19

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19 is a prime number. Any positive integer can be expressed as a sum of 19 or fewer fourth powers.

Membership in core sequences

Odd numbers	..., 13, 15, 17, 19, 21, 23, 25, ...	A005408
Prime numbers	..., 11, 13, 17, 19, 23, 29, 31, ...	A000040
Squarefree numbers	..., 14, 15, 17, 19, 21, 22, 23, ...	A005117
Mersenne exponents	2, 3, 5, 7, 13, 17, 19, 31, 61, ...	A000043
Central trinomial coefficients	1, 1, 3, 7, 19, 51, 141, 393, ...	A002426

In Pascal's triangle, 19 occurs twice. (In Lozanić's triangle, 19 occurs four times).

Sequences pertaining to 19

Multiples of 19	0, 19, 38, 57, 76, 95, 114, 133, 152, 171, 190, 209, 228, ...	A008601
19-gonal numbers	1, 19, 54, 106, 175, 261, 364, 484, 621, 775, 946, 1134, ...	A051871
$3x+1$ sequence starting at 33	33, 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, 17, ...	A008880
$3x-1$ sequence starting at 36	36, 18, 9, 26, 13, 38, 19, 56, 28, 14, 7, 20, 10, 5, 14, 7, 20, ...	A008894
19-rough numbers	1, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, ...	A166061

Partitions of 19

There are 490 partitions of 19.

The Goldbach representations of 19 are 2 + 17 = 3 + 5 + 11 = 19 (there's also 3 + 3 + 13 = 5 + 7 + 7 if it's not required that the primes be distinct).

Roots and powers of 19

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

$\sqrt{19}$	4.35889894	A010475	19 2	361
$\sqrt[3]{19}$	2.66840164	A010591	19 3	6859
$\sqrt[4]{19}$	2.08779762	A011015	19 4	130321
$\sqrt[5]{19}$	1.80198312	A011104	19 5	2476099
$\sqrt[6]{19}$	1.63352430	A011410	19 6	47045881
$\sqrt[7]{19}$	1.52292699	A011411	19 7	893871739
$\sqrt[8]{19}$	1.44492132	A011412	19 8	16983563041
$\sqrt[9]{19}$	1.38702322	A011413	19 9	322687697779
$\sqrt[10]{19}$	1.34237965	A011414	19 10	6131066257801
				A001029

Logarithms and nineteenth powers

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

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

TABLE GOES HERE

Values for number theoretic functions with 19 as an argument

$\mu (19)$	–1
$M(19)$	–3
$\pi (19)$	8
${\sigma }_1(19)$	20
${\sigma }_0(19)$	2
$\varphi (19)$	18
$\mathrm{\Omega }(19)$	1
$\omega (19)$	1
$\lambda (19)$	18	This is the Carmichael lambda function.
$\lambda (19)$	–1	This is the Liouville lambda function.
$\zeta (19)$	1.0000019...	See A013677.
19!	121645100408832000
$\mathrm{\Gamma }(19)$	6402373705728000

Factorization of some small integers in a quadratic integer ring adjoining the square roots of −19, 19

Both ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$ and $\mathbb{\mathbb{Z}}[\sqrt{19}]$ are unique factorization domains. There is something quite interesting about the way in which ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$ is a UFD. But first, we'll look at $\mathbb{\mathbb{Z}}[\sqrt{19}]$. In that domain, the norm function is $N(a+b\sqrt{19})=a^2-19b^2$.

Theorem Z19EUC. The domain $\mathbb{\mathbb{Z}}[\sqrt{19}]$ is a Euclidean domain in which the absolute value of the norm is a suitable Euclidean function. Given any two nonzero numbers $m,n\in \mathbb{\mathbb{Z}}[\sqrt{19}]$, it is always possible to find two other numbers $q,r\in \mathbb{\mathbb{Z}}[\sqrt{19}]$ such that $n=qm+r$, $q\ne 0$ and $0\le |N(r)|<|N(m)|$.

If this is true, then the Euclidean GCD algorithm can be implemented in this domain. Whatever two nonzero numbers in the domain are input into the algorithm, a sequence of quotients and remainders is produced, and the absolute values of the norms of the remainders form a decreasing sequence. The Euclidean algorithm may or may not always produce optimal results, but it should always produce an answer in a finite number of steps.

