3104412433 has 2 divisors, whose sum is σ = 3104412434. Its totient is φ = 3104412432.

The previous prime is 3104412401. The next prime is 3104412439. The reversal of 3104412433 is 3342144013.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 2854551184 + 249861249 = 53428^2 + 15807^2 .

It is a cyclic number.

It is not a de Polignac number, because 3104412433 - 2⁵ = 3104412401 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 3104412398 and 3104412407.

It is not a weakly prime, because it can be changed into another prime (3104412439) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1552206216 + 1552206217.

It is an arithmetic number, because the mean of its divisors is an integer number (1552206217).

Almost surely, 2^3104412433 is an apocalyptic number.

It is an amenable number.

3104412433 is a deficient number, since it is larger than the sum of its proper divisors (1).

3104412433 is an equidigital number, since it uses as much as digits as its factorization.

3104412433 is an odious number, because the sum of its binary digits is odd.

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

The square root of 3104412433 is about 55717.2543562584. The cubic root of 3104412433 is about 1458.7912108019.

Adding to 3104412433 its reverse (3342144013), we get a palindrome (6446556446).

The spelling of 3104412433 in words is "three billion, one hundred four million, four hundred twelve thousand, four hundred thirty-three".