3211140121 has 4 divisors (see below), whose sum is σ = 3669874432. Its totient is φ = 2752405812.

The previous prime is 3211140113. The next prime is 3211140143. The reversal of 3211140121 is 1210411123.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also an emirpimes, since its reverse is a distinct semiprime: 1210411123 = 941 ⋅1286303.

It is a cyclic number.

It is not a de Polignac number, because 3211140121 - 2³ = 3211140113 is a prime.

It is a super-2 number, since 2×3211140121² = 20622841753391789282, which contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 3211140095 and 3211140104.

It is not an unprimeable number, because it can be changed into a prime (3211140221) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 229367145 + ... + 229367158.

It is an arithmetic number, because the mean of its divisors is an integer number (917468608).

Almost surely, 2^3211140121 is an apocalyptic number.

It is an amenable number.

3211140121 is a deficient number, since it is larger than the sum of its proper divisors (458734311).

3211140121 is an equidigital number, since it uses as much as digits as its factorization.

3211140121 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 458734310.

The product of its (nonzero) digits is 48, while the sum is 16.

The square root of 3211140121 is about 56666.9226356964. The cubic root of 3211140121 is about 1475.3206421184.

Adding to 3211140121 its reverse (1210411123), we get a palindrome (4421551244).

The spelling of 3211140121 in words is "three billion, two hundred eleven million, one hundred forty thousand, one hundred twenty-one".