10014113 has 2 divisors, whose sum is σ = 10014114. Its totient is φ = 10014112.

The previous prime is 10014097. The next prime is 10014163. The reversal of 10014113 is 31141001.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 5121169 + 4892944 = 2263^2 + 2212^2 .

It is an emirp because it is prime and its reverse (31141001) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 10014113 - 2⁴ = 10014097 is a prime.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 10014094 and 10014103.

It is not a weakly prime, because it can be changed into another prime (10014163) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5007056 + 5007057.

It is an arithmetic number, because the mean of its divisors is an integer number (5007057).

Almost surely, 2^10014113 is an apocalyptic number.

It is an amenable number.

10014113 is a deficient number, since it is larger than the sum of its proper divisors (1).

10014113 is an equidigital number, since it uses as much as digits as its factorization.

10014113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12, while the sum is 11.

The square root of 10014113 is about 3164.5083346390. The cubic root of 10014113 is about 215.5447731505.

Adding to 10014113 its reverse (31141001), we get a palindrome (41155114).

The spelling of 10014113 in words is "ten million, fourteen thousand, one hundred thirteen".