13110000000101 has 2 divisors, whose sum is σ = 13110000000102. Its totient is φ = 13110000000100.

The previous prime is 13110000000091. The next prime is 13110000000103. The reversal of 13110000000101 is 10100000001131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 8324292484225 + 4785707515876 = 2885185^2 + 2187626^2 .

It is an emirp because it is prime and its reverse (10100000001131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13110000000101 - 2⁶ = 13110000000037 is a prime.

It is a Sophie Germain prime.

Together with 13110000000103, it forms a pair of twin primes.

It is a Chen prime.

It is a Curzon number.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13110000000103) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6555000000050 + 6555000000051.

It is an arithmetic number, because the mean of its divisors is an integer number (6555000000051).

Almost surely, 2^{13110000000101} is an apocalyptic number.

It is an amenable number.

13110000000101 is a deficient number, since it is larger than the sum of its proper divisors (1).

13110000000101 is an equidigital number, since it uses as much as digits as its factorization.

13110000000101 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 8.

Adding to 13110000000101 its reverse (10100000001131), we get a palindrome (23210000001232).

The spelling of 13110000000101 in words is "thirteen trillion, one hundred ten billion, one hundred one", and thus it is an aban number.