1010111300113 has 4 divisors (see below), whose sum is σ = 1010691492148. Its totient is φ = 1009531108080.

The previous prime is 1010111300017. The next prime is 1010111300117. The reversal of 1010111300113 is 3110031110101.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 518734133824 + 491377166289 = 720232^2 + 700983^2 .

It is a cyclic number.

It is not a de Polignac number, because 1010111300113 - 2²¹ = 1010109202961 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1010111300093 and 1010111300102.

It is not an unprimeable number, because it can be changed into a prime (1010111300117) 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 in 3 ways as a sum of consecutive naturals, for example, 290093406 + ... + 290096887.

It is an arithmetic number, because the mean of its divisors is an integer number (252672873037).

Almost surely, 2^{1010111300113} is an apocalyptic number.

It is an amenable number.

1010111300113 is a deficient number, since it is larger than the sum of its proper divisors (580192035).

1010111300113 is an equidigital number, since it uses as much as digits as its factorization.

1010111300113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 580192034.

The product of its (nonzero) digits is 9, while the sum is 13.

Adding to 1010111300113 its reverse (3110031110101), we get a palindrome (4120142410214).

The spelling of 1010111300113 in words is "one trillion, ten billion, one hundred eleven million, three hundred thousand, one hundred thirteen".