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# 69

Please do not rely on any information it contains.

69 is an integer.

## Membership in core sequences

 Odd numbers ..., 63, 65, 67, 69, 71, 73, 79, ... A005843 Squarefree numbers ..., 65, 66, 67, 69, 70, 71, 73, ... A005117

In Pascal's triangle, 69 occurs twice.

## Sequences pertaining to 69

 Multiples of 69 69, 138, 207, 276, 345, 414, 483, 552, 621, 690, 759, 828, ... 69-gonal numbers 1, 69, 204, 406, 675, 1011, 1414, 1884, 2421, 3025, 3696, ...

## Partitions of 69

There are 3554345 partitions of 69.

## Roots and powers of 69

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

 ${\displaystyle {\sqrt {69}}}$ 8.30662386 A010521 69 2 4761 ${\displaystyle {\sqrt[{3}]{69}}}$ 4.10156592 A010639 69 3 328509 ${\displaystyle {\sqrt[{4}]{69}}}$ 2.88212141 A011061 69 4 22667121 ${\displaystyle {\sqrt[{5}]{69}}}$ 2.33222162 A011154 69 5 1564031349 ${\displaystyle {\sqrt[{6}]{69}}}$ 2.02523231 69 6 107918163081 ${\displaystyle {\sqrt[{7}]{69}}}$ 1.83101847 69 7 7446353252589 ${\displaystyle {\sqrt[{8}]{69}}}$ 1.69768118 69 8 513798374428641 ${\displaystyle {\sqrt[{9}]{69}}}$ 1.60072440 69 9 35452087835576229 ${\displaystyle {\sqrt[{10}]{69}}}$ 1.52716129 69 10 2446194060654759801

## Logarithms and 69th 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.

PLACEHOLDER

## Values for number theoretic functions with 69 as an argument

 ${\displaystyle \mu (69)}$ 1 ${\displaystyle M(69)}$ –1 ${\displaystyle \pi (69)}$ 19 ${\displaystyle \sigma _{1}(69)}$ 96 ${\displaystyle \sigma _{0}(69)}$ 4 ${\displaystyle \phi (69)}$ 44 ${\displaystyle \Omega (69)}$ 2 ${\displaystyle \omega (69)}$ 2 ${\displaystyle \lambda (69)}$ 22 This is the Carmichael lambda function. ${\displaystyle \lambda (69)}$ 1 This is the Liouville lambda function. 69! 1.71122452... × 10 98 ${\displaystyle \Gamma (69)}$ 2.48003554... × 10 96

## Factorization of some small integers in a quadratic integer ring adjoining ${\displaystyle {\sqrt {-69}}}$, ${\displaystyle {\sqrt {69}}}$

The commutative quadratic integer ring with unity ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$, with units of the form ${\displaystyle \scriptstyle \pm \left({\frac {25}{2}}+{\frac {3{\sqrt {69}}}{2}}\right)^{n}\,}$ (${\displaystyle \scriptstyle n\,\in \,\mathbb {Z} \,}$), is a unique factorization domain. But it is not norm-Euclidean. If we picked numbers out of a hat to put through the Euclidean algorithm using the absolute value of the norm, we'd probably believe that the algorithm works. That is, unless the numbers we picked happened to be certain multiples of ${\displaystyle {\frac {23}{2}}+{\frac {3{\sqrt {69}}}{2}}}$ and ${\displaystyle 18+2{\sqrt {69}}}$. Let's call the former number ${\displaystyle m}$ and see if we can express the latter number as ${\displaystyle qm+r}$ such that ${\displaystyle q,r\in {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$, ${\displaystyle q\neq 0}$ and ${\displaystyle -23—or we could also say ${\displaystyle 0\leq |N(r)|<23}$.

 ${\displaystyle q}$ ${\displaystyle qm}$ ${\displaystyle r}$ ${\displaystyle N(r)}$ 1 ${\displaystyle {\frac {23}{2}}+{\frac {3{\sqrt {69}}}{2}}}$ ${\displaystyle {\frac {13}{2}}+{\frac {\sqrt {69}}{2}}}$ 25 ${\displaystyle {\frac {25}{2}}+{\frac {3{\sqrt {69}}}{2}}}$ ${\displaystyle 299+36{\sqrt {69}}}$ ${\displaystyle -281-34{\sqrt {69}}}$ −803 ${\displaystyle {\frac {25}{2}}-{\frac {3{\sqrt {69}}}{2}}}$ ${\displaystyle -{\frac {23}{2}}+{\frac {3{\sqrt {69}}}{2}}}$ ${\displaystyle {\frac {59}{2}}+{\frac {\sqrt {69}}{2}}}$ 853 2 ${\displaystyle 23+3{\sqrt {69}}}$ ${\displaystyle -5-{\sqrt {69}}}$ −44 −2 ${\displaystyle -23-3{\sqrt {69}}}$ ${\displaystyle 41+5{\sqrt {69}}}$ −44

Part of what makes this all so frustrating is that we're talking about numbers that don't seem to be that far apart on the number line. ${\displaystyle {\frac {23}{2}}+{\frac {3{\sqrt {69}}}{2}}}$ is approximately 23.959935794377 while ${\displaystyle 18+2{\sqrt {69}}}$ is approximately 34.61324772. And yet the norms of the remainders give such an impression of insurmountable distance.

