130440142031 has 2 divisors, whose sum is σ = 130440142032. Its totient is φ = 130440142030.

The previous prime is 130440142013. The next prime is 130440142033. The reversal of 130440142031 is 130241044031.

Together with previous prime (130440142013) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-130440142031 is a prime.

It is a super-2 number, since 2×130440142031² (a number of 23 digits) contains 22 as substring.

Together with 130440142033, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 130440141988 and 130440142006.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (130440142033) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65220071015 + 65220071016.

It is an arithmetic number, because the mean of its divisors is an integer number (65220071016).

Almost surely, 2^130440142031 is an apocalyptic number.

130440142031 is a deficient number, since it is larger than the sum of its proper divisors (1).

130440142031 is an equidigital number, since it uses as much as digits as its factorization.

130440142031 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1152, while the sum is 23.

Adding to 130440142031 its reverse (130241044031), we get a palindrome (260681186062).

The spelling of 130440142031 in words is "one hundred thirty billion, four hundred forty million, one hundred forty-two thousand, thirty-one".