131404012004113 has 4 divisors (see below), whose sum is σ = 131441652801648. Its totient is φ = 131366371206580.

The previous prime is 131404012004071. The next prime is 131404012004159. The reversal of 131404012004113 is 311400210404131.

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 a de Polignac number, because none of the positive numbers 2^k-131404012004113 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (131404019004113) 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, 18820393531 + ... + 18820400512.

It is an arithmetic number, because the mean of its divisors is an integer number (32860413200412).

Almost surely, 2^{131404012004113} is an apocalyptic number.

It is an amenable number.

131404012004113 is a deficient number, since it is larger than the sum of its proper divisors (37640797535).

131404012004113 is an equidigital number, since it uses as much as digits as its factorization.

131404012004113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 37640797534.

The product of its (nonzero) digits is 1152, while the sum is 25.

Adding to 131404012004113 its reverse (311400210404131), we get a palindrome (442804222408244).

The spelling of 131404012004113 in words is "one hundred thirty-one trillion, four hundred four billion, twelve million, four thousand, one hundred thirteen".