13001110212113 has 4 divisors (see below), whose sum is σ = 13100355328368. Its totient is φ = 12901865095860.

The previous prime is 13001110212107. The next prime is 13001110212121. The reversal of 13001110212113 is 31121201110031.

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 a cyclic number.

It is not a de Polignac number, because 13001110212113 - 2²⁴ = 13001093434897 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13001110212091 and 13001110212100.

It is not an unprimeable number, because it can be changed into a prime (13001110212143) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 49622557931 + ... + 49622558192.

It is an arithmetic number, because the mean of its divisors is an integer number (3275088832092).

Almost surely, 2^{13001110212113} is an apocalyptic number.

It is an amenable number.

13001110212113 is a deficient number, since it is larger than the sum of its proper divisors (99245116255).

13001110212113 is an equidigital number, since it uses as much as digits as its factorization.

13001110212113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 99245116254.

The product of its (nonzero) digits is 36, while the sum is 17.

Adding to 13001110212113 its reverse (31121201110031), we get a palindrome (44122311322144).

The spelling of 13001110212113 in words is "thirteen trillion, one billion, one hundred ten million, two hundred twelve thousand, one hundred thirteen".