2002414214113 has 4 divisors (see below), whose sum is σ = 2003651037360. Its totient is φ = 2001177390868.

The previous prime is 2002414214111. The next prime is 2002414214131. The reversal of 2002414214113 is 3114124142002.

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 2002414214113 - 2¹ = 2002414214111 is a prime.

It is a super-3 number, since 3×2002414214113³ (a number of 38 digits) contains 333 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (2002414214111) 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, 618409195 + ... + 618412432.

It is an arithmetic number, because the mean of its divisors is an integer number (500912759340).

Almost surely, 2^{2002414214113} is an apocalyptic number.

It is an amenable number.

2002414214113 is a deficient number, since it is larger than the sum of its proper divisors (1236823247).

2002414214113 is a wasteful number, since it uses less digits than its factorization.

2002414214113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1236823246.

The product of its (nonzero) digits is 1536, while the sum is 25.

Adding to 2002414214113 its reverse (3114124142002), we get a palindrome (5116538356115).

The spelling of 2002414214113 in words is "two trillion, two billion, four hundred fourteen million, two hundred fourteen thousand, one hundred thirteen".