13320232400041 has 4 divisors (see below), whose sum is σ = 15223122742912. Its totient is φ = 11417342057172.

The previous prime is 13320232400039. The next prime is 13320232400063. The reversal of 13320232400041 is 14000423202331.

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 13320232400041 - 2¹ = 13320232400039 is a prime.

It is a super-2 number, since 2×13320232400041² (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13320232399979 and 13320232400015.

It is not an unprimeable number, because it can be changed into a prime (13320232404041) 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, 951445171425 + ... + 951445171438.

It is an arithmetic number, because the mean of its divisors is an integer number (3805780685728).

Almost surely, 2^{13320232400041} is an apocalyptic number.

It is an amenable number.

13320232400041 is a deficient number, since it is larger than the sum of its proper divisors (1902890342871).

13320232400041 is an equidigital number, since it uses as much as digits as its factorization.

13320232400041 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1902890342870.

The product of its (nonzero) digits is 3456, while the sum is 25.

Adding to 13320232400041 its reverse (14000423202331), we get a palindrome (27320655602372).

The spelling of 13320232400041 in words is "thirteen trillion, three hundred twenty billion, two hundred thirty-two million, four hundred thousand, forty-one".