101101101113 has 2 divisors, whose sum is σ = 101101101114. Its totient is φ = 101101101112.

The previous prime is 101101101097. The next prime is 101101101151. The reversal of 101101101113 is 311101101101.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 86922959929 + 14178141184 = 294827^2 + 119072^2 .

It is a cyclic number.

It is not a de Polignac number, because 101101101113 - 2⁴ = 101101101097 is a prime.

It is a super-2 number, since 2×101101101113² (a number of 23 digits) contains 22 as substring.

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 = 101101101094 and 101101101103.

It is not a weakly prime, because it can be changed into another prime (101101101163) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50550550556 + 50550550557.

It is an arithmetic number, because the mean of its divisors is an integer number (50550550557).

Almost surely, 2^101101101113 is an apocalyptic number.

It is an amenable number.

101101101113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101101101113 is an equidigital number, since it uses as much as digits as its factorization.

101101101113 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 101101101113 its reverse (311101101101), we get a palindrome (412202202214).

The spelling of 101101101113 in words is "one hundred one billion, one hundred one million, one hundred one thousand, one hundred thirteen".