14540441113 has 4 divisors (see below), whose sum is σ = 15862299408. Its totient is φ = 13218582820.

The previous prime is 14540441101. The next prime is 14540441117. The reversal of 14540441113 is 31114404541.

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 14540441113 - 2⁵ = 14540441081 is a prime.

It is a super-2 number, since 2×14540441113² (a number of 21 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (14540441117) 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, 660929131 + ... + 660929152.

It is an arithmetic number, because the mean of its divisors is an integer number (3965574852).

Almost surely, 2^14540441113 is an apocalyptic number.

It is an amenable number.

14540441113 is a deficient number, since it is larger than the sum of its proper divisors (1321858295).

14540441113 is a wasteful number, since it uses less digits than its factorization.

14540441113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1321858294.

The product of its (nonzero) digits is 3840, while the sum is 28.

Adding to 14540441113 its reverse (31114404541), we get a palindrome (45654845654).

The spelling of 14540441113 in words is "fourteen billion, five hundred forty million, four hundred forty-one thousand, one hundred thirteen".