14300014287311 has 2 divisors, whose sum is σ = 14300014287312. Its totient is φ = 14300014287310.

The previous prime is 14300014287287. The next prime is 14300014287313. The reversal of 14300014287311 is 11378241000341.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-14300014287311 is a prime.

It is a super-2 number, since 2×14300014287311² (a number of 27 digits) contains 22 as substring.

Together with 14300014287313, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (14300014287313) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7150007143655 + 7150007143656.

It is an arithmetic number, because the mean of its divisors is an integer number (7150007143656).

Almost surely, 2^{14300014287311} is an apocalyptic number.

14300014287311 is a deficient number, since it is larger than the sum of its proper divisors (1).

14300014287311 is an equidigital number, since it uses as much as digits as its factorization.

14300014287311 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 16128, while the sum is 35.

Adding to 14300014287311 its reverse (11378241000341), we get a palindrome (25678255287652).

The spelling of 14300014287311 in words is "fourteen trillion, three hundred billion, fourteen million, two hundred eighty-seven thousand, three hundred eleven".