13121011102019 has 2 divisors, whose sum is σ = 13121011102020. Its totient is φ = 13121011102018.

The previous prime is 13121011101893. The next prime is 13121011102021. The reversal of 13121011102019 is 91020111012131.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13121011102019 is a prime.

It is a super-2 number, since 2×13121011102019² (a number of 27 digits) contains 22 as substring.

Together with 13121011102021, 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 = 13121011101985 and 13121011102003.

It is not a weakly prime, because it can be changed into another prime (13121011402019) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6560505551009 + 6560505551010.

It is an arithmetic number, because the mean of its divisors is an integer number (6560505551010).

Almost surely, 2^{13121011102019} is an apocalyptic number.

13121011102019 is a deficient number, since it is larger than the sum of its proper divisors (1).

13121011102019 is an equidigital number, since it uses as much as digits as its factorization.

13121011102019 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 108, while the sum is 23.

The spelling of 13121011102019 in words is "thirteen trillion, one hundred twenty-one billion, eleven million, one hundred two thousand, nineteen".