13124414404041 has 4 divisors (see below), whose sum is σ = 17499219205392. Its totient is φ = 8749609602692.

The previous prime is 13124414403937. The next prime is 13124414404043. The reversal of 13124414404041 is 14040441442131.

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 a de Polignac number, because none of the positive numbers 2^k-13124414404041 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 13124414403993 and 13124414404011.

It is not an unprimeable number, because it can be changed into a prime (13124414404043) 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, 2187402400671 + ... + 2187402400676.

It is an arithmetic number, because the mean of its divisors is an integer number (4374804801348).

Almost surely, 2^{13124414404041} is an apocalyptic number.

It is an amenable number.

13124414404041 is a deficient number, since it is larger than the sum of its proper divisors (4374804801351).

13124414404041 is an equidigital number, since it uses as much as digits as its factorization.

13124414404041 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4374804801350.

The product of its (nonzero) digits is 24576, while the sum is 33.

Adding to 13124414404041 its reverse (14040441442131), we get a palindrome (27164855846172).

The spelling of 13124414404041 in words is "thirteen trillion, one hundred twenty-four billion, four hundred fourteen million, four hundred four thousand, forty-one".