13000130113130 has 8 divisors (see below), whose sum is σ = 23400234203652. Its totient is φ = 5200052045248.

The previous prime is 13000130113073. The next prime is 13000130113153. The reversal of 13000130113130 is 3131103100031.

13000130113130 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It can be written as a sum of positive squares in 2 ways, for example, as 12935294185489 + 64835927641 = 3596567^2 + 254629^2 .

It is a sphenic number, since it is the product of 3 distinct primes.

It is a junction number, because it is equal to n+sod(n) for n = 13000130113099 and 13000130113108.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 650006505647 + ... + 650006505666.

Almost surely, 2^{13000130113130} is an apocalyptic number.

13000130113130 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

13000130113130 is a deficient number, since it is larger than the sum of its proper divisors (10400104090522).

13000130113130 is a wasteful number, since it uses less digits than its factorization.

13000130113130 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1300013011320.

The product of its (nonzero) digits is 81, while the sum is 17.

Adding to 13000130113130 its reverse (3131103100031), we get a palindrome (16131233213161).

The spelling of 13000130113130 in words is "thirteen trillion, one hundred thirty million, one hundred thirteen thousand, one hundred thirty".