13101000113 has 2 divisors, whose sum is σ = 13101000114. Its totient is φ = 13101000112.

The previous prime is 13101000109. The next prime is 13101000131. The reversal of 13101000113 is 31100010131.

It is a happy number.

Together with next prime (13101000131) it forms an Ormiston pair, because they use the same digits, order apart.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 10280540449 + 2820459664 = 101393^2 + 53108^2 .

It is an emirp because it is prime and its reverse (31100010131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13101000113 - 2² = 13101000109 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 13101000094 and 13101000103.

It is not a weakly prime, because it can be changed into another prime (13101000163) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6550500056 + 6550500057.

It is an arithmetic number, because the mean of its divisors is an integer number (6550500057).

Almost surely, 2^13101000113 is an apocalyptic number.

It is an amenable number.

13101000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

13101000113 is an equidigital number, since it uses as much as digits as its factorization.

13101000113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 9, while the sum is 11.

Adding to 13101000113 its reverse (31100010131), we get a palindrome (44201010244).

The spelling of 13101000113 in words is "thirteen billion, one hundred one million, one hundred thirteen", and thus it is an aban number.