14144013311433 has 4 divisors (see below), whose sum is σ = 18858684415248. Its totient is φ = 9429342207620.

The previous prime is 14144013311423. The next prime is 14144013311471. The reversal of 14144013311433 is 33411331044141.

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 14144013311433 - 2¹⁶ = 14144013245897 is a prime.

It is a super-2 number, since 2×14144013311433² (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 14144013311394 and 14144013311403.

It is not an unprimeable number, because it can be changed into a prime (14144013311423) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2357335551903 + ... + 2357335551908.

It is an arithmetic number, because the mean of its divisors is an integer number (4714671103812).

Almost surely, 2^{14144013311433} is an apocalyptic number.

It is an amenable number.

14144013311433 is a deficient number, since it is larger than the sum of its proper divisors (4714671103815).

14144013311433 is an equidigital number, since it uses as much as digits as its factorization.

14144013311433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4714671103814.

The product of its (nonzero) digits is 20736, while the sum is 33.

Adding to 14144013311433 its reverse (33411331044141), we get a palindrome (47555344355574).

The spelling of 14144013311433 in words is "fourteen trillion, one hundred forty-four billion, thirteen million, three hundred eleven thousand, four hundred thirty-three".