1440202214113 has 4 divisors (see below), whose sum is σ = 1550986999828. Its totient is φ = 1329417428400.

The previous prime is 1440202214107. The next prime is 1440202214119. The reversal of 1440202214113 is 3114122020441.

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

It is an interprime number because it is at equal distance from previous prime (1440202214107) and next prime (1440202214119).

It can be written as a sum of positive squares in 2 ways, for example, as 217689431184 + 1222512782929 = 466572^2 + 1105673^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1440202214113 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (1440202214119) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 55392392838 + ... + 55392392863.

It is an arithmetic number, because the mean of its divisors is an integer number (387746749957).

Almost surely, 2^{1440202214113} is an apocalyptic number.

1440202214113 is a gapful number since it is divisible by the number (13) formed by its first and last digit.

It is an amenable number.

1440202214113 is a deficient number, since it is larger than the sum of its proper divisors (110784785715).

1440202214113 is a wasteful number, since it uses less digits than its factorization.

1440202214113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 110784785714.

The product of its (nonzero) digits is 1536, while the sum is 25.

Adding to 1440202214113 its reverse (3114122020441), we get a palindrome (4554324234554).

The spelling of 1440202214113 in words is "one trillion, four hundred forty billion, two hundred two million, two hundred fourteen thousand, one hundred thirteen".