1014341313 has 4 divisors (see below), whose sum is σ = 1352455088. Its totient is φ = 676227540.

The previous prime is 1014341303. The next prime is 1014341369. The reversal of 1014341313 is 3131434101.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also an emirpimes, since its reverse is a distinct semiprime: 3131434101 = 3 ⋅1043811367.

It is not a de Polignac number, because 1014341313 - 2¹³ = 1014333121 is a prime.

It is a super-4 number, since 4×1014341313⁴ (a number of 37 digits) contains 4444 as substring.

It is a Duffinian number.

It is a Curzon number.

It is a self number, because there is not a number n which added to its sum of digits gives 1014341313.

It is not an unprimeable number, because it can be changed into a prime (1014341303) 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 3 ways as a sum of consecutive naturals, for example, 169056883 + ... + 169056888.

It is an arithmetic number, because the mean of its divisors is an integer number (338113772).

Almost surely, 2^1014341313 is an apocalyptic number.

It is an amenable number.

1014341313 is a deficient number, since it is larger than the sum of its proper divisors (338113775).

1014341313 is an equidigital number, since it uses as much as digits as its factorization.

1014341313 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 338113774.

The product of its (nonzero) digits is 432, while the sum is 21.

The square root of 1014341313 is about 31848.7254533050. The cubic root of 1014341313 is about 1004.7577654353.

Adding to 1014341313 its reverse (3131434101), we get a palindrome (4145775414).

The spelling of 1014341313 in words is "one billion, fourteen million, three hundred forty-one thousand, three hundred thirteen".