13010150113 has 4 divisors (see below), whose sum is σ = 13109464368. Its totient is φ = 12910835860.

The previous prime is 13010150093. The next prime is 13010150147. The reversal of 13010150113 is 31105101031.

13010150113 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: 31105101031 = 23 ⋅1352395697.

It is a cyclic number.

It is not a de Polignac number, because 13010150113 - 2⁵ = 13010150081 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (13010150183) 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, 49656931 + ... + 49657192.

It is an arithmetic number, because the mean of its divisors is an integer number (3277366092).

Almost surely, 2^13010150113 is an apocalyptic number.

It is an amenable number.

13010150113 is a deficient number, since it is larger than the sum of its proper divisors (99314255).

13010150113 is an equidigital number, since it uses as much as digits as its factorization.

13010150113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 99314254.

The product of its (nonzero) digits is 45, while the sum is 16.

Adding to 13010150113 its reverse (31105101031), we get a palindrome (44115251144).

The spelling of 13010150113 in words is "thirteen billion, ten million, one hundred fifty thousand, one hundred thirteen".