13540444033 has 4 divisors (see below), whose sum is σ = 13540832560. Its totient is φ = 13540055508.

The previous prime is 13540443977. The next prime is 13540444061. The reversal of 13540444033 is 33044404531.

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.

It is a cyclic number.

It is not a de Polignac number, because 13540444033 - 2¹³ = 13540435841 is a prime.

It is a super-2 number, since 2×13540444033² (a number of 21 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13540443986 and 13540444004.

It is not an unprimeable number, because it can be changed into a prime (13540444063) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 136203 + ... + 213616.

It is an arithmetic number, because the mean of its divisors is an integer number (3385208140).

Almost surely, 2^13540444033 is an apocalyptic number.

It is an amenable number.

13540444033 is a deficient number, since it is larger than the sum of its proper divisors (388527).

13540444033 is an equidigital number, since it uses as much as digits as its factorization.

13540444033 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 388526.

The product of its (nonzero) digits is 34560, while the sum is 31.

Adding to 13540444033 its reverse (33044404531), we get a palindrome (46584848564).

The spelling of 13540444033 in words is "thirteen billion, five hundred forty million, four hundred forty-four thousand, thirty-three".