121

Number
$121$ (one hundred and twenty-one) is:


 * $11^2$


 * The $11$th square number after $1$, $4$, $9$, $16$, $25$, $36$, $49$, $64$, $81$, $100$:
 * $121 = 11 \times 11$


 * The $39$th semiprime:
 * $121 = 11 \times 11$


 * The second square number after $25$ of the form $n! + 1$:
 * $121 = 5! + 1 = 11^2$
 * where $!$ denotes the factorial function.


 * The $16$th powerful number after $1$, $4$, $8$, $9$, $16$, $25$, $27$, $32$, $36$, $49$, $64$, $72$, $81$, $100$, $108$


 * The $2$nd Fermat pseudoprime to base $3$ after $91$:
 * $3^{121} \equiv 3 \pmod {121}$


 * A palindromic number whose square is also palindromic:


 * The $2$nd (with $4$) of the $2$ square numbers which are $4$ less than a cube:
 * $121 = 11^2 = 5^3 - 4$


 * The $6$th integer after $0$, $1$, $2$, $4$, $8$ which is palindromic in both decimal and ternary:
 * $121_{10} = 11 \, 111_3$


 * The $9$th positive integer which cannot be expressed as the sum of a square and a prime:
 * $1$, $10$, $25$, $34$, $58$, $64$, $85$, $91$, $121$, $\ldots$


 * The only square number which is the sum of consecutive powers of a positive integer:
 * $121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4$


 * The $7$th Smith number after $4$, $22$, $27$, $58$, $85$, $94$:
 * $1 + 2 + 1 = 1 + 1 + 1 + 1 = 4$

Also see

 * Brocard's Problem
 * Square of Small-Digit Palindromic Number is Palindromic
 * Squares which are 4 Less than Cubes
 * Square Number which is Sum of Consecutive Powers
 * Integer Greater than 121 is Sum of Distinct Primes of form 4 n + 1