311312431132033 has 4 divisors (see below), whose sum is σ = 311667405398740. Its totient is φ = 310957456865328.

The previous prime is 311312431132019. The next prime is 311312431132117. The reversal of 311312431132033 is 330231134213113.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 283804157934144 + 27508273197889 = 16846488^2 + 5244833^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-311312431132033 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 311312431131986 and 311312431132004.

It is not an unprimeable number, because it can be changed into a prime (311312431132333) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 177487132038 + ... + 177487133791.

It is an arithmetic number, because the mean of its divisors is an integer number (77916851349685).

Almost surely, 2^{311312431132033} is an apocalyptic number.

It is an amenable number.

311312431132033 is a deficient number, since it is larger than the sum of its proper divisors (354974266707).

311312431132033 is an equidigital number, since it uses as much as digits as its factorization.

311312431132033 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 354974266706.

The product of its (nonzero) digits is 11664, while the sum is 31.

Adding to 311312431132033 its reverse (330231134213113), we get a palindrome (641543565345146).

The spelling of 311312431132033 in words is "three hundred eleven trillion, three hundred twelve billion, four hundred thirty-one million, one hundred thirty-two thousand, thirty-three".