1310003000113 has 2 divisors, whose sum is σ = 1310003000114. Its totient is φ = 1310003000112.

The previous prime is 1310003000081. The next prime is 1310003000123. The reversal of 1310003000113 is 3110003000131.

It is a happy number.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1303508574369 + 6494425744 = 1141713^2 + 80588^2 .

It is an emirp because it is prime and its reverse (3110003000131) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1310003000113 - 2⁵ = 1310003000081 is a prime.

It is a super-2 number, since 2×1310003000113² (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1310003000093 and 1310003000102.

It is not a weakly prime, because it can be changed into another prime (1310003000123) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655001500056 + 655001500057.

It is an arithmetic number, because the mean of its divisors is an integer number (655001500057).

Almost surely, 2^{1310003000113} is an apocalyptic number.

It is an amenable number.

1310003000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1310003000113 is an equidigital number, since it uses as much as digits as its factorization.

1310003000113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 27, while the sum is 13.

Adding to 1310003000113 its reverse (3110003000131), we get a palindrome (4420006000244).

The spelling of 1310003000113 in words is "one trillion, three hundred ten billion, three million, one hundred thirteen", and thus it is an aban number.