Is it usually this difficult to find a suitable remainder? Let's for a moment consider a superficially similar situation: in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {73}})}}$, compute ${\displaystyle \gcd \left({\frac {23}{2}}+{\frac {3{\sqrt {73}}}{2}},18+2{\sqrt {73}}\right)}$. We're looking to solve ${\displaystyle 18+2{\sqrt {73}}=q\left({\frac {23}{2}}+{\frac {3{\sqrt {73}}}{2}}\right)+r}$ so that ${\displaystyle -32. If we try ${\displaystyle q=1}$, we get ${\displaystyle r={\frac {13}{2}}+{\frac {\sqrt {73}}{2}}}$ and ${\displaystyle N(r)=24}$. Easy.

Returning now to the vexed question of finding a suitable remainder in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$, what if we simply made another adjustment to the norm function? If ${\displaystyle p}$ is prime in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$, then ${\displaystyle {\overline {N}}(p)=|N(p)|+3\delta _{23}^{|N(p)}}$ And if ${\displaystyle n}$ is a product of primes, then ${\displaystyle {\overline {N}}(p_{1}p_{2}\ldots )={\overline {N}}(p_{1}){\overline {N}}(p_{2})\ldots }$. Essentially, if ${\displaystyle n}$ is coprime to 23, then ${\displaystyle {\overline {N}}(n)=|N(n)|}$.

The following table gives the factorizations of the integers from 2 to 22 in ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$.

 ${\displaystyle n}$ ${\displaystyle {\mathcal {O}}_{\mathbb {Q} ({\sqrt {69}})}}$ 2 Prime 3 ${\displaystyle \left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 4 2 2 5 ${\displaystyle (-1)\left({\frac {7}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {7}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 6 ${\displaystyle 2\left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 7 Prime 8 2 3 9 ${\displaystyle \left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)^{2}\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)^{2}}$ 10 ${\displaystyle (-1)2\left({\frac {7}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {7}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 11 ${\displaystyle (-1)\left({\frac {5}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {5}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 12 ${\displaystyle 2^{2}\left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 13 ${\displaystyle \left({\frac {11}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {11}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 14 2 × 7 15 ${\displaystyle (-1)\left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)\left({\frac {7}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {7}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 16 2 4 17 ${\displaystyle (-1)\left({\frac {1}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {1}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 18 ${\displaystyle 2\left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)^{2}\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)^{2}}$ 19 Prime 20 ${\displaystyle (-1)2^{2}\left({\frac {7}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {7}{2}}+{\frac {\sqrt {69}}{2}}\right)}$ 21 ${\displaystyle \left({\frac {9}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {9}{2}}+{\frac {\sqrt {69}}{2}}\right)7}$ 22 ${\displaystyle (-1)2\left({\frac {5}{2}}-{\frac {\sqrt {69}}{2}}\right)\left({\frac {5}{2}}+{\frac {\sqrt {69}}{2}}\right)}$

And so for 23, the primes that make it up each have a norm of −23, but with the specially adjusted norm function, that becomes 26. ${\displaystyle q=1}$ gives us ${\displaystyle r={\frac {13}{2}}+{\frac {\sqrt {69}}{2}}}$, which with its norm of 25 is now a suitable remainder. We can then move on to solving ${\displaystyle {\frac {23}{2}}+{\frac {3{\sqrt {69}}}{2}}=q\left({\frac {13}{2}}+{\frac {\sqrt {69}}{2}}\right)+r}$ then the of the [FINISH WRITING]

Now, if we have to know the prime factorizations of numbers before we can apply the adjusted norm function, doesn't that mean the Euclidean algorithm is hopelessly inefficient in this domain? Perhaps. But with the relatively recently proven fact that the adjusted norm function described above is a valid Euclidean function for this domain,[1] at least we know that this is a Euclidean domain after all.

In regards to Z[-69], the class number the of the [FINISH WRITING]

PLACEHOLDER

TABLE GOES HERE

## Representation of 69 in various bases

 Base 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 Representation 1000101 2220 1011 234 153 126 105 76 69 63 59 54 4D 49 45 41 3F 3C 39

REMARKS GO HERE

 ${\displaystyle -1}$ 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