411324153133 has 4 divisors (see below), whose sum is σ = 411364398528. Its totient is φ = 411283907740.

The previous prime is 411324153049. The next prime is 411324153143. The reversal of 411324153133 is 331351423114.

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, and also an emirpimes, since its reverse is a distinct semiprime: 331351423114 = 2 ⋅165675711557.

It is a cyclic number.

It is not a de Polignac number, because 411324153133 - 2¹³ = 411324144941 is a prime.

It is a Smith number, since the sum of its digits (31) coincides with the sum of the digits of its prime factors. Since it is squarefree, it is also a hoax number.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 411324153095 and 411324153104.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (411324153143) 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, 20107363 + ... + 20127808.

It is an arithmetic number, because the mean of its divisors is an integer number (102841099632).

Almost surely, 2^411324153133 is an apocalyptic number.

It is an amenable number.

411324153133 is a deficient number, since it is larger than the sum of its proper divisors (40245395).

411324153133 is a wasteful number, since it uses less digits than its factorization.

411324153133 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 40245394.

The product of its digits is 12960, while the sum is 31.

Adding to 411324153133 its reverse (331351423114), we get a palindrome (742675576247).

The spelling of 411324153133 in words is "four hundred eleven billion, three hundred twenty-four million, one hundred fifty-three thousand, one hundred thirty-three".