311003112113 has 2 divisors, whose sum is σ = 311003112114. Its totient is φ = 311003112112.

The previous prime is 311003112089. The next prime is 311003112127. The reversal of 311003112113 is 311211300113.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 271715115169 + 39287996944 = 521263^2 + 198212^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-311003112113 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 311003112091 and 311003112100.

It is not a weakly prime, because it can be changed into another prime (311003112193) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155501556056 + 155501556057.

It is an arithmetic number, because the mean of its divisors is an integer number (155501556057).

Almost surely, 2^311003112113 is an apocalyptic number.

It is an amenable number.

311003112113 is a deficient number, since it is larger than the sum of its proper divisors (1).

311003112113 is an equidigital number, since it uses as much as digits as its factorization.

311003112113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 54, while the sum is 17.

Adding to 311003112113 its reverse (311211300113), we get a palindrome (622214412226).

The spelling of 311003112113 in words is "three hundred eleven billion, three million, one hundred twelve thousand, one hundred thirteen".