14053103141081 has 2 divisors, whose sum is σ = 14053103141082. Its totient is φ = 14053103141080.

The previous prime is 14053103141023. The next prime is 14053103141083. The reversal of 14053103141081 is 18014130135041.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 9311139605056 + 4741963536025 = 3051416^2 + 2177605^2 .

It is an emirp because it is prime and its reverse (18014130135041) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 14053103141081 - 2⁶ = 14053103141017 is a prime.

It is a super-2 number, since 2×14053103141081² (a number of 27 digits) contains 22 as substring.

Together with 14053103141083, 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 (14053103141083) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7026551570540 + 7026551570541.

It is an arithmetic number, because the mean of its divisors is an integer number (7026551570541).

Almost surely, 2^{14053103141081} is an apocalyptic number.

It is an amenable number.

14053103141081 is a deficient number, since it is larger than the sum of its proper divisors (1).

14053103141081 is an equidigital number, since it uses as much as digits as its factorization.

14053103141081 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 5760, while the sum is 32.

The spelling of 14053103141081 in words is "fourteen trillion, fifty-three billion, one hundred three million, one hundred forty-one thousand, eighty-one".