13012100113 has 2 divisors, whose sum is σ = 13012100114. Its totient is φ = 13012100112.

The previous prime is 13012100099. The next prime is 13012100141. The reversal of 13012100113 is 31100121031.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 13003297024 + 8803089 = 114032^2 + 2967^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13012100113 is a prime.

It is a super-3 number, since 3×13012100113³ (a number of 31 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a junction number, because it is equal to n+sod(n) for n = 13012100093 and 13012100102.

It is not a weakly prime, because it can be changed into another prime (13012100173) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6506050056 + 6506050057.

It is an arithmetic number, because the mean of its divisors is an integer number (6506050057).

Almost surely, 2^13012100113 is an apocalyptic number.

It is an amenable number.

13012100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

13012100113 is an equidigital number, since it uses as much as digits as its factorization.

13012100113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 18, while the sum is 13.

Adding to 13012100113 its reverse (31100121031), we get a palindrome (44112221144).

The spelling of 13012100113 in words is "thirteen billion, twelve million, one hundred thousand, one hundred thirteen".