14113140245473 has 2 divisors, whose sum is σ = 14113140245474. Its totient is φ = 14113140245472.

The previous prime is 14113140245471. The next prime is 14113140245531. The reversal of 14113140245473 is 37454204131141.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11298956177664 + 2814184067809 = 3361392^2 + 1677553^2 .

It is a cyclic number.

It is not a de Polignac number, because 14113140245473 - 2¹ = 14113140245471 is a prime.

Together with 14113140245471, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (14113140245471) 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, 7056570122736 + 7056570122737.

It is an arithmetic number, because the mean of its divisors is an integer number (7056570122737).

Almost surely, 2^{14113140245473} is an apocalyptic number.

It is an amenable number.

14113140245473 is a deficient number, since it is larger than the sum of its proper divisors (1).

14113140245473 is an equidigital number, since it uses as much as digits as its factorization.

14113140245473 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 161280, while the sum is 40.

The spelling of 14113140245473 in words is "fourteen trillion, one hundred thirteen billion, one hundred forty million, two hundred forty-five thousand, four hundred seventy-three".