3113401231 has 4 divisors (see below), whose sum is σ = 3113516232. Its totient is φ = 3113286232.

The previous prime is 3113401229. The next prime is 3113401253. The reversal of 3113401231 is 1321043113.

It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length.

It is a cyclic number.

It is not a de Polignac number, because 3113401231 - 2¹ = 3113401229 is a prime.

It is a super-3 number, since 3×3113401231³ (a number of 29 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 3113401199 and 3113401208.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (3113401211) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 7920 + ... + 79306.

It is an arithmetic number, because the mean of its divisors is an integer number (778379058).

Almost surely, 2^3113401231 is an apocalyptic number.

3113401231 is a deficient number, since it is larger than the sum of its proper divisors (115001).

3113401231 is an equidigital number, since it uses as much as digits as its factorization.

3113401231 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 115000.

The product of its (nonzero) digits is 216, while the sum is 19.

The square root of 3113401231 is about 55797.8604518130. The cubic root of 3113401231 is about 1460.1978262932.

Adding to 3113401231 its reverse (1321043113), we get a palindrome (4434444344).

The spelling of 3113401231 in words is "three billion, one hundred thirteen million, four hundred one thousand, two hundred thirty-one".