14040123525121 has 4 divisors (see below), whose sum is σ = 14040343574272. Its totient is φ = 14039903475972.

The previous prime is 14040123525061. The next prime is 14040123525139. The reversal of 14040123525121 is 12152532104041.

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 14040123525121 - 2¹¹ = 14040123523073 is a prime.

It is a super-2 number, since 2×14040123525121² (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (14040123525191) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 109928841 + ... + 110056486.

It is an arithmetic number, because the mean of its divisors is an integer number (3510085893568).

Almost surely, 2^{14040123525121} is an apocalyptic number.

It is an amenable number.

14040123525121 is a deficient number, since it is larger than the sum of its proper divisors (220049151).

14040123525121 is an equidigital number, since it uses as much as digits as its factorization.

14040123525121 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 220049150.

The product of its (nonzero) digits is 9600, while the sum is 31.

Adding to 14040123525121 its reverse (12152532104041), we get a palindrome (26192655629162).

The spelling of 14040123525121 in words is "fourteen trillion, forty billion, one hundred twenty-three million, five hundred twenty-five thousand, one hundred twenty-one".