104034113 has 2 divisors, whose sum is σ = 104034114. Its totient is φ = 104034112.

The previous prime is 104034103. The next prime is 104034127. The reversal of 104034113 is 311430401.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 62457409 + 41576704 = 7903^2 + 6448^2 .

It is an emirp because it is prime and its reverse (311430401) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 104034113 - 2¹⁰ = 104033089 is a prime.

It is a Sophie Germain prime.

It is a Chen prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 104034091 and 104034100.

It is not a weakly prime, because it can be changed into another prime (104034103) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 52017056 + 52017057.

It is an arithmetic number, because the mean of its divisors is an integer number (52017057).

Almost surely, 2^104034113 is an apocalyptic number.

It is an amenable number.

104034113 is a deficient number, since it is larger than the sum of its proper divisors (1).

104034113 is an equidigital number, since it uses as much as digits as its factorization.

104034113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 144, while the sum is 17.

The square root of 104034113 is about 10199.7114174863. The cubic root of 104034113 is about 470.3183492825.

Adding to 104034113 its reverse (311430401), we get a palindrome (415464514).

It can be divided in two parts, 1040 and 34113, that added together give a palindrome (35153).

The spelling of 104034113 in words is "one hundred four million, thirty-four thousand, one hundred thirteen".