11413141122433 has 4 divisors (see below), whose sum is σ = 11655973912320. Its totient is φ = 11170308332548.

The previous prime is 11413141122419. The next prime is 11413141122463. The reversal of 11413141122433 is 33422114131411.

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 11413141122433 - 2²⁵ = 11413107568001 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 11413141122395 and 11413141122404.

It is not an unprimeable number, because it can be changed into a prime (11413141122463) 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, 121416394873 + ... + 121416394966.

It is an arithmetic number, because the mean of its divisors is an integer number (2913993478080).

Almost surely, 2^{11413141122433} is an apocalyptic number.

It is an amenable number.

11413141122433 is a deficient number, since it is larger than the sum of its proper divisors (242832789887).

11413141122433 is an equidigital number, since it uses as much as digits as its factorization.

11413141122433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 242832789886.

The product of its digits is 6912, while the sum is 31.

Adding to 11413141122433 its reverse (33422114131411), we get a palindrome (44835255253844).

The spelling of 11413141122433 in words is "eleven trillion, four hundred thirteen billion, one hundred forty-one million, one hundred twenty-two thousand, four hundred thirty-three".