53040401 has 4 divisors (see below), whose sum is σ = 53055012. Its totient is φ = 53025792.

The previous prime is 53040397. The next prime is 53040443. The reversal of 53040401 is 10404035.

It is a semiprime because it is the product of two primes, and also a brilliant number, because the two primes have the same length, and also an emirpimes, since its reverse is a distinct semiprime: 10404035 = 5 ⋅2080807.

It can be written as a sum of positive squares in 2 ways, for example, as 11566801 + 41473600 = 3401^2 + 6440^2 .

It is a cyclic number.

It is not a de Polignac number, because 53040401 - 2² = 53040397 is a prime.

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 53040401.

It is not an unprimeable number, because it can be changed into a prime (53040601) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (11) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2801 + ... + 10673.

It is an arithmetic number, because the mean of its divisors is an integer number (13263753).

Almost surely, 2^53040401 is an apocalyptic number.

It is an amenable number.

53040401 is a deficient number, since it is larger than the sum of its proper divisors (14611).

53040401 is an equidigital number, since it uses as much as digits as its factorization.

53040401 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 14610.

The product of its (nonzero) digits is 240, while the sum is 17.

The square root of 53040401 is about 7282.8841127674. The cubic root of 53040401 is about 375.7239962751.

Adding to 53040401 its reverse (10404035), we get a palindrome (63444436).

The spelling of 53040401 in words is "fifty-three million, forty thousand, four hundred one".