1300402120163 has 2 divisors, whose sum is σ = 1300402120164. Its totient is φ = 1300402120162.

The previous prime is 1300402120117. The next prime is 1300402120177. The reversal of 1300402120163 is 3610212040031.

It is a strong prime.

It is an emirp because it is prime and its reverse (3610212040031) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1300402120163 is a prime.

It is a super-2 number, since 2×1300402120163² (a number of 25 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (1300402125163) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 650201060081 + 650201060082.

It is an arithmetic number, because the mean of its divisors is an integer number (650201060082).

Almost surely, 2^{1300402120163} is an apocalyptic number.

1300402120163 is a deficient number, since it is larger than the sum of its proper divisors (1).

1300402120163 is an equidigital number, since it uses as much as digits as its factorization.

1300402120163 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 864, while the sum is 23.

Adding to 1300402120163 its reverse (3610212040031), we get a palindrome (4910614160194).

The spelling of 1300402120163 in words is "one trillion, three hundred billion, four hundred two million, one hundred twenty thousand, one hundred sixty-three".