13040211204031 has 4 divisors (see below), whose sum is σ = 14225684949864. Its totient is φ = 11854737458200.

The previous prime is 13040211203993. The next prime is 13040211204037.

13040211204031 is nontrivially palindromic in base 10.

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 13040211204031 - 2²¹ = 13040209106879 is a prime.

It is a super-2 number, since 2×13040211204031² (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 = 13040211203993 and 13040211204011.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (13040211204037) 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, 592736872900 + ... + 592736872921.

It is an arithmetic number, because the mean of its divisors is an integer number (3556421237466).

Almost surely, 2^{13040211204031} is an apocalyptic number.

13040211204031 is a gapful number since it is divisible by the number (11) formed by its first and last digit.

13040211204031 is a deficient number, since it is larger than the sum of its proper divisors (1185473745833).

13040211204031 is a wasteful number, since it uses less digits than its factorization.

13040211204031 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1185473745832.

The product of its (nonzero) digits is 576, while the sum is 22.

Multiplying 13040211204031 by its sum of digits (22), we get a palindrome (286884646488682).

It can be divided in two parts, 1304021 and 1204031, that added together give a palindrome (2508052).

The spelling of 13040211204031 in words is "thirteen trillion, forty billion, two hundred eleven million, two hundred four thousand, thirty-one".