14032412113 has 4 divisors (see below), whose sum is σ = 14038670448. Its totient is φ = 14026153780.

The previous prime is 14032412111. The next prime is 14032412131. The reversal of 14032412113 is 31121423041.

14032412113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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: 31121423041 = 263 ⋅118332407.

It is a cyclic number.

It is not a de Polignac number, because 14032412113 - 2¹ = 14032412111 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (14032412111) 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, 3125803 + ... + 3130288.

It is an arithmetic number, because the mean of its divisors is an integer number (3509667612).

Almost surely, 2^14032412113 is an apocalyptic number.

It is an amenable number.

14032412113 is a deficient number, since it is larger than the sum of its proper divisors (6258335).

14032412113 is an equidigital number, since it uses as much as digits as its factorization.

14032412113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6258334.

The product of its (nonzero) digits is 576, while the sum is 22.

Adding to 14032412113 its reverse (31121423041), we get a palindrome (45153835154).

The spelling of 14032412113 in words is "fourteen billion, thirty-two million, four hundred twelve thousand, one hundred thirteen".