1010021000113 has 4 divisors (see below), whose sum is σ = 1010560831488. Its totient is φ = 1009481168740.

The previous prime is 1010021000069. The next prime is 1010021000143. The reversal of 1010021000113 is 3110001200101.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 1010021000113 - 2¹³ = 1010020991921 is a prime.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 1010021000113.

It is not an unprimeable number, because it can be changed into a prime (1010021000143) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 269912881 + ... + 269916622.

It is an arithmetic number, because the mean of its divisors is an integer number (252640207872).

Almost surely, 2^{1010021000113} is an apocalyptic number.

It is an amenable number.

1010021000113 is a deficient number, since it is larger than the sum of its proper divisors (539831375).

1010021000113 is an equidigital number, since it uses as much as digits as its factorization.

1010021000113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 539831374.

The product of its (nonzero) digits is 6, while the sum is 10.

Adding to 1010021000113 its reverse (3110001200101), we get a palindrome (4120022200214).

The spelling of 1010021000113 in words is "one trillion, ten billion, twenty-one million, one hundred thirteen", and thus it is an aban number.