1340533451219 has 2 divisors, whose sum is σ = 1340533451220. Its totient is φ = 1340533451218.

The previous prime is 1340533451203. The next prime is 1340533451221. The reversal of 1340533451219 is 9121543350431.

It is a strong prime.

It is an emirp because it is prime and its reverse (9121543350431) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1340533451219 - 2⁴ = 1340533451203 is a prime.

It is a super-2 number, since 2×1340533451219² (a number of 25 digits) contains 22 as substring.

Together with 1340533451221, 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 (1340533421219) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 670266725609 + 670266725610.

It is an arithmetic number, because the mean of its divisors is an integer number (670266725610).

Almost surely, 2^{1340533451219} is an apocalyptic number.

1340533451219 is a deficient number, since it is larger than the sum of its proper divisors (1).

1340533451219 is an equidigital number, since it uses as much as digits as its factorization.

1340533451219 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 194400, while the sum is 41.

The spelling of 1340533451219 in words is "one trillion, three hundred forty billion, five hundred thirty-three million, four hundred fifty-one thousand, two hundred nineteen".