14233340131 has 2 divisors, whose sum is σ = 14233340132. Its totient is φ = 14233340130.

The previous prime is 14233340113. The next prime is 14233340159. The reversal of 14233340131 is 13104333241.

14233340131 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

Together with previous prime (14233340113) it forms an Ormiston pair, because they use the same digits, order apart.

It is a weak prime.

It is an emirp because it is prime and its reverse (13104333241) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 14233340131 - 2⁷ = 14233340003 is a prime.

It is a super-2 number, since 2×14233340131² (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 14233340096 and 14233340105.

It is not a weakly prime, because it can be changed into another prime (14233380131) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7116670065 + 7116670066.

It is an arithmetic number, because the mean of its divisors is an integer number (7116670066).

Almost surely, 2^14233340131 is an apocalyptic number.

14233340131 is a deficient number, since it is larger than the sum of its proper divisors (1).

14233340131 is an equidigital number, since it uses as much as digits as its factorization.

14233340131 is an odious number, because the sum of its binary digits is odd.

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

Adding to 14233340131 its reverse (13104333241), we get a palindrome (27337673372).

The spelling of 14233340131 in words is "fourteen billion, two hundred thirty-three million, three hundred forty thousand, one hundred thirty-one".