133312100000113 has 4 divisors (see below), whose sum is σ = 133355651816308. Its totient is φ = 133268548183920.

The previous prime is 133312100000081. The next prime is 133312100000137. The reversal of 133312100000113 is 311000001213331.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 122770987162849 + 10541112837264 = 11080207^2 + 3246708^2 .

It is a cyclic number.

It is not a de Polignac number, because 133312100000113 - 2⁵ = 133312100000081 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (133312100003113) 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, 21775903506 + ... + 21775909627.

It is an arithmetic number, because the mean of its divisors is an integer number (33338912954077).

Almost surely, 2^{133312100000113} is an apocalyptic number.

It is an amenable number.

133312100000113 is a deficient number, since it is larger than the sum of its proper divisors (43551816195).

133312100000113 is an equidigital number, since it uses as much as digits as its factorization.

133312100000113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 43551816194.

The product of its (nonzero) digits is 162, while the sum is 19.

Adding to 133312100000113 its reverse (311000001213331), we get a palindrome (444312101213444).

The spelling of 133312100000113 in words is "one hundred thirty-three trillion, three hundred twelve billion, one hundred million, one hundred thirteen", and thus it is an aban number.