1313403655042 has 4 divisors (see below), whose sum is σ = 1970105482566. Its totient is φ = 656701827520.

The previous prime is 1313403654949. The next prime is 1313403655051. The reversal of 1313403655042 is 2405563043131.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in only one way, i.e., 1301563821321 + 11839833721 = 1140861^2 + 108811^2 .

It is a super-2 number, since 2×1313403655042² (a number of 25 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1313403654992 and 1313403655010.

It is an unprimeable number.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 328350913759 + ... + 328350913762.

Almost surely, 2^{1313403655042} is an apocalyptic number.

1313403655042 is a deficient number, since it is larger than the sum of its proper divisors (656701827524).

1313403655042 is an equidigital number, since it uses as much as digits as its factorization.

1313403655042 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 656701827523.

The product of its (nonzero) digits is 129600, while the sum is 37.

Adding to 1313403655042 its reverse (2405563043131), we get a palindrome (3718966698173).

The spelling of 1313403655042 in words is "one trillion, three hundred thirteen billion, four hundred three million, six hundred fifty-five thousand, forty-two".