321213130000041 has 4 divisors (see below), whose sum is σ = 428284173333392. Its totient is φ = 214142086666692.

The previous prime is 321213130000027. The next prime is 321213130000069. The reversal of 321213130000041 is 140000031312123.

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 not a de Polignac number, because 321213130000041 - 2¹⁵ = 321213129967273 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 321213129999972 and 321213130000017.

It is not an unprimeable number, because it can be changed into a prime (321213130005041) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 53535521666671 + ... + 53535521666676.

It is an arithmetic number, because the mean of its divisors is an integer number (107071043333348).

Almost surely, 2^{321213130000041} is an apocalyptic number.

It is an amenable number.

321213130000041 is a deficient number, since it is larger than the sum of its proper divisors (107071043333351).

321213130000041 is a wasteful number, since it uses less digits than its factorization.

321213130000041 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 107071043333350.

The product of its (nonzero) digits is 432, while the sum is 21.

Adding to 321213130000041 its reverse (140000031312123), we get a palindrome (461213161312164).

The spelling of 321213130000041 in words is "three hundred twenty-one trillion, two hundred thirteen billion, one hundred thirty million, forty-one", and thus it is an aban number.