14011303241041 has 4 divisors (see below), whose sum is σ = 14011341049584. Its totient is φ = 14011265432500.

The previous prime is 14011303241027. The next prime is 14011303241063. The reversal of 14011303241041 is 14014230311041.

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 14011303241041 - 2¹¹ = 14011303238993 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 14011303240997 and 14011303241015.

It is not an unprimeable number, because it can be changed into a prime (14011303241011) 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, 18342835 + ... + 19091416.

It is an arithmetic number, because the mean of its divisors is an integer number (3502835262396).

Almost surely, 2^{14011303241041} is an apocalyptic number.

It is an amenable number.

14011303241041 is a deficient number, since it is larger than the sum of its proper divisors (37808543).

14011303241041 is an equidigital number, since it uses as much as digits as its factorization.

14011303241041 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 37808542.

The product of its (nonzero) digits is 1152, while the sum is 25.

Adding to 14011303241041 its reverse (14014230311041), we get a palindrome (28025533552082).

The spelling of 14011303241041 in words is "fourteen trillion, eleven billion, three hundred three million, two hundred forty-one thousand, forty-one".