1311333314433 has 4 divisors (see below), whose sum is σ = 1748444419248. Its totient is φ = 874222209620.

The previous prime is 1311333314393. The next prime is 1311333314443. The reversal of 1311333314433 is 3344133331131.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 1311333314433 - 2¹⁴ = 1311333298049 is a prime.

It is a super-2 number, since 2×1311333314433² (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1311333314394 and 1311333314403.

It is not an unprimeable number, because it can be changed into a prime (1311333314443) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 218555552403 + ... + 218555552408.

It is an arithmetic number, because the mean of its divisors is an integer number (437111104812).

Almost surely, 2^{1311333314433} is an apocalyptic number.

It is an amenable number.

1311333314433 is a deficient number, since it is larger than the sum of its proper divisors (437111104815).

1311333314433 is an equidigital number, since it uses as much as digits as its factorization.

1311333314433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 437111104814.

The product of its digits is 34992, while the sum is 33.

Adding to 1311333314433 its reverse (3344133331131), we get a palindrome (4655466645564).

The spelling of 1311333314433 in words is "one trillion, three hundred eleven billion, three hundred thirty-three million, three hundred fourteen thousand, four hundred thirty-three".