13343404000201 has 4 divisors (see below), whose sum is σ = 14556440727504. Its totient is φ = 12130367272900.

The previous prime is 13343404000187. The next prime is 13343404000217. The reversal of 13343404000201 is 10200040434331.

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 a de Polignac number, because none of the positive numbers 2^k-13343404000201 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (13343404007201) 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, 606518363635 + ... + 606518363656.

It is an arithmetic number, because the mean of its divisors is an integer number (3639110181876).

Almost surely, 2^{13343404000201} is an apocalyptic number.

13343404000201 is a gapful number since it is divisible by the number (11) formed by its first and last digit.

It is an amenable number.

13343404000201 is a deficient number, since it is larger than the sum of its proper divisors (1213036727303).

13343404000201 is a wasteful number, since it uses less digits than its factorization.

13343404000201 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1213036727302.

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

Adding to 13343404000201 its reverse (10200040434331), we get a palindrome (23543444434532).

The spelling of 13343404000201 in words is "thirteen trillion, three hundred forty-three billion, four hundred four million, two hundred one", and thus it is an aban number.