13111414331 has 2 divisors, whose sum is σ = 13111414332. Its totient is φ = 13111414330.

The previous prime is 13111414307. The next prime is 13111414333. The reversal of 13111414331 is 13341411131.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 13111414331 - 2¹⁰ = 13111413307 is a prime.

It is a super-2 number, since 2×13111414331² (a number of 21 digits) contains 22 as substring.

Together with 13111414333, 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 = 13111414297 and 13111414306.

It is not a weakly prime, because it can be changed into another prime (13111414333) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555707165 + 6555707166.

It is an arithmetic number, because the mean of its divisors is an integer number (6555707166).

Almost surely, 2^13111414331 is an apocalyptic number.

13111414331 is a deficient number, since it is larger than the sum of its proper divisors (1).

13111414331 is an equidigital number, since it uses as much as digits as its factorization.

13111414331 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 432, while the sum is 23.

Adding to 13111414331 its reverse (13341411131), we get a palindrome (26452825462).

The spelling of 13111414331 in words is "thirteen billion, one hundred eleven million, four hundred fourteen thousand, three hundred thirty-one".