443112121245541 has 4 divisors (see below), whose sum is σ = 483395041358784. Its totient is φ = 402829201132300.

The previous prime is 443112121245539. The next prime is 443112121245631. The reversal of 443112121245541 is 145542121211344.

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 443112121245541 - 2¹ = 443112121245539 is a prime.

It is a Smith number, since the sum of its digits (40) 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 = 443112121245494 and 443112121245503.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (443112121245641) 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, 20141460056605 + ... + 20141460056626.

It is an arithmetic number, because the mean of its divisors is an integer number (120848760339696).

Almost surely, 2^{443112121245541} is an apocalyptic number.

It is an amenable number.

443112121245541 is a deficient number, since it is larger than the sum of its proper divisors (40282920113243).

443112121245541 is a wasteful number, since it uses less digits than its factorization.

443112121245541 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 40282920113242.

The product of its digits is 153600, while the sum is 40.

Adding to 443112121245541 its reverse (145542121211344), we get a palindrome (588654242456885).

The spelling of 443112121245541 in words is "four hundred forty-three trillion, one hundred twelve billion, one hundred twenty-one million, two hundred forty-five thousand, five hundred forty-one".