1001000113 has 2 divisors, whose sum is σ = 1001000114. Its totient is φ = 1001000112.

The previous prime is 1001000111. The next prime is 1001000131. The reversal of 1001000113 is 3110001001.

It is a happy number.

1001000113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

Together with next prime (1001000131) it forms an Ormiston pair, because they use the same digits, order apart.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 973876849 + 27123264 = 31207^2 + 5208^2 .

It is a cyclic number.

It is not a de Polignac number, because 1001000113 - 2¹ = 1001000111 is a prime.

Together with 1001000111, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 1001000096 and 1001000105.

It is not a weakly prime, because it can be changed into another prime (1001000111) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 500500056 + 500500057.

It is an arithmetic number, because the mean of its divisors is an integer number (500500057).

Almost surely, 2^1001000113 is an apocalyptic number.

It is an amenable number.

1001000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1001000113 is an equidigital number, since it uses as much as digits as its factorization.

1001000113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3, while the sum is 7.

The square root of 1001000113 is about 31638.5858249069. The cubic root of 1001000113 is about 1000.3332599255. Note that the first 3 decimals are identical.

Adding to 1001000113 its reverse (3110001001), we get a palindrome (4111001114).

The spelling of 1001000113 in words is "one billion, one million, one hundred thirteen", and thus it is an aban number.