13110111214113 has 4 divisors (see below), whose sum is σ = 17480148285488. Its totient is φ = 8740074142740.

The previous prime is 13110111214111. The next prime is 13110111214117. The reversal of 13110111214113 is 31141211101131.

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 13110111214113 - 2¹ = 13110111214111 is a prime.

It is a super-3 number, since 3×13110111214113³ (a number of 40 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 self number, because there is not a number n which added to its sum of digits gives 13110111214113.

It is not an unprimeable number, because it can be changed into a prime (13110111214111) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2185018535683 + ... + 2185018535688.

It is an arithmetic number, because the mean of its divisors is an integer number (4370037071372).

Almost surely, 2^{13110111214113} is an apocalyptic number.

It is an amenable number.

13110111214113 is a deficient number, since it is larger than the sum of its proper divisors (4370037071375).

13110111214113 is an equidigital number, since it uses as much as digits as its factorization.

13110111214113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 4370037071374.

The product of its (nonzero) digits is 72, while the sum is 21.

Adding to 13110111214113 its reverse (31141211101131), we get a palindrome (44251322315244).

The spelling of 13110111214113 in words is "thirteen trillion, one hundred ten billion, one hundred eleven million, two hundred fourteen thousand, one hundred thirteen".