13032011100113 has 2 divisors, whose sum is σ = 13032011100114. Its totient is φ = 13032011100112.

The previous prime is 13032011100077. The next prime is 13032011100151. The reversal of 13032011100113 is 31100111023031.

13032011100113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 12159922203664 + 872088896449 = 3487108^2 + 933857^2 .

It is a cyclic number.

It is not a de Polignac number, because 13032011100113 - 2⁴⁰ = 11932499472337 is a prime.

It is a super-2 number, since 2×13032011100113² (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13032011100091 and 13032011100100.

It is not a weakly prime, because it can be changed into another prime (13032011102113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6516005550056 + 6516005550057.

It is an arithmetic number, because the mean of its divisors is an integer number (6516005550057).

Almost surely, 2^{13032011100113} is an apocalyptic number.

It is an amenable number.

13032011100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

13032011100113 is an equidigital number, since it uses as much as digits as its factorization.

13032011100113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 54, while the sum is 17.

Adding to 13032011100113 its reverse (31100111023031), we get a palindrome (44132122123144).

The spelling of 13032011100113 in words is "thirteen trillion, thirty-two billion, eleven million, one hundred thousand, one hundred thirteen".