441

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Number

$441$ (four hundred and forty-one) is:

$3^2 \times 7^2$


The sum of the first $6$ cubes:
$441 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3$


The $13$th square after $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $121$, $144$, $225$ which has no more than $2$ distinct digits and does not end in $0$:
$441 = 21^2$


The $21$st square number after $1$, $4$, $9$, $16$, $25$, $36$, $\ldots$, $225$, $256$, $289$, $324$, $361$, $400$:
$441 = 21 \times 21$


The $36$th powerful number after $1$, $4$, $8$, $9$, $16$, $25$, $\ldots$, $289$, $324$, $343$, $361$, $392$, $400$, $432$:
$441 = 3^2 \times 7^2$


Also see