13044112113 has 4 divisors (see below), whose sum is σ = 17392149488. Its totient is φ = 8696074740.

The previous prime is 13044112111. The next prime is 13044112117. The reversal of 13044112113 is 31121144031.

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.

It is not a de Polignac number, because 13044112113 - 2¹ = 13044112111 is a prime.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 13044112113.

It is not an unprimeable number, because it can be changed into a prime (13044112111) 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, 2174018683 + ... + 2174018688.

It is an arithmetic number, because the mean of its divisors is an integer number (4348037372).

Almost surely, 2^13044112113 is an apocalyptic number.

It is an amenable number.

13044112113 is a deficient number, since it is larger than the sum of its proper divisors (4348037375).

13044112113 is an equidigital number, since it uses as much as digits as its factorization.

13044112113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 4348037374.

The product of its (nonzero) digits is 288, while the sum is 21.

Adding to 13044112113 its reverse (31121144031), we get a palindrome (44165256144).

The spelling of 13044112113 in words is "thirteen billion, forty-four million, one hundred twelve thousand, one hundred thirteen".