130040406113 has 4 divisors (see below), whose sum is σ = 133064601648. Its totient is φ = 127016210580.

The previous prime is 130040406109. The next prime is 130040406137. The reversal of 130040406113 is 311604040031.

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 not a de Polignac number, because 130040406113 - 2² = 130040406109 is a prime.

It is a Duffinian number.

It is a Curzon number.

It is not an unprimeable number, because it can be changed into a prime (130040406103) 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, 1512097703 + ... + 1512097788.

It is an arithmetic number, because the mean of its divisors is an integer number (33266150412).

Almost surely, 2^130040406113 is an apocalyptic number.

It is an amenable number.

130040406113 is a deficient number, since it is larger than the sum of its proper divisors (3024195535).

130040406113 is an equidigital number, since it uses as much as digits as its factorization.

130040406113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3024195534.

The product of its (nonzero) digits is 864, while the sum is 23.

Adding to 130040406113 its reverse (311604040031), we get a palindrome (441644446144).

The spelling of 130040406113 in words is "one hundred thirty billion, forty million, four hundred six thousand, one hundred thirteen".