110001300113 has 2 divisors, whose sum is σ = 110001300114. Its totient is φ = 110001300112.

The previous prime is 110001300103. The next prime is 110001300119. The reversal of 110001300113 is 311003100011.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 96862867984 + 13138432129 = 311228^2 + 114623^2 .

It is an emirp because it is prime and its reverse (311003100011) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 110001300113 - 2⁸ = 110001299857 is a prime.

It is a Sophie Germain prime.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 110001300094 and 110001300103.

It is not a weakly prime, because it can be changed into another prime (110001300119) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55000650056 + 55000650057.

It is an arithmetic number, because the mean of its divisors is an integer number (55000650057).

Almost surely, 2^110001300113 is an apocalyptic number.

It is an amenable number.

110001300113 is a deficient number, since it is larger than the sum of its proper divisors (1).

110001300113 is an equidigital number, since it uses as much as digits as its factorization.

110001300113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 9, while the sum is 11.

Adding to 110001300113 its reverse (311003100011), we get a palindrome (421004400124).

The spelling of 110001300113 in words is "one hundred ten billion, one million, three hundred thousand, one hundred thirteen".