13104401205281 has 2 divisors, whose sum is σ = 13104401205282. Its totient is φ = 13104401205280.

The previous prime is 13104401205277. The next prime is 13104401205283. The reversal of 13104401205281 is 18250210440131.

13104401205281 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 9665290299025 + 3439110906256 = 3108905^2 + 1854484^2 .

It is a cyclic number.

It is not a de Polignac number, because 13104401205281 - 2² = 13104401205277 is a prime.

It is a super-2 number, since 2×13104401205281² (a number of 27 digits) contains 22 as substring.

Together with 13104401205283, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (13104401205283) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6552200602640 + 6552200602641.

It is an arithmetic number, because the mean of its divisors is an integer number (6552200602641).

Almost surely, 2^{13104401205281} is an apocalyptic number.

It is an amenable number.

13104401205281 is a deficient number, since it is larger than the sum of its proper divisors (1).

13104401205281 is an equidigital number, since it uses as much as digits as its factorization.

13104401205281 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 7680, while the sum is 32.

The spelling of 13104401205281 in words is "thirteen trillion, one hundred four billion, four hundred one million, two hundred five thousand, two hundred eighty-one".