13100441113 has 2 divisors, whose sum is σ = 13100441114. Its totient is φ = 13100441112.

The previous prime is 13100441087. The next prime is 13100441129. The reversal of 13100441113 is 31114400131.

13100441113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11433597184 + 1666843929 = 106928^2 + 40827^2 .

It is an emirp because it is prime and its reverse (31114400131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13100441113 is a prime.

It is not a weakly prime, because it can be changed into another prime (13100441143) 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 as a sum of consecutive naturals, namely, 6550220556 + 6550220557.

It is an arithmetic number, because the mean of its divisors is an integer number (6550220557).

Almost surely, 2^13100441113 is an apocalyptic number.

It is an amenable number.

13100441113 is a deficient number, since it is larger than the sum of its proper divisors (1).

13100441113 is an equidigital number, since it uses as much as digits as its factorization.

13100441113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 144, while the sum is 19.

Adding to 13100441113 its reverse (31114400131), we get a palindrome (44214841244).

The spelling of 13100441113 in words is "thirteen billion, one hundred million, four hundred forty-one thousand, one hundred thirteen".