231522001131433 has 4 divisors (see below), whose sum is σ = 231524751569040. Its totient is φ = 231519250693828.

The previous prime is 231522001131413. The next prime is 231522001131461. The reversal of 231522001131433 is 334131100225132.

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 231522001131433 - 2²⁹ = 231521464260521 is a prime.

It is a super-3 number, since 3×231522001131433³ (a number of 44 digits) contains 333 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 231522001131395 and 231522001131404.

It is not an unprimeable number, because it can be changed into a prime (231522001131413) 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, 1375092535 + ... + 1375260892.

It is an arithmetic number, because the mean of its divisors is an integer number (57881187892260).

Almost surely, 2^{231522001131433} is an apocalyptic number.

It is an amenable number.

231522001131433 is a deficient number, since it is larger than the sum of its proper divisors (2750437607).

231522001131433 is an equidigital number, since it uses as much as digits as its factorization.

231522001131433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 2750437606.

The product of its (nonzero) digits is 12960, while the sum is 31.

Adding to 231522001131433 its reverse (334131100225132), we get a palindrome (565653101356565).

The spelling of 231522001131433 in words is "two hundred thirty-one trillion, five hundred twenty-two billion, one million, one hundred thirty-one thousand, four hundred thirty-three".