13413004044131 has 4 divisors (see below), whose sum is σ = 13418458704600. Its totient is φ = 13407549383664.

The previous prime is 13413004044119. The next prime is 13413004044211. The reversal of 13413004044131 is 13144040031431.

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

It is a cyclic number.

It is not a de Polignac number, because 13413004044131 - 2¹⁰ = 13413004043107 is a prime.

It is a super-2 number, since 2×13413004044131² (a number of 27 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 = 13413004044094 and 13413004044103.

It is not an unprimeable number, because it can be changed into a prime (13413004049131) 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, 2727326546 + ... + 2727331463.

It is an arithmetic number, because the mean of its divisors is an integer number (3354614676150).

Almost surely, 2^{13413004044131} is an apocalyptic number.

13413004044131 is a deficient number, since it is larger than the sum of its proper divisors (5454660469).

13413004044131 is an equidigital number, since it uses as much as digits as its factorization.

13413004044131 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 5454660468.

The product of its (nonzero) digits is 6912, while the sum is 29.

Adding to 13413004044131 its reverse (13144040031431), we get a palindrome (26557044075562).

The spelling of 13413004044131 in words is "thirteen trillion, four hundred thirteen billion, four million, forty-four thousand, one hundred thirty-one".