10004411340113 has 2 divisors, whose sum is σ = 10004411340114. Its totient is φ = 10004411340112.

The previous prime is 10004411340101. The next prime is 10004411340191. The reversal of 10004411340113 is 31104311440001.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 5641275768769 + 4363135571344 = 2375137^2 + 2088812^2 .

It is an emirp because it is prime and its reverse (31104311440001) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 10004411340113 - 2²⁶ = 10004344231249 is a prime.

It is not a weakly prime, because it can be changed into another prime (10004411349113) 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 as a sum of consecutive naturals, namely, 5002205670056 + 5002205670057.

It is an arithmetic number, because the mean of its divisors is an integer number (5002205670057).

Almost surely, 2^{10004411340113} is an apocalyptic number.

It is an amenable number.

10004411340113 is a deficient number, since it is larger than the sum of its proper divisors (1).

10004411340113 is an equidigital number, since it uses as much as digits as its factorization.

10004411340113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 576, while the sum is 23.

Adding to 10004411340113 its reverse (31104311440001), we get a palindrome (41108722780114).

The spelling of 10004411340113 in words is "ten trillion, four billion, four hundred eleven million, three hundred forty thousand, one hundred thirteen".