1431043320031 has 4 divisors (see below), whose sum is σ = 1541123575432. Its totient is φ = 1320963064632.

The previous prime is 1431043319983. The next prime is 1431043320053. The reversal of 1431043320031 is 1300233401341.

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 1431043320031 - 2¹³ = 1431043311839 is a prime.

It is a super-2 number, since 2×1431043320031² (a number of 25 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 = 1431043319978 and 1431043320005.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (1431043320071) 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, 55040127681 + ... + 55040127706.

It is an arithmetic number, because the mean of its divisors is an integer number (385280893858).

Almost surely, 2^{1431043320031} is an apocalyptic number.

1431043320031 is a deficient number, since it is larger than the sum of its proper divisors (110080255401).

1431043320031 is a wasteful number, since it uses less digits than its factorization.

1431043320031 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 110080255400.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 1431043320031 its reverse (1300233401341), we get a palindrome (2731276721372).

The spelling of 1431043320031 in words is "one trillion, four hundred thirty-one billion, forty-three million, three hundred twenty thousand, thirty-one".