14011313411 has 2 divisors, whose sum is σ = 14011313412. Its totient is φ = 14011313410.

The previous prime is 14011313377. The next prime is 14011313413. The reversal of 14011313411 is 11431311041.

It is a strong prime.

It is an emirp because it is prime and its reverse (11431311041) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 14011313411 - 2⁶ = 14011313347 is a prime.

Together with 14011313413, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (14011313413) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7005656705 + 7005656706.

It is an arithmetic number, because the mean of its divisors is an integer number (7005656706).

Almost surely, 2^14011313411 is an apocalyptic number.

14011313411 is a deficient number, since it is larger than the sum of its proper divisors (1).

14011313411 is an equidigital number, since it uses as much as digits as its factorization.

14011313411 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 144, while the sum is 20.

Adding to 14011313411 its reverse (11431311041), we get a palindrome (25442624452).

The spelling of 14011313411 in words is "fourteen billion, eleven million, three hundred thirteen thousand, four hundred eleven".