101100101010113 has 2 divisors, whose sum is σ = 101100101010114. Its totient is φ = 101100101010112.

The previous prime is 101100101010091. The next prime is 101100101010149. The reversal of 101100101010113 is 311010101001101.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 101081970926209 + 18130083904 = 10053953^2 + 134648^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-101100101010113 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 101100101010094 and 101100101010103.

It is not a weakly prime, because it can be changed into another prime (101100101016113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50550050505056 + 50550050505057.

It is an arithmetic number, because the mean of its divisors is an integer number (50550050505057).

Almost surely, 2^{101100101010113} is an apocalyptic number.

It is an amenable number.

101100101010113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101100101010113 is an equidigital number, since it uses as much as digits as its factorization.

101100101010113 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 11.

Adding to 101100101010113 its reverse (311010101001101), we get a palindrome (412110202011214).

The spelling of 101100101010113 in words is "one hundred one trillion, one hundred billion, one hundred one million, ten thousand, one hundred thirteen".