131001412393931 has 2 divisors, whose sum is σ = 131001412393932. Its totient is φ = 131001412393930.

The previous prime is 131001412393909. The next prime is 131001412393933. The reversal of 131001412393931 is 139393214100131.

It is a strong prime.

It is an emirp because it is prime and its reverse (139393214100131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-131001412393931 is a prime.

It is a super-2 number, since 2×131001412393931² (a number of 29 digits) contains 22 as substring.

Together with 131001412393933, 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 (131001412393933) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 65500706196965 + 65500706196966.

It is an arithmetic number, because the mean of its divisors is an integer number (65500706196966).

Almost surely, 2^{131001412393931} is an apocalyptic number.

131001412393931 is a deficient number, since it is larger than the sum of its proper divisors (1).

131001412393931 is an equidigital number, since it uses as much as digits as its factorization.

131001412393931 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 52488, while the sum is 41.

The spelling of 131001412393931 in words is "one hundred thirty-one trillion, one billion, four hundred twelve million, three hundred ninety-three thousand, nine hundred thirty-one".