1801

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Number

$1801$ (one thousand, eight hundred and one) is:

The $279$th prime number


The $4$th prime number after $1009$, $1129$, $1201$ which can be expressed in the form $x^2 + n y^2$ for all values of $n$ from $1$ to $10$:
\(\ds 1801\) \(=\) \(\ds 35^2 + 1 \times 24^2\)
\(\ds \) \(=\) \(\ds 1^2 + 2 \times 30^2\)
\(\ds \) \(=\) \(\ds 37^2 + 3 \times 12^2\)
\(\ds \) \(=\) \(\ds 35^2 + 4 \times 12^2\)
\(\ds \) \(=\) \(\ds 26^2 + 5 \times 15^2\)
\(\ds \) \(=\) \(\ds 25^2 + 6 \times 14^2\)
\(\ds \) \(=\) \(\ds 3^2 + 7 \times 16^2\)
\(\ds \) \(=\) \(\ds 1^2 + 8 \times 15^2\)
\(\ds \) \(=\) \(\ds 35^2 + 9 \times 8^2\)
\(\ds \) \(=\) \(\ds 19^2 + 10 \times 12^2\)


The $25$th centered hexagonal number after $1$, $7$, $19$, $37$, $61$, $91$, $127$, $\ldots$, $817$, $919$, $1027$, $1141$, $1261$, $1387$, $1519$, $1657$:
$1801 = \ds 1 + \sum_{k \mathop = 1}^{25 - 1} 6 k = 25^3 - 24^3$


Also see