101001100101113 has 2 divisors, whose sum is σ = 101001100101114. Its totient is φ = 101001100101112.

The previous prime is 101001100101037. The next prime is 101001100101137. The reversal of 101001100101113 is 311101001100101.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 80421921101929 + 20579178999184 = 8967827^2 + 4536428^2 .

It is an emirp because it is prime and its reverse (311101001100101) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 101001100101113 - 2³⁰ = 101000026359289 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 101001100101094 and 101001100101103.

It is not a weakly prime, because it can be changed into another prime (101001100121113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50500550050556 + 50500550050557.

It is an arithmetic number, because the mean of its divisors is an integer number (50500550050557).

Almost surely, 2^{101001100101113} is an apocalyptic number.

It is an amenable number.

101001100101113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101001100101113 is an equidigital number, since it uses as much as digits as its factorization.

101001100101113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3, while the sum is 11.

Adding to 101001100101113 its reverse (311101001100101), we get a palindrome (412102101201214).

The spelling of 101001100101113 in words is "one hundred one trillion, one billion, one hundred million, one hundred one thousand, one hundred thirteen".