1340131031231 has 2 divisors, whose sum is σ = 1340131031232. Its totient is φ = 1340131031230.

The previous prime is 1340131031203. The next prime is 1340131031257. The reversal of 1340131031231 is 1321301310431.

It is a strong prime.

It is an emirp because it is prime and its reverse (1321301310431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1340131031231 - 2³⁰ = 1339057289407 is a prime.

It is a super-2 number, since 2×1340131031231² (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1340131031197 and 1340131031206.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1340131031131) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670065515615 + 670065515616.

It is an arithmetic number, because the mean of its divisors is an integer number (670065515616).

Almost surely, 2^{1340131031231} is an apocalyptic number.

1340131031231 is a deficient number, since it is larger than the sum of its proper divisors (1).

1340131031231 is an equidigital number, since it uses as much as digits as its factorization.

1340131031231 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 648, while the sum is 23.

Adding to 1340131031231 its reverse (1321301310431), we get a palindrome (2661432341662).

The spelling of 1340131031231 in words is "one trillion, three hundred forty billion, one hundred thirty-one million, thirty-one thousand, two hundred thirty-one".