Proof. If $m$ is a divisor of $n$, or the other way around, then $q=\frac{n}{m}$ or $\frac{m}{n}$ as needed, and $r=0$. If this is not the case, it does not automatically mean $m$ and $n$ are coprime, but it does mean that $\frac{n}{m}\in ̸\mathbb{\mathbb{Z}}[\sqrt{19}]$ but $\frac{n}{m}\in \mathbb{\mathbb{Q}}(\sqrt{19})$. Notate $n=a+b\sqrt{19}$ and $m=c+d\sqrt{19}$. The we have:

$\frac{a+b\sqrt{19}}{c+d\sqrt{19}}=\frac{(a+b\sqrt{19})(c+d\sqrt{19})}{c^2-19d^2}=s+t\sqrt{19}$

where $s,t\in \mathbb{\mathbb{Q}}(\sqrt{19})$. Now choose $u,v\in \mathbb{\mathbb{Z}}$ such that $|s-u|,|t-v|\le \frac{1}{2}$ and then set $q=u+v\sqrt{19}$ and $r=n-qm$. Since the norm is multiplicative, it follows that $N(r)=N(m)N((s-u)+(t-v)\sqrt{19})$ and therefore

???$-\frac{3}{4}N(m)<N(r)<\frac{3}{4}N(m)$????

as specified by the theorem. ENDOFPROOFMARK

We then say that $\mathbb{\mathbb{Z}}[\sqrt{19}]$ is norm-Euclidean, and from this it automatically follows that it is a principal ideal domain and therefore a unique factorization domain. ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$, on the other hand, is a principal ideal domain but not norm-Euclidean, and a single example example suffices to demonstrate that it is not norm-Euclidean.

Given $m=\frac{3}{2}+\frac{\sqrt{-19}}{2}$ and $n=10$, can we find numbers $q,r\in {{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$ such that $n=qm+r$, $q\ne 0$ and $0\le N(r)<N(m)$? In this domain, the norm is never negative, and since the numbers are spread out throughout the complex plane rather than crammed on the real number line, it is easy to make diagrams to help us visualize these things. In the following diagram, 0 is represented by a black dot, the nonzero numbers of $⟨\frac{3}{2}+\frac{\sqrt{-19}}{2}⟩$ by green dots, and the nonzero numbers of $⟨10⟩$ by red dots.

As you can see, 10 is a red dot that almost smack dab in the middle of a lozenge-like shape formed by green dots. It stands to reason that, as values of $qm$, those green dots closest to 10 correspond to smaller $r$ than green dots farther away. We need to find just one dot such that $N(r)<7$.

$qm$	$r$	$N(r)$
7	3	9
$\frac{17}{2}+\frac{\sqrt{-19}}{2}$	$\frac{3}{2}+\frac{\sqrt{-19}}{2}$	7
$10+\sqrt{-19}$	$-\sqrt{-19}$	19
$11+\sqrt{-19}$	$-1-\sqrt{-19}$	20
$\frac{25}{2}+\frac{\sqrt{-19}}{2}$	$-\frac{5}{2}-\frac{\sqrt{-19}}{2}$	11
14	–4	16

It's not necessary to continue this with $\frac{25}{2}-\frac{\sqrt{-19}}{2}$ and so on to see that there is no value of $r$ suitable for the Euclidean algorithm to be found.

We do need a theorem to prove that ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$ is not Euclidean for any suitable function whatsoever, and it becomes a little more involved to show that it is a principal ideal domain. But we will not include those theorems here at this time.

Of course the Euclidean algorithm is not the only way to find GCDs. From prime factorizaton alone it would be obvious that $\gcd (\frac{3}{2}+\frac{\sqrt{-19}}{2},10)=1$. Now follows a table of factorization of some small integers in ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$ and $\mathbb{\mathbb{Z}}[\sqrt{19}]$.

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

Since both of these domains are UFDs, looking at the factorizations of ideals does not give us much insight. The factorization of an ideal is pretty much the same as that of the generating number. Often it's just a matter of changing parentheses to angle brackets.

However, there are a few subtleties to consider, which concern 2 and 19 (and, of course, also their negative counterparts).

First, although $2=(-1)(13-3\sqrt{19})(13+3\sqrt{19})$ (as noted above), the ideal $⟨2⟩$ in $\mathbb{\mathbb{Z}}[\sqrt{19}]$ ramifies, since $13-3\sqrt{19}$ and $13+3\sqrt{19}$ are associated by multiplication by a unit. And so, $⟨2⟩=⟨13+3\sqrt{19}{⟩}^2$.

In ${{𝒪}}_{\mathbb{\mathbb{Q}}(\sqrt{-19})}$, 2 is prime, as is the ideal generated by it. The equation $x^2+76y^2=2^z$ has no solutions in nonzero integers. This means that if a number in this domain has nonzero imaginary part and an even norm, then that norm is divisible by an odd number greater than 1. For example, $1+\sqrt{-19}$ and $3+\sqrt{-19}$ have norms that are divisible by 5 and by 7, respectively.

Lastly, note that although $(\sqrt{-19}{)}^2=-19$, as ideals go, we still have $⟨19⟩=⟨\sqrt{-19}{⟩}^2$.

Factorization of 19 in some quadratic integer rings

As was mentioned above, 19 is a prime number in $\mathbb{\mathbb{Z}}$. But it is composite in some quadratic integer rings.

TABLE GOES HERE

Representation of 19 in various bases

Base	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20 through 36
Representation	10011	201	103	34	31	25	23	21	19	18	17	16	15	14	13	12	11	10	J

See also

Some integers

							$-1$
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