1001

Number
$1001$ (one thousand and one) is:


 * $7 \times 11 \times 13$


 * The $26$th pentagonal number after $1$, $5$, $12$, $22$, $35$, $\ldots$, $477$, $532$, $590$, $651$, $715$, $782$, $852$, $925$:
 * $1001 = \displaystyle \sum_{k \mathop = 1}^{26} \left({3 k - 2}\right) = \dfrac {26 \left({3 \times 26 - 1}\right)} 2$


 * The $51$st generalized pentagonal number after $1$, $2$, $5$, $7$, $12$, $15$, $\ldots$, $610$, $651$, $672$, $715$, $737$, $782$, $805$, $852$, $876$, $925$, $950$:
 * $1001 = \displaystyle \sum_{k \mathop = 1}^{26} \left({3 k - 2}\right) = \dfrac {26 \left({3 \times 26 - 1}\right)} 2$


 * The $11$th pentatope number after $1$, $5$, $15$, $35$, $70$, $126$, $210$, $330$, $495$, $715$:
 * $1001 = \displaystyle \sum_{k \mathop = 1}^{11} \dfrac {k \left({k + 1}\right) \left({k + 2}\right)} 6 = \dfrac {11 \left({11 + 1}\right) \left({11 + 2}\right) \left({11 + 3}\right)} {24}$


 * The $4$th pentagonal number after $1$, $5$, $22$ which is also palindromic:
 * $1001 = \displaystyle \sum_{k \mathop = 1}^{26} \left({3 k - 2}\right) = \dfrac {26 \left({3 \times 26 - 1}\right)} 2$


 * The $7$th positive integer after $1$, $2$, $7$, $11$, $101$, $111$ whose cube is palindromic:
 * $1001^3 = 1 \, 003 \, 003 \, 001$