^{4}+3(2i)^{2}+(2i)^{0}=16+3\times -4+1}$. In the factorial numeral system, 5 is 21, since ${\displaystyle 2\times 2!+1=5}$.

 ${\displaystyle -1}$ 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 1729