13121040113 has 8 divisors (see below), whose sum is σ = 14048967240. Its totient is φ = 12210457600.

The previous prime is 13121040109. The next prime is 13121040131. The reversal of 13121040113 is 31104012131.

13121040113 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 4 ways, for example, as 2795025424 + 10326014689 = 52868^2 + 101617^2 .

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

It is a cyclic number.

It is not a de Polignac number, because 13121040113 - 2² = 13121040109 is a prime.

It is a Harshad number since it is a multiple of its sum of digits (17).

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13121040091 and 13121040100.

It is not an unprimeable number, because it can be changed into a prime (13121040103) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 4334588 + ... + 4337613.

It is an arithmetic number, because the mean of its divisors is an integer number (1756120905).

Almost surely, 2^13121040113 is an apocalyptic number.

It is an amenable number.

13121040113 is a deficient number, since it is larger than the sum of its proper divisors (927927127).

13121040113 is an equidigital number, since it uses as much as digits as its factorization.

13121040113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 8672307.

The product of its (nonzero) digits is 72, while the sum is 17.

Adding to 13121040113 its reverse (31104012131), we get a palindrome (44225052244).

The spelling of 13121040113 in words is "thirteen billion, one hundred twenty-one million, forty thousand, one hundred thirteen".