13044000131 has 2 divisors, whose sum is σ = 13044000132. Its totient is φ = 13044000130.

The previous prime is 13044000127. The next prime is 13044000133. The reversal of 13044000131 is 13100044031.

13044000131 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 is a cyclic number.

It is not a de Polignac number, because 13044000131 - 2² = 13044000127 is a prime.

It is a super-2 number, since 2×13044000131² (a number of 21 digits) contains 22 as substring.

Together with 13044000133, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (13044000133) 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, 6522000065 + 6522000066.

It is an arithmetic number, because the mean of its divisors is an integer number (6522000066).

Almost surely, 2^13044000131 is an apocalyptic number.

13044000131 is a deficient number, since it is larger than the sum of its proper divisors (1).

13044000131 is an equidigital number, since it uses as much as digits as its factorization.

13044000131 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 17.

Adding to 13044000131 its reverse (13100044031), we get a palindrome (26144044162).

The spelling of 13044000131 in words is "thirteen billion, forty-four million, one hundred thirty-one", and thus it is an aban number.