13122022110047 has 2 divisors, whose sum is σ = 13122022110048. Its totient is φ = 13122022110046.

The previous prime is 13122022110019. The next prime is 13122022110049. The reversal of 13122022110047 is 74001122022131.

It is a happy number.

It is a strong prime.

It is an emirp because it is prime and its reverse (74001122022131) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13122022110047 is a prime.

Together with 13122022110049, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 13122022109998 and 13122022110025.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (13122022110049) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (31) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6561011055023 + 6561011055024.

It is an arithmetic number, because the mean of its divisors is an integer number (6561011055024).

Almost surely, 2^{13122022110047} is an apocalyptic number.

13122022110047 is a deficient number, since it is larger than the sum of its proper divisors (1).

13122022110047 is an equidigital number, since it uses as much as digits as its factorization.

13122022110047 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1344, while the sum is 26.

Adding to 13122022110047 its reverse (74001122022131), we get a palindrome (87123144132178).

The spelling of 13122022110047 in words is "thirteen trillion, one hundred twenty-two billion, twenty-two million, one hundred ten thousand, forty-seven".