130440000000001 has 4 divisors (see below), whose sum is σ = 130457283696432. Its totient is φ = 130422716303572.

The previous prime is 130439999999917. The next prime is 130440000000017. The reversal of 130440000000001 is 100000000044031.

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 130440000000001 - 2¹³ = 130439999991809 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (130440000000061) 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, 8641836895 + ... + 8641851988.

It is an arithmetic number, because the mean of its divisors is an integer number (32614320924108).

Almost surely, 2^{130440000000001} is an apocalyptic number.

It is an amenable number.

130440000000001 is a deficient number, since it is larger than the sum of its proper divisors (17283696431).

130440000000001 is an equidigital number, since it uses as much as digits as its factorization.

130440000000001 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 17283696430.

The product of its (nonzero) digits is 48, while the sum is 13.

Adding to 130440000000001 its reverse (100000000044031), we get a palindrome (230440000044032).

The spelling of 130440000000001 in words is "one hundred thirty trillion, four hundred forty billion, one", and thus it is an aban number.